Wish to measure the resistance of a tube amp, (Rg).
The reason is that, when I model the loudspeaker to be used with this amp, I will add a series resistor equivalent to Rg in the louspeaker circuit. We also plan to include the DC reistance of any inductors in the crossover in the model.
Is there a simple way, preferably just using a meter, that I can measure the output resistance of a tube amplifier?
PS: The builder has just finished refurbishing a Dynaco ST-70, in addition ot other tube amps he already has. So the spec for Rg for the ST-70 would be helpful.
I include a pic of Small's Closed Box model to show which resistance I mean.
Thank you for any help you can give on this issue. 🙂
The reason is that, when I model the loudspeaker to be used with this amp, I will add a series resistor equivalent to Rg in the louspeaker circuit. We also plan to include the DC reistance of any inductors in the crossover in the model.
Is there a simple way, preferably just using a meter, that I can measure the output resistance of a tube amplifier?
PS: The builder has just finished refurbishing a Dynaco ST-70, in addition ot other tube amps he already has. So the spec for Rg for the ST-70 would be helpful.
I include a pic of Small's Closed Box model to show which resistance I mean.
Thank you for any help you can give on this issue. 🙂
Attachments
Hi Keltic,
Do you have a some BIG resistors lying around? Likw 5-10 Watt or so... around 10 - 50 Ohms?
Do you have a signal generator or something like a test tone CD or???
Do you have a some BIG resistors lying around? Likw 5-10 Watt or so... around 10 - 50 Ohms?
Do you have a signal generator or something like a test tone CD or???
Poobah:
thanks for responding.
The testing will be done by another member whom I am helping to design a loudspeaker. I don't even know which state he is in. Since he tinkers with tube amps, solid state amps and loudspeakers, I assume he has a multimeter or will be willing to obtain one.
The large resistors required are available at Radio Shack, or a combo can be made of Radio Shack resistors to suit.
Tone generator? I like to use this freeware online:
Freeware tone generator:
www.satsignal.net => Audio Tools => SweepGen
From there, it is a question of if his computer sound card has a Speaker Out, and what it's output resistance is. Less than 0.1 ohm would probably pass muster.
Or else, he can just hook up the computer to one of his solid state units for testing purposes.
I plan to ask him to please Right Mark his sound card to make sure it is functioning properly, (the first time I ran Right Mark I was shocked at how bad the response was-it was all right after I reinstalled).
Is that sufficient to test?
thanks for responding.
The testing will be done by another member whom I am helping to design a loudspeaker. I don't even know which state he is in. Since he tinkers with tube amps, solid state amps and loudspeakers, I assume he has a multimeter or will be willing to obtain one.
The large resistors required are available at Radio Shack, or a combo can be made of Radio Shack resistors to suit.
Tone generator? I like to use this freeware online:
Freeware tone generator:
www.satsignal.net => Audio Tools => SweepGen
From there, it is a question of if his computer sound card has a Speaker Out, and what it's output resistance is. Less than 0.1 ohm would probably pass muster.
Or else, he can just hook up the computer to one of his solid state units for testing purposes.
I plan to ask him to please Right Mark his sound card to make sure it is functioning properly, (the first time I ran Right Mark I was shocked at how bad the response was-it was all right after I reinstalled).
Is that sufficient to test?
Hey Keltic,
Yeah that will work... anything that will produce a steady sine wave to feed into the amp. You want to have pretty low level to avoid needing a 100 Watt resistor on the output of the amp.
Now if this is SS amp it may have have really low output impedance so it might pretty hard to measure accurately
with a low-cost setup.
Lemme think about formula/procedure so I don't accidentally post a bunch of hooey.
Back at you in a while...
Yeah that will work... anything that will produce a steady sine wave to feed into the amp. You want to have pretty low level to avoid needing a 100 Watt resistor on the output of the amp.
Now if this is SS amp it may have have really low output impedance so it might pretty hard to measure accurately
with a low-cost setup.
Lemme think about formula/procedure so I don't accidentally post a bunch of hooey.
Back at you in a while...
Hey Keltic,
A good place to start wold be about a 10 ohm 5 watt (bigger wattage is better) resistor. You can use anything in the ballpark.
1. Measure the resistor and record the value as "RL".
2. Hook up the resistor to the output of the amp, apply 1 kHz to the input at a VERY low value.
3. Read the AC RMS voltage across the resistor, adjust the input signal until you get about 6 Volts RMS on the output. (This will give you about 3.6 Watts on a 10 Ohm resistor... the trick here is not to fry the resistor... do the test quick so you don't get the resistor real hot as that will change its value) now record that voltage as "VL".
4. Now, the input signal EXACTLY the same, remove the resistor and measure the AC RMS voltage at the output again. Record this voltage as "VO"
5. Plug all these numbers into the formula below; and there you go.
Rout = ((RL*VO)/VL) - RL
Now, don't panic if there is almost no difference between VO & VL, that just means your amp has a real low Rout. If that is the case; this method won't be very accurate... but if your Rout is that low... you won't care anyway.
Also, this only measures the resistive aspects of the output impedance... to measure all the reactive stuff takes more toys and $.
A good place to start wold be about a 10 ohm 5 watt (bigger wattage is better) resistor. You can use anything in the ballpark.
1. Measure the resistor and record the value as "RL".
2. Hook up the resistor to the output of the amp, apply 1 kHz to the input at a VERY low value.
3. Read the AC RMS voltage across the resistor, adjust the input signal until you get about 6 Volts RMS on the output. (This will give you about 3.6 Watts on a 10 Ohm resistor... the trick here is not to fry the resistor... do the test quick so you don't get the resistor real hot as that will change its value) now record that voltage as "VL".
4. Now, the input signal EXACTLY the same, remove the resistor and measure the AC RMS voltage at the output again. Record this voltage as "VO"
5. Plug all these numbers into the formula below; and there you go.
Rout = ((RL*VO)/VL) - RL
Now, don't panic if there is almost no difference between VO & VL, that just means your amp has a real low Rout. If that is the case; this method won't be very accurate... but if your Rout is that low... you won't care anyway.
Also, this only measures the resistive aspects of the output impedance... to measure all the reactive stuff takes more toys and $.
The way I read this, measurements would be taken with and without a resistor on the outputs. Wouldn't that be risky for the OPT's of a tube amp? - specifically the no load test?
Sheldon
Sheldon
Sheldon,
It would certainly be possible to do the test with two resistors; one that goes in and out of circuit, and one that remains connected at all times to prevent an open circuit.
I am somewhat of a newbie to tubes though... is there a problem with an unloaded tube circuit?
Are different topologies affected differently?
It would certainly be possible to do the test with two resistors; one that goes in and out of circuit, and one that remains connected at all times to prevent an open circuit.
I am somewhat of a newbie to tubes though... is there a problem with an unloaded tube circuit?
Are different topologies affected differently?
poobah said:
Yes, to be safe you'd need a load at all times. The inductance of the secondaries can lead to very high voltages across the output with no load. This can cause the insulation to fail with internal shorting in the secondary windings. And the equation would be (my math skills are rusty so this would take a bit of doodling on my part)? Your measurement delta would be best with the common resistor of high value and you might be able to ignore its value in the calcs for rough estimates of impedence. How high? Maybe some of the experienced tube guys can tell us what a typical transformer would be happy with at your test levels - a couple hundred ohms?
Sheldon
I gotta think about this one... I keep forgetting they run these things open loop...
A delta v / delta I test would be the same... and would solve the problem.
Hey Keltic... hold up here until someone else responds... be nice if Eli Duttman would read this...
A delta v / delta I test would be the same... and would solve the problem.
Hey Keltic... hold up here until someone else responds... be nice if Eli Duttman would read this...
kw, there's a good, dirty, approximate way. Let's say you're interested in the 8 ohm terminal. Connect a 10 ohm power resistor across it (the resistor can be a wirewound) and drive the input with a 60 Hz sine wave to roughly 5V; note the exact output voltage. If you don't have a signal generator, you can use a filament transformer and a potentiometer as the source.
Now, without changing the drive level, parallel the 10 ohm load with another power resistor to take the impedance a bit below 8 ohms. Say it's a 20 ohm resistor. This drops the load resistance to 6.7 ohm. Note the exact voltage. About 15 seconds scratching the back of an envelope with the Thevenin equivalent equations will give you the source resistance. Understand, of course, that it's not a simple number, it varies with frequency and level.
It is not a good idea to run a tube amp without a load. It can easily tolerate a short. Exactly the opposite of solid state...
Now, without changing the drive level, parallel the 10 ohm load with another power resistor to take the impedance a bit below 8 ohms. Say it's a 20 ohm resistor. This drops the load resistance to 6.7 ohm. Note the exact voltage. About 15 seconds scratching the back of an envelope with the Thevenin equivalent equations will give you the source resistance. Understand, of course, that it's not a simple number, it varies with frequency and level.
It is not a good idea to run a tube amp without a load. It can easily tolerate a short. Exactly the opposite of solid state...
SY,
Is this because the output tubes are acting like current sources?
Slamming around into infinte impedance?
And this applies across all but cathode follower style outputs right?
Is this because the output tubes are acting like current sources?
Slamming around into infinte impedance?
And this applies across all but cathode follower style outputs right?
SY said:
Comprehension level on this end was entirely adequate until we hit this sentence. Please give step-by-step instructions on this.
I got Ohm's law down pat.
Kirchoff's Law? Memorized.
Anything else? Murky at best.
SY said:
Would this apply to tube amps only, or to solid state amps as well? Of course, at the moment I am primarily interested in tube amps, since Rg is negligible in most solid state amps.
SY said:
Thanks for the advice. Sheldon, too. I did not know that.
Keltic,
If you trust me at all at this point... I will feed some equations for a 2 resistor method. The key is that you cannot run your tube amp open.
I have blown up a lot of tube amps... always with something fun plugged in, but never un-loaded.
We need to make a 2 resistor test, I can help with the math...
For now...
If you trust me at all at this point... I will feed some equations for a 2 resistor method. The key is that you cannot run your tube amp open.
I have blown up a lot of tube amps... always with something fun plugged in, but never un-loaded.
We need to make a 2 resistor test, I can help with the math...
For now...
Thevenin's Theorem: Any combination of voltage sources and impedances can be represented as a single voltage source in series with a single impedance.
To make life simple, let's make the (bad) assumption that the amp's output impedance is a single number, i.e., doesn't vary with level or frequency. That means the source impedance is a plain vanilla resistance. So the amp can be modeled as a voltage source (controlled by the input voltage) in series with a source resistance, which I'll call Ro. We connect a load resistor of R1 and apply a voltage. Now, the amp's fictitious Thevenin voltage source will be some multiple of the input voltage, but whatever it is, it's a constant. Call it V.
What voltage do we measure across the load? Well, the Thevenin equivalent voltage, V, is divided by the series combination of Ro and R1. So the voltage across the load is V times R1/(Ro + R1), which is what we get in the first measurement.
We do the same with a new load, R2. The Thevenin voltage doesn't change since we didn't change the drive voltage. The new voltage across the load, then, is V times R2/(Ro + R2), which we get in the second measurement.
OK, we know R1, R2, and the two corresponding divided down voltages. A few seconds of algebra to eliminate the Thevenin voltage, V, yields the equation for Ro in terms of all the known variables.
To make life simple, let's make the (bad) assumption that the amp's output impedance is a single number, i.e., doesn't vary with level or frequency. That means the source impedance is a plain vanilla resistance. So the amp can be modeled as a voltage source (controlled by the input voltage) in series with a source resistance, which I'll call Ro. We connect a load resistor of R1 and apply a voltage. Now, the amp's fictitious Thevenin voltage source will be some multiple of the input voltage, but whatever it is, it's a constant. Call it V.
What voltage do we measure across the load? Well, the Thevenin equivalent voltage, V, is divided by the series combination of Ro and R1. So the voltage across the load is V times R1/(Ro + R1), which is what we get in the first measurement.
We do the same with a new load, R2. The Thevenin voltage doesn't change since we didn't change the drive voltage. The new voltage across the load, then, is V times R2/(Ro + R2), which we get in the second measurement.
OK, we know R1, R2, and the two corresponding divided down voltages. A few seconds of algebra to eliminate the Thevenin voltage, V, yields the equation for Ro in terms of all the known variables.
All these methods are fine, and work well. But if you want a very simple method with even less math, here's one that works well.
Try this. Let's say you have an 8R (or close!) power resistor you can use as a purely resistive load.
Using a 1KHz signal input, raise the output of the amp to say about 20% max power. If it's a 30W amp, take it to 6W, which corresponds to 20Vpp. [Don't go over this as the tube could flash over at the socket without loading].
Measure this UNLOADED with a good DMM or CRO. We'll assume 20Vpp.
Now connect an 8R (or 10R) load, resistive.
Note the voltage reading. Let's assume it drops to 16Vpp.
That's all you need. Divide loaded by unloaded, to give 16/20 or 0.8, subtract from 1 to give 0.2, and multiply by the load (8R, or maybe 10R) to give 1.6R (or 2R).
This is the source impedance at that power and frequency, with a resistive load. For other powers and frequencies, and with reactive loads, this figure will change, but this test is extremely useful.
Cheers,
Hugh
Try this. Let's say you have an 8R (or close!) power resistor you can use as a purely resistive load.
Using a 1KHz signal input, raise the output of the amp to say about 20% max power. If it's a 30W amp, take it to 6W, which corresponds to 20Vpp. [Don't go over this as the tube could flash over at the socket without loading].
Measure this UNLOADED with a good DMM or CRO. We'll assume 20Vpp.
Now connect an 8R (or 10R) load, resistive.
Note the voltage reading. Let's assume it drops to 16Vpp.
That's all you need. Divide loaded by unloaded, to give 16/20 or 0.8, subtract from 1 to give 0.2, and multiply by the load (8R, or maybe 10R) to give 1.6R (or 2R).
This is the source impedance at that power and frequency, with a resistive load. For other powers and frequencies, and with reactive loads, this figure will change, but this test is extremely useful.
Cheers,
Hugh
Hugh,
I'll take your word that this level is safe, but what if we inadvertently drive too hard? Would it be safe to use something like a 200 ohm resistor (or?) as the "open" test? It shouldn't affect the calculation that much, as we are only approximating here anyway. It would prevent the voltage on the secondaries from going too high.
Sheldon
I'll take your word that this level is safe, but what if we inadvertently drive too hard? Would it be safe to use something like a 200 ohm resistor (or?) as the "open" test? It shouldn't affect the calculation that much, as we are only approximating here anyway. It would prevent the voltage on the secondaries from going too high.
Sheldon
this impedance test set is extremely easy to assemble and will measure amplifier impedance (all you need is a pair of opamps and a resistor of known value)- was described in September/October QEX:
http://www.arrl.org/qex/2005/Steber.pdf
there is a little Visual Basic program which you can download from the ARRL website
http://www.arrl.org/files/qst-binaries/
http://www.arrl.org/qex/2005/Steber.pdf
there is a little Visual Basic program which you can download from the ARRL website
http://www.arrl.org/files/qst-binaries/
Attachments
Hey Keltic,
SY makes a good point, you should test for Rout using relatively small changes in Rload... and the value(s) of Rload should be close to the actual impedance the amp will see when running.
You could make it one step better and use a frequency that matches the passband in your crossover. You could do this test two or three times to get Rout for each different passband in your x-over
So. A 20 Ohm (2 Watt) resistor and 10 Ohm (5 Watt at least)
1. Measure the 20 Ohm resistor and record this as RH
2. Measure the 20 & 10 ohm resistors in parallell and record this as RL
3. Put the 20 Ohm resistor on the amp and leave it there.
4. Fiddle with the input & frequency until you get a reasonable AC RMS voltage out... 5-6 Volts... record this voltage as VH.
5. Leave the input EXACTLY the same, add the 10 Ohm res. in parallel to 20 Ohm already there and record the output AC RMS as VL.
Rout = (VH-VL) / [(VL/RL) - (VH/RH)]
This is really easy to grasp... the change in Voltage / the change in Current. I flipped the terms in the bottom of the equation around so you won't come out with negative resistance (this happens because as the voltage goes down due to loading, the current goes up).
😉
SY makes a good point, you should test for Rout using relatively small changes in Rload... and the value(s) of Rload should be close to the actual impedance the amp will see when running.You could make it one step better and use a frequency that matches the passband in your crossover. You could do this test two or three times to get Rout for each different passband in your x-over
So. A 20 Ohm (2 Watt) resistor and 10 Ohm (5 Watt at least)
1. Measure the 20 Ohm resistor and record this as RH
2. Measure the 20 & 10 ohm resistors in parallell and record this as RL
3. Put the 20 Ohm resistor on the amp and leave it there.
4. Fiddle with the input & frequency until you get a reasonable AC RMS voltage out... 5-6 Volts... record this voltage as VH.
5. Leave the input EXACTLY the same, add the 10 Ohm res. in parallel to 20 Ohm already there and record the output AC RMS as VL.
Rout = (VH-VL) / [(VL/RL) - (VH/RH)]
This is really easy to grasp... the change in Voltage / the change in Current. I flipped the terms in the bottom of the equation around so you won't come out with negative resistance (this happens because as the voltage goes down due to loading, the current goes up).
😉
Thank you all for taking the time to respond, and for your explanations.
I think I will first recommend that the builder, (I am not the one with the tube amps, I am just assisting the guy who has them), to use Poobah's method, as he has worked all the equations out and all anyone has to do is substitute the values.
Radio Shack has a 10W 10 ohm resistor, $1.79 for a two pack, part # 271-132, which seems ideal for this test.
I will recommend that he test around the crossover frequency, (if he doesn't go full range), and especially around the bass cutoff frequency, to adjust the Thiele-Small numbers.
Again, thank you all.
I think I will first recommend that the builder, (I am not the one with the tube amps, I am just assisting the guy who has them), to use Poobah's method, as he has worked all the equations out and all anyone has to do is substitute the values.
Radio Shack has a 10W 10 ohm resistor, $1.79 for a two pack, part # 271-132, which seems ideal for this test.
I will recommend that he test around the crossover frequency, (if he doesn't go full range), and especially around the bass cutoff frequency, to adjust the Thiele-Small numbers.
Again, thank you all.
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