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An antenna question--43 ft vertical
On 7/6/2015 4:50 PM, Ian Jackson wrote:
In message , rickman writes The only case I am aware of that will give total reflection is when the terminal is open circuit with infinite impedance absorbing *no* signal. Also when it is a zero impedance (short circuit). I'm trying to picture this. In the case of an open circuit a matched driver drives the transmission line to 50% of the driving voltage. The wave reaches the open termination and is reflected with the same polarity resulting in a return wave that reaches 100% of the driving voltage. In the same vein, if the wave hits the short circuit the reflected wave will be the opposite polarity making the reflected wave 0% of the driving voltage resulting in the short circuit eventually showing to the drive circuit. -- Rick |
An antenna question--43 ft vertical
The only case I am aware of that will give total reflection is when
the terminal is open circuit with infinite impedance absorbing *no* signal. Also when it is a zero impedance (short circuit). I'm trying to picture this. In the case of an open circuit a matched driver drives the transmission line to 50% of the driving voltage. The wave reaches the open termination and is reflected with the same polarity resulting in a return wave that reaches 100% of the driving voltage. In the same vein, if the wave hits the short circuit the reflected wave will be the opposite polarity making the reflected wave 0% of the driving voltage resulting in the short circuit eventually showing to the drive circuit. Yup. You can see this happen on a time-domain reflectometer (an o'scope and a pulse generator will do). (although, to pick nits, I'd clarify your latter paragraph to read "will be of the opposite polarity, making the sum of the forward and reflected wave 0% of the driving voltage...") |
An antenna question--43 ft vertical
Jerry Stuckle wrote:
On 7/6/2015 12:41 PM, John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yup, AC has reactance. DC does not. Big difference. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. That is demonstrably false. Please demonstrate this for us as we wish to learn. OK, take your amateur transmitter. Connect it through a 1:1 balun to 300 ohm feedline. Connect that to a 300 ohm antenna. The VSWR would be near to 1:1. Of course, you would need a 300 ohm meter to measure it. It would be the same whether or not you connected your amateur transmitter According to you, you should get full power output at the antenna. In reality, you will get a 6:1 SWR and about 49% of the power at the antenna, minus transmission line loss (assuming, of course, your transmitter hasn't cut it's power back). You probably won't get full power, because the transmitter would have to produce 2.4 times it's usual output voltage to achieve it. But the system would be gratifyingly low in SWR and transmission line loss. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/6/2015 7:59 PM, Roger Hayter wrote:
Jerry Stuckle wrote: On 7/6/2015 12:41 PM, John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yup, AC has reactance. DC does not. Big difference. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. That is demonstrably false. Please demonstrate this for us as we wish to learn. OK, take your amateur transmitter. Connect it through a 1:1 balun to 300 ohm feedline. Connect that to a 300 ohm antenna. The VSWR would be near to 1:1. Of course, you would need a 300 ohm meter to measure it. It would be the same whether or not you connected your amateur transmitter That's right. It's a 1:1 SWR on the transmission line. And it matches your requirements. According to you, you should get full power output at the antenna. In reality, you will get a 6:1 SWR and about 49% of the power at the antenna, minus transmission line loss (assuming, of course, your transmitter hasn't cut it's power back). You probably won't get full power, because the transmitter would have to produce 2.4 times it's usual output voltage to achieve it. But the system would be gratifyingly low in SWR and transmission line loss. Not according to you. Since the transmitter has a comparatively low impedance, it should dissipate very little power and virtually all of the power should go to the antenna. After all, you do have a 1:1 VSWR on the transmission line/antenna, fed by a low impedance transmitter. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On Mon, 6 Jul 2015 20:33:29 +0100, Ian Jackson
wrote: In message , Jeff Liebermann writes One problem with the thermocouple ammeter. It's slow. The other problem with the thermocouple ammeter is that it's easy to burn out the thermocouple. Yep. All mine were guaranteed to be fried when I obtained them for free. However, they can be repaired. I've only done one successfully so far: http://www.tuberadio.com/robinson/Thermocouples/ But if you've been unfortunate to buy a duffer, don't totally despair. With a simple internal rewire, at least you'll have a usable* 500uA or 1mA FSD moving coil meter. *Convert to diode type? That's the most common solution: http://k4che.com/GO-9%20Transmitter/RF%20Ammeters/RF%20Ammeters.htm I had problems with RF leakage being rectified by the diode resulting in inaccurate readings. At least the typical WWII ammeter is well shielded. Calibrating the scale is impossible but replacing the meter scale is easy: http://www.tonnesoftware.com/meter2.html -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/6/2015 12:02 PM, rickman wrote:
On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. |
An antenna question--43 ft vertical
Jerry Stuckle wrote:
On 7/6/2015 7:59 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/6/2015 12:41 PM, John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yup, AC has reactance. DC does not. Big difference. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. That is demonstrably false. Please demonstrate this for us as we wish to learn. OK, take your amateur transmitter. Connect it through a 1:1 balun to 300 ohm feedline. Connect that to a 300 ohm antenna. The VSWR would be near to 1:1. Of course, you would need a 300 ohm meter to measure it. It would be the same whether or not you connected your amateur transmitter That's right. It's a 1:1 SWR on the transmission line. And it matches your requirements. According to you, you should get full power output at the antenna. In reality, you will get a 6:1 SWR and about 49% of the power at the antenna, minus transmission line loss (assuming, of course, your transmitter hasn't cut it's power back). You probably won't get full power, because the transmitter would have to produce 2.4 times it's usual output voltage to achieve it. But the system would be gratifyingly low in SWR and transmission line loss. Not according to you. Since the transmitter has a comparatively low impedance, it should dissipate very little power and virtually all of the power should go to the antenna. After all, you do have a 1:1 VSWR on the transmission line/antenna, fed by a low impedance transmitter. True. I agree. But the power it can generate depends on its load impedance. Consider a very high or even open circuit load. It can put little or no power into it, while having its normal voltage output. -- Roger Hayter |
An antenna question--43 ft vertical
In message , Jerry Stuckle
writes So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? They probably would - if they could (at least for some applications). That would then enable you to step up the TX output voltage (using a transformer), so that you could drive more power into a higher (eg 50 ohm) load. But of course, the overall output impedance would then become correspondingly higher. You would also be drawing correspondingly more current from the original 1 ohm source, and if you used too high a step-up, you would risk exceeding the permitted internal power dissipation (and other performance parameters). So yes, you are getting more power output when you match* the source impedance to the load - but it doesn't necessarily mean you always can (or should) go the whole hog. *Or, at least, partially match. -- Ian |
An antenna question--43 ft vertical
In message , Jerry Stuckle
writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! -- Ian |
An antenna question--43 ft vertical
In message , Jerry Stuckle
writes But if what you say is correct, then I should be able to get a lot of power out of my 100 watt transmitter feeding a 1 ohm antenna. But probably not for long! -- Ian |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , Jerry Stuckle writes So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? They probably would - if they could (at least for some applications). That would then enable you to step up the TX output voltage (using a transformer), so that you could drive more power into a higher (eg 50 ohm) load. But of course, the overall output impedance would then become correspondingly higher. You would also be drawing correspondingly more current from the original 1 ohm source, and if you used too high a step-up, you would risk exceeding the permitted internal power dissipation (and other performance parameters). So yes, you are getting more power output when you match* the source impedance to the load - but it doesn't necessarily mean you always can (or should) go the whole hog. *Or, at least, partially match. No, honestly, you're not getting more power output when you match the load to the source. *If* you have a *given* voltage generator with a *given* source impedance, then yes: that situation arises, for instance, if you have a very low noise amplifier with given output characteristics and you want to extract the maximum signal power in order to maintain the best noise factor through stages of amplification. But when you are designing a PA you start with a pile of components (or a catalogue of same) and you choose your voltage swing and current capacity to put as much power in the load as you want to (limited largely by the heat dissipation of the output devices in a practical circuit) and design the circuit to dissipate as little power in the amplifier as you can. You are *not* interested in transferring as much power as you can from a given circuit. It may only be tenth of the power output that you could get (ignoring practical dissipation limits) from a certain voltage with a different load, or a higher voltage with the same load (which you could achieve with a transformer), but that is irrelevant. Among other things, you are likely to end up with a low source impedance compared with the load and that makes no difference to the operation of the transmission line or aerial. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/7/2015 3:19 AM, John S wrote:
On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? -- Rick |
An antenna question--43 ft vertical
On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Why do you say that? If there is no reflection the voltage on the line is purely due to the forward signal and so the VSWR is 1:1. What's wrong with that? -- Rick |
An antenna question--43 ft vertical
On 7/7/2015 7:19 AM, rickman wrote:
On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? I will not lead you by the hand. I have supplied the requested references. If you can't find the referenced material, then you need to learn how to research, buy books, go to the library, etc. It is your loss. Education is more than online chatter. |
An antenna question--43 ft vertical
rickman wrote:
On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Why do you say that? If there is no reflection the voltage on the line is purely due to the forward signal and so the VSWR is 1:1. What's wrong with that? You are, of course, right. I suspect that VSWR was defined to give technicians a nice easy number to aim for, rather than infinite return loss, to indicate no reflections. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/7/2015 8:52 AM, John S wrote:
On 7/7/2015 7:19 AM, rickman wrote: On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? I will not lead you by the hand. I have supplied the requested references. If you can't find the referenced material, then you need to learn how to research, buy books, go to the library, etc. It is your loss. Education is more than online chatter. Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. -- Rick |
An antenna question--43 ft vertical
On 06/07/15 01:21, wrote:
John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. That is true but is doesn't address the point. There should still be somewhere to represent the source impedance, albeit normalised. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Likewise, that is a sweeping statement which evades the point. |
An antenna question--43 ft vertical
On 06/07/15 17:48, wrote:
John S wrote: On 7/6/2015 12:19 AM, wrote: John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. So, it can't be used in a 50 ohm environment? What does that have to do with anything? The chart has a SWR graph and nowhere does it need source impedance. If you disagree, please link to one. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Please show the EZNEC statement that "assumes the transmission line matches the transmitter". Look in the help section if you have EZNEC and can cut and paste or just refer me to the chapter and verse. Also, if you have EZNEC, you can insert a transmission line with arbitrary characteristic impedance, put a load on the far end matching the line, and look at the SWR. It will still be 1:1 because the LOAD matches the LINE. Not because EZNEC assumes a source impedance. Try it with and report back here. There is no way that a source initiates reflections. That is a property of the line and load only. It may re-reflect a wave reflected from the load, but that is all. You can also verify this in LTSPICE if you wish. What happens if you take any off the shelf commercial amateur radio transmitter that does not have a built in tuner and: Attach a 10 Ohm load. Attach a 200 Ohm load. Attach a 1,000 Ohm load. Attach a 1 Ohm load. Attach a 50 Ohm load. Please address my questions first before setting up another strawman. Start with Electromagnetics by Kraus and Carver, Chapter 13. Do the experiment. |
An antenna question--43 ft vertical
On 07/07/15 00:36, Dave Platt wrote:
The only case I am aware of that will give total reflection is when the terminal is open circuit with infinite impedance absorbing *no* signal. Also when it is a zero impedance (short circuit). I'm trying to picture this. In the case of an open circuit a matched driver drives the transmission line to 50% of the driving voltage. The wave reaches the open termination and is reflected with the same polarity resulting in a return wave that reaches 100% of the driving voltage. In the same vein, if the wave hits the short circuit the reflected wave will be the opposite polarity making the reflected wave 0% of the driving voltage resulting in the short circuit eventually showing to the drive circuit. Yup. You can see this happen on a time-domain reflectometer (an o'scope and a pulse generator will do). (although, to pick nits, I'd clarify your latter paragraph to read "will be of the opposite polarity, making the sum of the forward and reflected wave 0% of the driving voltage...") There is a good demo on YouTube of this. The presenter built simple little, rather neat, pulse generator to demonstrate such things. I built one for demos. It works very well. It is quite useful for measurements etc. https://www.youtube.com/watch?v=9cP6w2odGUc |
An antenna question--43 ft vertical
On 7/7/2015 6:17 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? They probably would - if they could (at least for some applications). That would then enable you to step up the TX output voltage (using a transformer), so that you could drive more power into a higher (eg 50 ohm) load. Oh, it's completely possible. It's just a matching network, anyway - one which has to be in place anyway, because the output of a tube amp is relatively high impedance, and the output of a transistor amp is relatively low impedance. In fact, a 144W transistor amp running on 12V wouldn't even need a matching network. It's output would have 1 ohm impedance. But of course, the overall output impedance would then become correspondingly higher. You would also be drawing correspondingly more current from the original 1 ohm source, and if you used too high a step-up, you would risk exceeding the permitted internal power dissipation (and other performance parameters). And why would the output impedance change just because the load impedance changes? They are two separate things. But according to you, no step-up is required - you can drive any (comparatively) high impedance source most efficiently from a low impedance. So yes, you are getting more power output when you match* the source impedance to the load - but it doesn't necessarily mean you always can (or should) go the whole hog. *Or, at least, partially match. So, which is it? First you say the output impedance of the transmitter should be very low for maximum power to the antenna. Now you say it should be matched. Then you say it shouldn't be matched. Which is it? -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/7/2015 8:33 AM, rickman wrote:
On 7/7/2015 8:52 AM, John S wrote: On 7/7/2015 7:19 AM, rickman wrote: On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? I will not lead you by the hand. I have supplied the requested references. If you can't find the referenced material, then you need to learn how to research, buy books, go to the library, etc. It is your loss. Education is more than online chatter. Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. Yes, there is. I did NOT say that a wire was dissipationless. Please reread the above. I said "There can be a lossless resistive part of source impedance". Source impedance may or may not contain resistance which dissipates. A 50 ohm piece of coax has an impedance of approximately 50+j0 ohms. So, does the 50 ohms dissipate? Look at the specifications of the coax. How much loss? If you are unwilling to go to the library or do research or buy a book on you own, then you are beyond help. What is your ignorance worth to you? |
An antenna question--43 ft vertical
On 7/7/2015 3:05 AM, John S wrote:
On 7/6/2015 1:03 PM, wrote: John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of current. How does the AC change the law? You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps flowing (open circuit). You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing, with the current leading the voltage by 90 degrees. You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps flowing (short circuit). You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing with the voltage leading the current by 90 degrees. You place the capacitor and inductor in series. Fed with DC, you get 0 amps flowing (open circuit). Fed with 1MHz AC, you get infinite current flowing (short circuit). You place the capacitor and inductor in parallel. Fed with DC, you get infinite current flowing (short circuit). Fed with 1MHz AC you get 0 amps flowing (open circuit). There is a huge difference between ac and dc! -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/7/2015 3:37 AM, Roger Hayter wrote:
Jerry Stuckle wrote: On 7/6/2015 7:59 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/6/2015 12:41 PM, John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yup, AC has reactance. DC does not. Big difference. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. That is demonstrably false. Please demonstrate this for us as we wish to learn. OK, take your amateur transmitter. Connect it through a 1:1 balun to 300 ohm feedline. Connect that to a 300 ohm antenna. The VSWR would be near to 1:1. Of course, you would need a 300 ohm meter to measure it. It would be the same whether or not you connected your amateur transmitter That's right. It's a 1:1 SWR on the transmission line. And it matches your requirements. According to you, you should get full power output at the antenna. In reality, you will get a 6:1 SWR and about 49% of the power at the antenna, minus transmission line loss (assuming, of course, your transmitter hasn't cut it's power back). You probably won't get full power, because the transmitter would have to produce 2.4 times it's usual output voltage to achieve it. But the system would be gratifyingly low in SWR and transmission line loss. Not according to you. Since the transmitter has a comparatively low impedance, it should dissipate very little power and virtually all of the power should go to the antenna. After all, you do have a 1:1 VSWR on the transmission line/antenna, fed by a low impedance transmitter. True. I agree. But the power it can generate depends on its load impedance. Consider a very high or even open circuit load. It can put little or no power into it, while having its normal voltage output. No, the power the transmitter can generate has nothing to do with the load impedance. The transmitter generates what it can generate. The amount of power transferred to the load is dependent on the load impedance, with the maximum power transfer occurring when the impedances match. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
In message , rickman
writes On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Why do you say that? If there is no reflection the voltage on the line is purely due to the forward signal and so the VSWR is 1:1. What's wrong with that? A standing wave is caused by a reflection. If there IS no reflection, there is NO standing wave. So while you can have an SWR of 1.00000000000001-to-1 (because a standing wave DOES exist), you can't really have one of 1-to-1 (because there IS no standing wave). ;o)) [Just a bit of pedantic, lateral thinking on my part. Don't worry too much about it. It has absolutely no bearing whatsoever on the current discussions.] -- Ian |
An antenna question--43 ft vertical
On 7/5/2015 7:21 PM, wrote:
John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. The EZNEC help file is very comprehensive. Please find any reference to your assertion that there is an assumption of source impedance there and provide information for us to verify your assertion. |
An antenna question--43 ft vertical
"John S" wrote in message ... On 7/5/2015 7:08 PM, Wayne wrote: "Jeff Liebermann" wrote in message ... On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote: As for a simpler way, I'd recommend a remote auto-matcher like an SGC at the antenna base. It will minimise coax losses and should give you a good match, at least for most bands. I've used a similar set up (with radials) and achieved a good match even on 80m. If your radio has a built in tuner, then it can be used to 'tweak' the match in the event the radio isn't 'seeing' 1.5:1. Turn it off initially. Let the SGC find a match. If it isn't ideal, use the local ATU for a final tweak. I never found this was required but YMMV. Not everyone is a true believer in antenna tuners: http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html Interesting. I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. Hey, Wayne - As a matter of curiosity on my part, can you find a way to measure the ammeter's resistance and let me know the full-scale value? No, I don't have enough test equipment to easily do that. With a DVM it measures 0.4 ohms and with a VOM measures 28 ohms. And the VOM gives no needle movement. It is a O. D. McClintock Signal Corp typs I S-III with full scale of 2.5 amps. Since it was salvaged from some WW II equipment back in the 1950s, it probably isn't calibrated. But, it gives a useable relative reading. |
An antenna question--43 ft vertical
On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Wrong. An SWR of 1:1 indicates a perfect match, with no reflected power. It is recognized by all electronics texts and experts. My suggestion would be for you to learn some transmission line theory. Your statement here just showed you have no knowledge of it at all. Even when I took my novice test many years ago I had to understand SWR better than that. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/7/2015 6:34 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes But if what you say is correct, then I should be able to get a lot of power out of my 100 watt transmitter feeding a 1 ohm antenna. But probably not for long! Why not? The 100 W transmitter will only put out 100 watts! As a matter of fact, due to the 50:1 SWR, I will only get about 4 watts to the antenna, assuming, of course, the transmitter doesn't shut down from having to dissipate and additional 96 watts. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
"rickman" wrote in message ... On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? Just a comment.... Independent of discussions on this thread, Walt's book is an excellent reference to have. It's available on Amazon. |
An antenna question--43 ft vertical
On 7/7/2015 8:52 AM, John S wrote:
On 7/7/2015 7:19 AM, rickman wrote: On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? I will not lead you by the hand. I have supplied the requested references. If you can't find the referenced material, then you need to learn how to research, buy books, go to the library, etc. It is your loss. Education is more than online chatter. What he wants is a Wikipedia article. If something can't be found there, it doesn't exist. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/7/2015 10:14 AM, Wayne wrote:
"John S" wrote in message ... On 7/5/2015 7:08 PM, Wayne wrote: "Jeff Liebermann" wrote in message ... On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote: As for a simpler way, I'd recommend a remote auto-matcher like an SGC at the antenna base. It will minimise coax losses and should give you a good match, at least for most bands. I've used a similar set up (with radials) and achieved a good match even on 80m. If your radio has a built in tuner, then it can be used to 'tweak' the match in the event the radio isn't 'seeing' 1.5:1. Turn it off initially. Let the SGC find a match. If it isn't ideal, use the local ATU for a final tweak. I never found this was required but YMMV. Not everyone is a true believer in antenna tuners: http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html Interesting. I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. Hey, Wayne - As a matter of curiosity on my part, can you find a way to measure the ammeter's resistance and let me know the full-scale value? No, I don't have enough test equipment to easily do that. With a DVM it measures 0.4 ohms and with a VOM measures 28 ohms. And the VOM gives no needle movement. It is a O. D. McClintock Signal Corp typs I S-III with full scale of 2.5 amps. Since it was salvaged from some WW II equipment back in the 1950s, it probably isn't calibrated. But, it gives a useable relative reading. Many thanks for the info. The reason I asked was that I thought it might be possible to build one. I need a starting point. Again, thanks. |
An antenna question--43 ft vertical
On 7/5/2015 11:12 AM, rickman wrote:
On 7/5/2015 11:35 AM, John S wrote: I encourage all of you to read Walter Maxwell's (W2DU) book "Reflections III". It will explain everything about this. Everything you are discussing has been put to bed. Of course it has. We are not inventing anything here, we are trying to understand it. Does this writing have a Cliff Notes version? I don't know. I bought the book. |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , rickman writes On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Why do you say that? If there is no reflection the voltage on the line is purely due to the forward signal and so the VSWR is 1:1. What's wrong with that? A standing wave is caused by a reflection. If there IS no reflection, there is NO standing wave. So while you can have an SWR of 1.00000000000001-to-1 (because a standing wave DOES exist), you can't really have one of 1-to-1 (because there IS no standing wave). ;o)) [Just a bit of pedantic, lateral thinking on my part. Don't worry too much about it. It has absolutely no bearing whatsoever on the current discussions.] Quite so: a voltage standing wave *ratio* of 1 means no standing wave. But in the name of the unit the "standing wave" is adjectival, and it is still a valid name even when there is no standing wave. And, anyway, you can still colloquially have a resistance of zero ohms even for a superconductor! -- Roger Hayter |
An antenna question--43 ft vertical
In article , rickman wrote:
Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. Yah. It's a question of terminology. Unfortunately, one term has come to be used for two (related but different) concepts. There is "resistance", as in the E=I^2*R sort. If I recall correctly, Maxwell refers to this as "dissipative" impedance. If you put current through a dissipative resistance, a voltage drop develops across the resistance, and power is dissipated. There are plenty of examples of this, with which I'm sure you're familiar. There is also "resistance", as in "the 'real', non-reactive component of a complex impedance, in which current is in phase with voltage." This type of "resistance" is fundamentally non-dissipative - that is, you can run power through it without dissipating the power as heat. There are also good examples of this. One "textbook" example would be a perfectly-lossless transmission line... say, one made out of a wire and tube of a superconductor, cooled to below the superconducting temperature. You can (in principle) build such a superconducting coax to have almost any convenient impedance... 50 or 75 ohms, for example. Since we're theorizing, let's assume we can built one a few trillion miles long... so long that the far end is light-years away. If you hook a transmitter to one end of this and start transmitting, it will "look" to the transmitter like a 50-ohm dummy load. The transmitter itself won't be able to tell the difference. The transmitter puts out an RF voltage, and the line "takes current" exactly in phase with the voltage, in a ratio of one RF ampere per 50 RF volts. But, there's a fundamental difference between this "resistance" and that of a dummy load. A 50-ohm dummy load's resistance is dissipative... all of the power going into it turns into heat, and is dissipated in accordance with the fundamental laws of thermodynamics. *None* of the power being fed into the superconducting coax, is dissipated as heat in the coax. All of the power still exists, in its original RF form. It's being stored/propagated down the coax without loss. When it hits a load at the other end, it may be dissipated as heat there. Or, perhaps not. What if what's at the other end of the superconducting coax is a superconducting antenna, tweaked to present an impedance of exactly 50 ohms? The RF will be radiated into space. And, "free space" is another great example of a medium that has a well-defined "resistance" (in the non-dissipated sense). https://en.wikipedia.org/wiki/Impedance_of_free_space One of the fundamental jobs of an antenna, is to match the impedance of its feedline to the impedance of free space. Now, any coax you can buy at the store has *both* types of "resistance", of course. It has a dissipative component, and a non-dissipative component. Typically, the more you spend and the more you have to strain your back carrying it around, the lower the amount of dissipative resistance (which is only good for keeping the pigeons' feet warm) and the more predictable and precisely-defined the non-dissipative part. |
An antenna question--43 ft vertical
In message , Jerry Stuckle
writes On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Wrong. An SWR of 1:1 indicates a perfect match, with no reflected power. It is recognized by all electronics texts and experts. My suggestion would be for you to learn some transmission line theory. Your statement here just showed you have no knowledge of it at all. Even when I took my novice test many years ago I had to understand SWR better than that. The point I'm trying to make is not technical. It's simply one of verbal logic. Without the presence of a standing wave, you can't possibly have something called a "standing wave ratio". But, like all RF engineers, an SWR of 1-to-1 is something I too strive to achieve! "Yesterday, upon the stair, I met a man who wasn't there. He wasn't there again today, I wish, I wish he'd go away..." -- Ian |
An antenna question--43 ft vertical
John S wrote:
On 7/6/2015 1:03 PM, wrote: John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of current. How does the AC change the law? What part of "you get a whole new set of laws" was it you failed to understand? Here's a clue for you; at DC the reactive components of a length of wire are irrelevant but at AC they are not. -- Jim Pennino |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Despite the name, VSWR is defined in terms of complex impedances and wavelengths, not "waves" of any kind. -- Jim Pennino |
An antenna question--43 ft vertical
Brian Reay wrote:
Do the experiment. Did it decades ago in electromagnetics lab with calibrated test equipmemnt, not with amateur radio equipment. -- Jim Pennino |
An antenna question--43 ft vertical
Brian Reay wrote:
On 06/07/15 01:21, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. That is true but is doesn't address the point. There should still be somewhere to represent the source impedance, albeit normalised. The purpose of a Smith chart it to match a SOURCE to a LOAD. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Likewise, that is a sweeping statement which evades the point. The main purpose of EZNEC is to design an antenna for amateur radio use and all commercial amateur radio transmitters have an output impedance of 50 Ohms. -- Jim Pennino |
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