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Tuna Tin (II) output impedance
Over the Xmas break I constructed a simple 2 transistor QRP transmitter using
the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Richard Gary Morton wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Richard Gary Morton wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote: As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage I'd be interested to see the derivation of this. It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement I did the same thing (using different resistor values and then calculating by V^2/R like the OP has done. It worked for me. The Zout of my particular design was 140 ohms, though - just right for a folded dipole. I used 2n3904 transistors; they appear to give a better gain than 2n2222s. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote: As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage I'd be interested to see the derivation of this. It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement I did the same thing (using different resistor values and then calculating by V^2/R like the OP has done. It worked for me. The Zout of my particular design was 140 ohms, though - just right for a folded dipole. I used 2n3904 transistors; they appear to give a better gain than 2n2222s. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
If you want to obtain the maximum power transfer between a generator and a
load, and if all the pieces behave like ideal linear components, then you need to match the impedance of the two. But this is not what you are doing here. First, you want to obtain the maximum power out without burning up you output device, and second, a class-C output stage isn't a linear circuit! What this means is that in general your transmitter, like your wall socket, can deliver more than it's rated power if you put the right load on it. Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. Your circuit doesn't act exactly like that, even if it doesn't follow the same curve for a generator with a linear impedance. I suspect that with your circuit the 7-pole filter is evening things out quite a bit. There's also a good chance that the "strange results" you see with a load below 10 ohms are the final amplifier going unstable and oscillating, or your output transistor heating up and changing characteristics on you. So what it boils down to is that it is very important that your output stage _sees_ the right impedance, but you shouldn't expect the output stage to _deliver_ the same impedance that it needs to see. As long as you're getting power out and your output transistor isn't letting all the smoke out then you're fine. "Gary Morton" wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
If you want to obtain the maximum power transfer between a generator and a
load, and if all the pieces behave like ideal linear components, then you need to match the impedance of the two. But this is not what you are doing here. First, you want to obtain the maximum power out without burning up you output device, and second, a class-C output stage isn't a linear circuit! What this means is that in general your transmitter, like your wall socket, can deliver more than it's rated power if you put the right load on it. Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. Your circuit doesn't act exactly like that, even if it doesn't follow the same curve for a generator with a linear impedance. I suspect that with your circuit the 7-pole filter is evening things out quite a bit. There's also a good chance that the "strange results" you see with a load below 10 ohms are the final amplifier going unstable and oscillating, or your output transistor heating up and changing characteristics on you. So what it boils down to is that it is very important that your output stage _sees_ the right impedance, but you shouldn't expect the output stage to _deliver_ the same impedance that it needs to see. As long as you're getting power out and your output transistor isn't letting all the smoke out then you're fine. "Gary Morton" wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Tim Wescott wrote:
snip What this means is that in general your transmitter, like your wall socket, can deliver more than it's rated power if you put the right load on it. Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. Your circuit doesn't act exactly like that, even if it doesn't follow the same curve for a generator with a linear impedance. I suspect that with your circuit the 7-pole filter is evening things out quite a bit. There's also a good chance that the "strange results" you see with a load below 10 ohms are the final amplifier going unstable and oscillating, or your output transistor heating up and changing characteristics on you. Yes there were bad things going on with very low value loads, the output was collapsing and re-establishing. The srtange results may also be caused by the loads with were high wattage resistors - no idea what inductance they might have had. So what it boils down to is that it is very important that your output stage _sees_ the right impedance, but you shouldn't expect the output stage to _deliver_ the same impedance that it needs to see. As long as you're getting power out and your output transistor isn't letting all the smoke out then you're fine. So perhaps the fact that the transmitter is connected to a 50 ohm load and the output transformer is 2:1 turns ratio means that the transistor is "seeing" 200 ohms? Nevertheless I am still interested in how to verify that any circuit which I design (copy!) does have a 50 ohm output impedance. There must be some way to verify this figure. --Gary |
Tim Wescott wrote:
snip What this means is that in general your transmitter, like your wall socket, can deliver more than it's rated power if you put the right load on it. Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. Your circuit doesn't act exactly like that, even if it doesn't follow the same curve for a generator with a linear impedance. I suspect that with your circuit the 7-pole filter is evening things out quite a bit. There's also a good chance that the "strange results" you see with a load below 10 ohms are the final amplifier going unstable and oscillating, or your output transistor heating up and changing characteristics on you. Yes there were bad things going on with very low value loads, the output was collapsing and re-establishing. The srtange results may also be caused by the loads with were high wattage resistors - no idea what inductance they might have had. So what it boils down to is that it is very important that your output stage _sees_ the right impedance, but you shouldn't expect the output stage to _deliver_ the same impedance that it needs to see. As long as you're getting power out and your output transistor isn't letting all the smoke out then you're fine. So perhaps the fact that the transmitter is connected to a 50 ohm load and the output transformer is 2:1 turns ratio means that the transistor is "seeing" 200 ohms? Nevertheless I am still interested in how to verify that any circuit which I design (copy!) does have a 50 ohm output impedance. There must be some way to verify this figure. --Gary |
Richard Hosking wrote:
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load). This value is certainly in the right ballpark. I guess that I should also measure the current taken by the final stage and see if this ties in too. I connected various 1/4w resistors across the output using a couple of adaptors, but being aware to keep wire lengths as short as possible. I then simply used a scope probe and scope to make the measurement. Scope probe was a standard x10 10Mohm impedance and presumably small C. Interestingly I was surprised at the magnitude of the voltages for such a tiny power. regards.. --Gary |
Richard Hosking wrote:
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load). This value is certainly in the right ballpark. I guess that I should also measure the current taken by the final stage and see if this ties in too. I connected various 1/4w resistors across the output using a couple of adaptors, but being aware to keep wire lengths as short as possible. I then simply used a scope probe and scope to make the measurement. Scope probe was a standard x10 10Mohm impedance and presumably small C. Interestingly I was surprised at the magnitude of the voltages for such a tiny power. regards.. --Gary |
Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Leon -- Leon Heller, G1HSM Email: My low-cost Philips LPC210x ARM development system: http://www.geocities.com/leon_heller/lpc2104.html |
Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Leon -- Leon Heller, G1HSM Email: My low-cost Philips LPC210x ARM development system: http://www.geocities.com/leon_heller/lpc2104.html |
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote: Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Hi, Leon, Do you know Brian (G4BCO) by any chance? My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote: Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Hi, Leon, Do you know Brian (G4BCO) by any chance? My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Paul Burridge wrote: On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller wrote: Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Hi, Leon, Do you know Brian (G4BCO) by any chance? My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. No, that isn't the case. From AN721: "The load value is primarily dictated by the required output power and the peak voltage; it is not matched to the output impedance of the device." Leon -- Leon Heller, G1HSM Email: My low-cost Philips LPC210x ARM development system: http://www.geocities.com/leon_heller/lpc2104.html |
Paul Burridge wrote: On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller wrote: Paul Burridge wrote: On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott" wrote: Normally you'd expect the final amplifier to be fairly "stiff" - the output power should be inversely proportional to the resistor you put on it. ??? Shirley, some mistake. Both AF and HF power amplifiers generally have a very low output impedance. Motorola's AN721 app note on impedance matching networks for RF transistors explains this. This doesn't apply to QRP output stages, though. Hi, Leon, Do you know Brian (G4BCO) by any chance? My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. No, that isn't the case. From AN721: "The load value is primarily dictated by the required output power and the peak voltage; it is not matched to the output impedance of the device." Leon -- Leon Heller, G1HSM Email: My low-cost Philips LPC210x ARM development system: http://www.geocities.com/leon_heller/lpc2104.html |
OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can fool yourself into thinking it is, but it's not. (2) the typical RF output stage, sans output filter, which consists of a transistor fed by a choke or transformer and which pulls the collector voltage nearly down to the emitter acts like an RF voltage source with a fairly low series resistance. I've _done_ this. I've _measured_ this. I've _burnt_ transistors. You _can_ get much more power from your output stage than it's "normal" output power, for that brief (and often shining) moment before the output transistor lets the smoke out and reverts to the sand and basic petrochemicals from which it was originally made. This is how you know that you aren't running it at the Rs = Rl point -- rather, you're running it at the point where the junction temperature of the transistor is kept to a level that will let it stay a transistor for a satisfying amount of time. This is why in my original reply I compared the output amp to a wall socket. The equivalent Rs of your average wall socket is probably on the order 1/2 to 4 ohms (on the low end for 120V, high for 230). Since I live in the US and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit is only is only good for 15 amps and I'm asking for 120?!? Dang! What happened to the lights? So you see, every time you plug anything into the wall, you're failing to match Rs with Rl, and that's a _very_ good thing. It's for exactly the same reason that you don't match Rs with Rl in your output stage. "Paul Burridge" wrote in message ... On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller wrote: -- snip -- My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can fool yourself into thinking it is, but it's not. (2) the typical RF output stage, sans output filter, which consists of a transistor fed by a choke or transformer and which pulls the collector voltage nearly down to the emitter acts like an RF voltage source with a fairly low series resistance. I've _done_ this. I've _measured_ this. I've _burnt_ transistors. You _can_ get much more power from your output stage than it's "normal" output power, for that brief (and often shining) moment before the output transistor lets the smoke out and reverts to the sand and basic petrochemicals from which it was originally made. This is how you know that you aren't running it at the Rs = Rl point -- rather, you're running it at the point where the junction temperature of the transistor is kept to a level that will let it stay a transistor for a satisfying amount of time. This is why in my original reply I compared the output amp to a wall socket. The equivalent Rs of your average wall socket is probably on the order 1/2 to 4 ohms (on the low end for 120V, high for 230). Since I live in the US and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit is only is only good for 15 amps and I'm asking for 120?!? Dang! What happened to the lights? So you see, every time you plug anything into the wall, you're failing to match Rs with Rl, and that's a _very_ good thing. It's for exactly the same reason that you don't match Rs with Rl in your output stage. "Paul Burridge" wrote in message ... On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller wrote: -- snip -- My - admittedly little - understanding of the situation is that the power transferred into a resistor (regardless of power level) will be maximised when Rs=Rl,. whereupon half the power is dissipated in the final device and half in the load. Tim's posts seems to refute this basic truism. Clearly the statement that the output power should be inversely proportional to the load resistance *must* be false. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
"Gary Morton" wrote in message ... -- snip -- So perhaps the fact that the transmitter is connected to a 50 ohm load and the output transformer is 2:1 turns ratio means that the transistor is "seeing" 200 ohms? Exactly. Nevertheless I am still interested in how to verify that any circuit which I design (copy!) does have a 50 ohm output impedance. There must be some way to verify this figure. Please see my reply to Paul Burrage in the next branch over -- hopefully I'm clarifying that for a power output stage you _don't_ want to see matched impedances. For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! --Gary |
"Gary Morton" wrote in message ... -- snip -- So perhaps the fact that the transmitter is connected to a 50 ohm load and the output transformer is 2:1 turns ratio means that the transistor is "seeing" 200 ohms? Exactly. Nevertheless I am still interested in how to verify that any circuit which I design (copy!) does have a 50 ohm output impedance. There must be some way to verify this figure. Please see my reply to Paul Burrage in the next branch over -- hopefully I'm clarifying that for a power output stage you _don't_ want to see matched impedances. For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! --Gary |
Here's an experament: Build a power output stage _only_. Don't put a
filter on it. You should have a transistor with the emitter grounded, a choke to +12V, and some minimal network on the base to keep it biased. You should be able to pick this out of your Tuna Tin II design. Capacitor couple an appropriate resistor to the collector (200 ohms if you use your Tuna Tin II). Now feed it with enough RF so it's really into class C, and measure the RF voltage on the resistor. It should be around 8V (supply * 0.35) give or take a bit. Now change the output resistor by a factor of two and measure the RF voltage again. It should be about the same as before assuming that you've sized your choke correctly (4x the design resistance). This shows that your output stage is stiff (i.e. has a low equivalent Rs). What this all means is that your average well-designed output stage does _not_ have a matched Rs! Why? Because when Rs = Rl the power in Rl is maximized _for that Rs_, but the power burnt into heat inside of Rs is also maximized! Since you want the power in Rl to be much higher than the power used to heat up your shack (unless it's really cold inside) this is a good and desireable thing. "Gary Morton" wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
Here's an experament: Build a power output stage _only_. Don't put a
filter on it. You should have a transistor with the emitter grounded, a choke to +12V, and some minimal network on the base to keep it biased. You should be able to pick this out of your Tuna Tin II design. Capacitor couple an appropriate resistor to the collector (200 ohms if you use your Tuna Tin II). Now feed it with enough RF so it's really into class C, and measure the RF voltage on the resistor. It should be around 8V (supply * 0.35) give or take a bit. Now change the output resistor by a factor of two and measure the RF voltage again. It should be about the same as before assuming that you've sized your choke correctly (4x the design resistance). This shows that your output stage is stiff (i.e. has a low equivalent Rs). What this all means is that your average well-designed output stage does _not_ have a matched Rs! Why? Because when Rs = Rl the power in Rl is maximized _for that Rs_, but the power burnt into heat inside of Rs is also maximized! Since you want the power in Rl to be much higher than the power used to heat up your shack (unless it's really cold inside) this is a good and desireable thing. "Gary Morton" wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott"
wrote: For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott"
wrote: For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period. Your arrogant ignorance overwhelms me!! Listen to those who know more than you. Read a few engineering texts. I am not willing to explain this again to a fool. John - K6QQ |
Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period. Your arrogant ignorance overwhelms me!! Listen to those who know more than you. Read a few engineering texts. I am not willing to explain this again to a fool. John - K6QQ |
Wow. Paul...have you ever thought about trying caffeine free?
"Paul Burridge" wrote in message ... Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. |
Wow. Paul...have you ever thought about trying caffeine free?
"Paul Burridge" wrote in message ... Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. |
Paul, please look over my various posts on this issue and tell me where I
failed to point out that you can get more power from a bare RF output stage by loading it differently. I'm gonna go fertilize my garden! "Paul Burridge" wrote in message ... On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott" wrote: For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Paul, please look over my various posts on this issue and tell me where I
failed to point out that you can get more power from a bare RF output stage by loading it differently. I'm gonna go fertilize my garden! "Paul Burridge" wrote in message ... On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott" wrote: For a stage (such as an intermediate, class A stage) where you _do_ want to verify the output impedance you can either find the resistance _and reactance_ that maximizes the power output, or you can build an RF impedance bridge (I use a noise bridge) and measure the stage output impedance directly. Just don't let it break into oscillation on the output of that poor impedance bridge! Bull****. In any class of operation, maximum power will be transferred when the impedances are matched. Period. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Mon, 5 Jan 2004 08:48:35 -0800, "Tim Wescott"
wrote: Paul, please look over my various posts on this issue and tell me where I failed to point out that you can get more power from a bare RF output stage by loading it differently. Sorry, Tim. I've no idea what you mean. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Mon, 5 Jan 2004 08:48:35 -0800, "Tim Wescott"
wrote: Paul, please look over my various posts on this issue and tell me where I failed to point out that you can get more power from a bare RF output stage by loading it differently. Sorry, Tim. I've no idea what you mean. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Well, you're objecting rather strongly to an assertion that I never made.
In theory if you want to get the maximum available power out of a generator you _do_ match it's output impedance. In practice if you do this with most RF final amplifiers you will reduce your final element (whether transistor or tube) to slag -- that's why modern commercial radios have SWR protection circuitry. Please see my reply to your post starting with "Hi Leon" for a full explanation, and please actually read it before replying. Remember also that everything I say (including the attempt to match output impedances with the wall socket -- I was only 8 but that's no excuse) has been backed up by experament. "Paul Burridge" wrote in message ... On Mon, 5 Jan 2004 08:48:35 -0800, "Tim Wescott" wrote: Paul, please look over my various posts on this issue and tell me where I failed to point out that you can get more power from a bare RF output stage by loading it differently. Sorry, Tim. I've no idea what you mean. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Well, you're objecting rather strongly to an assertion that I never made.
In theory if you want to get the maximum available power out of a generator you _do_ match it's output impedance. In practice if you do this with most RF final amplifiers you will reduce your final element (whether transistor or tube) to slag -- that's why modern commercial radios have SWR protection circuitry. Please see my reply to your post starting with "Hi Leon" for a full explanation, and please actually read it before replying. Remember also that everything I say (including the attempt to match output impedances with the wall socket -- I was only 8 but that's no excuse) has been backed up by experament. "Paul Burridge" wrote in message ... On Mon, 5 Jan 2004 08:48:35 -0800, "Tim Wescott" wrote: Paul, please look over my various posts on this issue and tell me where I failed to point out that you can get more power from a bare RF output stage by loading it differently. Sorry, Tim. I've no idea what you mean. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Mon, 5 Jan 2004 15:35:24 -0800, "Tim Wescott"
wrote: Well, you're objecting rather strongly to an assertion that I never made. In theory if you want to get the maximum available power out of a generator you _do_ match it's output impedance. In practice if you do this with most RF final amplifiers you will reduce your final element (whether transistor or tube) to slag -- that's why modern commercial radios have SWR protection circuitry. Please see my reply to your post starting with "Hi Leon" for a full explanation, and please actually read it before replying. Remember also that everything I say (including the attempt to match output impedances with the wall socket -- I was only 8 but that's no excuse) has been backed up by experament. Okay, so if I understand you correctly you're saying that whilst you should try to match impedances in Class A; it's inapplicable in Class C? It's just an operation class distinction? -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
On Mon, 5 Jan 2004 15:35:24 -0800, "Tim Wescott"
wrote: Well, you're objecting rather strongly to an assertion that I never made. In theory if you want to get the maximum available power out of a generator you _do_ match it's output impedance. In practice if you do this with most RF final amplifiers you will reduce your final element (whether transistor or tube) to slag -- that's why modern commercial radios have SWR protection circuitry. Please see my reply to your post starting with "Hi Leon" for a full explanation, and please actually read it before replying. Remember also that everything I say (including the attempt to match output impedances with the wall socket -- I was only 8 but that's no excuse) has been backed up by experament. Okay, so if I understand you correctly you're saying that whilst you should try to match impedances in Class A; it's inapplicable in Class C? It's just an operation class distinction? -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
Sorta. The distinction has more to do with whether you're following your
output stage with something that needs to see a specific impedance (like a filter in a receiver), and whether you're building a power stage. I couldn't think of a good generalization, so I listed some examples below. Then I realized that it really boils down to the fact that the impedance matching rule is a tool for predicing circuit behavior, not a guideline for life. RF stage in a receiver: you want to match the output impedance to the following filter or mixer for maximum power transfer and good filter/mixer performance. You usually _don't_ want to match input impedance because the best power transfer impedance is not the same as the best noise figure impedance. IF preamplifier (between a passive mixer and a crystal or mechanical filter): Match both input and output impedances, both to maximize power transfer and to get good performance from the attached stages. 1st buffer in a VFO: _don't_ match impedances with the oscillator! You want to purposely refrain from taking much power from the oscillator tank, because a heavily loaded oscillator is a poor oscillator. You probably _do_ want to match impedances on the output so the following stage won't need so much gain. 1-transistor transmitter: It's an oscillator, but a power stage too. Load it (match impedances) enough to get useful power on the antenna, but not so much that you screw up performance. Power output (usually): match impedances to the input, don't match impedance to the output (for all the reasons given earlier). This applies to any class of stage: A, B, AB, C, D and E (yes, there are class D and E stages). "Paul Burridge" wrote in message ... On Mon, 5 Jan 2004 15:35:24 -0800, "Tim Wescott" wrote: Well, you're objecting rather strongly to an assertion that I never made. In theory if you want to get the maximum available power out of a generator you _do_ match it's output impedance. In practice if you do this with most RF final amplifiers you will reduce your final element (whether transistor or tube) to slag -- that's why modern commercial radios have SWR protection circuitry. Please see my reply to your post starting with "Hi Leon" for a full explanation, and please actually read it before replying. Remember also that everything I say (including the attempt to match output impedances with the wall socket -- I was only 8 but that's no excuse) has been backed up by experament. Okay, so if I understand you correctly you're saying that whilst you should try to match impedances in Class A; it's inapplicable in Class C? It's just an operation class distinction? -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
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