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21st January 2013, 06:50 AM  #11 
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Join Date: Jan 2013

re: impedance calculation....ok, I think I know what you guys mean...instead of reading current in series with a variable resistor, I should put a voltmeter across the variable resistor and then adjust till the volts max out....but wouldn't the current max out in a similiar fashion? Or would there be a seperate current peak?

21st January 2013, 10:49 AM  #12 
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Join Date: May 2007

No. If your amp can cope with it and you really want to measure output/source impedance then you adjust the load for maximum power. Not maximum current (which would be a short circuit). Not maximum voltage (an open circuit).
Google maximum power transfer theorem. I am tempted to ask why you want to measure the output impedance? What will you do with this information? If you understand output impedance, then you will know how to measure it. 
21st January 2013, 11:25 AM  #13 
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21st January 2013, 11:37 AM  #14 
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Other question to ask is it small signal source impedance that you want to measure? That's usually the case, for which Chris's suggestion is the right way to go.
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21st January 2013, 02:01 PM  #15 
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Load your amplifier first with 8 ohms load and adjust the output voltage to, say 8.0 Vrms (=U1). Now you have 1.0 A output current (=I1).
Next put 4 ohms load at the output and keep the input voltage unchanged. Now read the output voltage which is obviously less than 8.0 Vrms (=U2). Then calculate the present output current with 4 ohms load (I2=U2/4 ohms). Now you have got two voltage and two current readings. Finally calculate the output impedance as Zout = (U1U2) : (I2I1) 
21st January 2013, 05:38 PM  #16 
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Join Date: Sep 2006

A much better technique is to use an auxiliary amplifier: the AUT is left without signal, and a test current is injected into the output via say a voltage of 10V and a resistor of 10 ohm.
The voltage appearing at the output is directly the impedance value in ohm (unless it is very large, in which case a larger resistor has to be used). This avoids the obvious pitfall of subtracting very similar quantities, and the more subtle one (but equally important) of comparing two similar voltages differing mostly by their harmonic content (that is important in the case of tube amplifiers). One may object that the AUT may not like being driven in such a way. In that case, this simply means the amplifier is not worth being tested.
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21st January 2013, 06:50 PM  #17 
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Since it is unlikely that the OP has any need for a very precise determination of the output impedance anyway, using a second amplifier to drive into the output of the first one seems like an overly complex procedure under the circumstances. Especially, bearing in mind that the OP's amplifier is a circlotron, so neither of the speaker output terminals is grounded. Special arrangements would be needed to ensure that the driving amplifier was "floating."
The method of adjusting a rheostat load for maximum power would also be a rather tedious one, requiring measuring the output voltage and current, calculating the product, then adjusting the rheostat, remeasuring, recalculating, seeing whether it is larger or smaller, then adjusting again, etc., etc. By comparison, hooking up a load resistor, measuring the voltage, paralleling with another and remeasuring the voltage is such a simple procedure, and a DVM will give quite adequate accuracy for the voltage comparison, I think. Chris Last edited by cnpope; 21st January 2013 at 06:52 PM. 
22nd January 2013, 03:04 PM  #18 
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Join Date: Jan 2013

ok, I'm convinced about the correct way to check the impedance of the circlotron....a volt meter across a varying resistance load....instead of the ampmeter in series...seems obvious now....thanks....I have the enclosure built and a 2 new tricks on how to lower the output impedance I'll be trying....mostly by increasing neg. feedback locally in the output stage....besides from plate to grid, which is already worked out.... plate to screen and cathode to cathode....I will post a schematic when things get worked out....midmoe

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