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Output Impedance
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? |
Output Impedance
On 16/05/14 20:44, gareth wrote:
One thing that puzzled me for years You seem to have spent quite a lot of your time puzzled by some, very basic, problems; have you ever thought that you may have chosen a route that wasn't 'right' for you? |
Output Impedance
"gareth" wrote in message ... One thing that puzzled me for years was why do I keep posting this crap? |
Output Impedance
On 16/05/14 20:49, Osama Bean Laden wrote:
One thing that puzzled me for years You seem to have spent quite a lot of your time puzzled by some, very basic, problems; have you ever thought that you may have chosen a route that wasn't 'right' for you? It isn't a good advent for his old Uni. Nor will it inspire confidence in anyone accepting his offers of assistance. |
Output Impedance
"Brian Reay" wrote in message
... It isn't a good advent for his old Uni. Nor will it inspire confidence in anyone accepting his offers of assistance. 39 |
Output Impedance
"gareth" wrote in message
... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. ***** And there are some who are simply too pig-ignorant even to bluster but they tailgate with glee any abusive remarks in order to justify their own personality defects. |
Output Impedance
"gareth" wrote in message
... "gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. You would need to increase the supply volts to at least 200V to give headroom, unless you want a square wave output. -- ;-) .. 73 de Frank Turner-Smith G3VKI - mine's a pint. .. http://turner-smith.co.uk |
Output Impedance
"gareth" wrote in message
... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Consider for a moment the equivalent circuit of a current generator with a shunt impedance feeding a load. The voltage across the load is the same as that across the current generator, and measuring the current coming out of the generator at any one time will tell you about the load impedance, but not the shunt impedance. However, by changing the load impedance and again measuring the voltage and the current, then you will ahve sufficient information to calcualte the source impedance. |
Output Impedance
FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message ... "gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. You would need to increase the supply volts to at least 200V to give headroom, unless you want a square wave output. It is a widely distributed misunderstanding that a linear that has been designed to delever its output in a 50 ohm load always has a 50 ohm output impedance. In fact it almost never has. |
Output Impedance
On 17/05/2014 07:07, gareth wrote:
So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. And this is why valved PAs in older radios (FT-101Es, for example) had variable impedance matching (the Pi Tank circuit), and why it's good practice to tweak that matching after increasing the drive to full power (say 250mA anode current) having first matched at a lower power (100mA anode current). |
Output Impedance
On 17/05/2014 08:12, FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message ... "gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. You would need to increase the supply volts to at least 200V to give headroom, unless you want a square wave output. Matthew 7:6 |
Output Impedance
On 17/05/14 09:29, Kafkaesque wrote:
On 17/05/2014 07:07, gareth wrote: So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. And this is why valved PAs in older radios (FT-101Es, for example) had variable impedance matching (the Pi Tank circuit), and why it's good practice to tweak that matching after increasing the drive to full power (say 250mA anode current) having first matched at a lower power (100mA anode current). Or use a tone-[pulser. -- Spike |
Output Impedance
On 17/05/2014 09:35, Spike wrote:
On 17/05/14 09:29, Kafkaesque wrote: On 17/05/2014 07:07, gareth wrote: So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. And this is why valved PAs in older radios (FT-101Es, for example) had variable impedance matching (the Pi Tank circuit), and why it's good practice to tweak that matching after increasing the drive to full power (say 250mA anode current) having first matched at a lower power (100mA anode current). Or use a tone-[pulser. Or a bug key sending a train of dits. 12wpm would NOT be fast enough. |
Output Impedance
In message , Rob
writes FranK Turner-Smith G3VKI wrote: "gareth" wrote in message ... "gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. You would need to increase the supply volts to at least 200V to give headroom, unless you want a square wave output. It is a widely distributed misunderstanding that a linear that has been designed to delever its output in a 50 ohm load always has a 50 ohm output impedance. In fact it almost never has. My understanding of RF design doesn't include how to calculate the output impedance of PA stages. However, it's obvious that of the output impedance was 50 ohms, the PA efficiency would be only 50% at best - and clearly this is not so (certainly for class-C operation, which is notionally 66%-ish). I've always understood that, in practice, it's usually resistively low-ish (around 25 ohms?) and quite capacitive. An expert opinion is needed! -- Ian |
Output Impedance
"Kafkaesque" wrote in message
... On 17/05/2014 08:12, FranK Turner-Smith G3VKI wrote: "gareth" wrote in message ... "gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Interesting how the great blusterers display their ignorance, either by not entering the discussion by the device of side-stepping with abusive remarks*****, or trumpeting complete nonsense .... Consider the case of your homebrewed 1.5kW linear being set up with QRP for safety. Let's say 6W output on a 12V supply. Some claim that this means an impedance of 24 ohms that has to be matched to the 50 ohm antenna. So, having done your matching, you now crank up the output to 1.44kW by increasing the drive, and those some claimers will now say that the ouput impedance is now 0.1 ohm. Hang on there! By their argument, you will now be in danger of severely damaging your pride and joy by driving into what you say is a complete mismatch, set, not for 0.1 ohm but for 24 ohms. You would need to increase the supply volts to at least 200V to give headroom, unless you want a square wave output. Matthew 7:6 Poor old Beanie, he still hasn't grasped what happens when you over-drive an amplifier. Not so long ago he proposed a spectrum analyser with a linear input and refused to see the problem there. The worrying thing is his offer to teach others. -- ;-) .. 73 de Frank Turner-Smith G3VKI - mine's a pint. .. http://turner-smith.co.uk |
Output Impedance
On 16/05/2014 20:44, gareth wrote:
One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? One is a static value the other is dynamic, you seem to be wondering why a real calculation should be different to a complex one. Once the signal passes through any phasing element the vector will shift off the real axis and the V=IR. That's why it is measured in impedance and not resistance. As for calculation - for simple stuff it is a matter of number crunching with complex numbers - for more complicated circuits then you need to look at some network analysis over the whole frequency range you will be using. Andy |
Output Impedance
"Generic" wrote in message
... why do I keep posting this crap? Because, Duncam, despite your years, you display the emotions of a child. |
Output Impedance
On 5/17/2014 3:12 AM, gareth wrote:
"gareth" wrote in message ... One thing that puzzled me for years was that the output impedance of an amplifying stage could be much higher than the voltage drop across the active element divided by the current through it. Now, clearly as current sources, this would be true, and it could be measured., by changing the loads. But, how could you calculate such a value up front? Consider for a moment the equivalent circuit of a current generator with a shunt impedance feeding a load. The voltage across the load is the same as that across the current generator, and measuring the current coming out of the generator at any one time will tell you about the load impedance, but not the shunt impedance. However, by changing the load impedance and again measuring the voltage and the current, then you will ahve sufficient information to calcualte the source impedance. Yes. What is your point? The output impedance of an amplifier depends on the circuit topology and the component values. I have designed amplifiers with "synthetic" impedance where a small series resistor was used with positive feedback to create an output impedance of a larger value. I think the actual numbers were 12.5 Ohm resistor and 50 Ohm output. -- Rick |
Output Impedance
On 5/17/2014 5:14 AM, Ian Jackson wrote:
In message , Rob writes It is a widely distributed misunderstanding that a linear that has been designed to delever its output in a 50 ohm load always has a 50 ohm output impedance. In fact it almost never has. My understanding of RF design doesn't include how to calculate the output impedance of PA stages. However, it's obvious that of the output impedance was 50 ohms, the PA efficiency would be only 50% at best - and clearly this is not so (certainly for class-C operation, which is notionally 66%-ish). I've always understood that, in practice, it's usually resistively low-ish (around 25 ohms?) and quite capacitive. An expert opinion is needed! I may be jumping in where I don't understand all the issues as I know little about RF... but... The output impedance of an amplifier does not dictate the power dissipation of that amplifier. In fact the one amp design I have done used synthetic output impedance for the sole purpose of reducing the power consumption of the amp circuit. As I mentioned in another post, it used a small series resistor and positive feedback to synthesize a larger output impedance. I only had 12 volts power supply available and needed to swing over 8 Vpp at the output. This circuit worked well for that. Is this never done in an RF amp? -- Rick |
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