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Measuring preamp output impedance

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Hi,

I have aikido preamp and I would like to measure its output impedance. I am using 6SN7 output tubes. According to the table provided by John Broskie, output impedance at 300V and 10mA (240R cathode resistors) should be around 500 Ohm.

So I load the preamp with 15k resistor, generate 300Hz sinewave. Then I measure AC voltages at output with 15k load resistor and without and calculate output impedance. But I do get different results (almost 1k of difference) with different input voltages and/or different load resistor. I am measuring just with DMM. Why do I get such different results?

Thanks.
 
Here are two sets of measurements:

15k - 16,54 ac volts
no load - 17,58 ac volts

and for higher input voltage

15k - 33 ac volts
no load - 36,8 ac volts

Calculating output impedance like this:
Output impedance=(Load resistance*(Vunloaded minus Vloaded))/Vloaded

Ok, calculating you have an output impedance of 943Ohm and litle more than 1mA.

Fot higher output you have probably reached the current capacity of the circuit, therefor it cannot supply more current and impedance seams to increse a lot.
15k is quite a high load for tube amp. Try using 100k for impedance measurement.
 
I think 15k should be ok, because there is cathode follower at output with about 500 Ohm impedance. And I am using 15k because aikido will drive 15k impedance of amplifier.

Cathode follower is bias at 10mA, so even with 40V input voltage it should not be current limited.

Also when I use lower input voltage (say 6V), then calculated output impedance is again different.
 
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Agreed. I was a little surprised to see the test voltage listed above. It's nice that tube preamps can do this, but they aren't often asked to. Keep the output below 5V rms for a better idea of what you have.

If you have a DMM that will measure small AC currents, you can directly measure the output current into a load. That's how I do it.
 
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Well, the preamp is intended to drive 100W class AB unity gain power buffer, so preamp output should be around 40V at full 100W output.

I do not know how to measure it properly, I mean why so different results? Is it the problem with DMM?

You might want to check with Broskie, I don't think the Aikido was really intended to provide such large amplitudes into such a low load impedance. You are heavily modulating the transconductance of the tubes in the white follower stage by asking for so much output current, and make no mistake 3mA is a lot for a follower running at 10mA, more than it can handle and stay linear in all probability. This probably explains the inconsistent output impedance measurements, I would not ask for much more than 1mA in this scenario. Broskie should be able to advise you as to the validity of my comments. (His should be the last word)

Can you change the input impedance of your power amplifier to a more benign 30 - 50K ?
 
Aikido output is not configures as white cathode follower, but single ended cathode follower which should deliver max its idle current into the load.

I can change the input impedance of the power buffer to say 50k.

Aikido should drive 15k without problem. Look for example at this hybrid amplifier by Broskie: "http://www.tubecad.com/2007/06/blog0111.htm"
There you have some power buffers too (one with IRFP has even 11k input impedance).
 
If you want to drive unity buffer amps as you indicate, consider frequent contributor SY's ImPasse preamp, which was designed for exactly this kind of thing. SYclotron.com for more details. Also boards available from Company Store.

I don´t how Impasse preamp could solve my problem. There is no problem in aikido driving unity gain power buffer (for example F4).
 
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The basic problem here is that the quoted 500 ohm output impedance is the small signal output impedance which is derived assuming the signal level is very small. As soon as you get to 2V or 5V rms you are no longer in the small signal regime and the 500 ohms therefore no longer applies. As you have seen, at higher signal levels the output impedance rises a lot.

Cheers

Ian
 
Why should output impedance very with output voltage? I don´t get it.

Here is the schematics, it is just plain aikido preamplifier.
 

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To a first approximation output impedance will be 1/gm. gm varies with current, and therefore will vary with signal. All circuits have this problem to some extent. Good design minimises it, but cannot eliminate it, therefore output impedance will vary with signal.

An alternative point of view is that AC impedance is a small-signal phenomenon (strictly, for infinitesimally small signals only) and is undefined for large signals where non-linearity is an issue.
 
My reasoning was quite simple: if two values exist for Rout (else than the sign), measured at differing operating points, the only explication can be you're measuring a variable and not a constant.
What I understand from Wikipedia the small signal model is an analysis technique while we (ought to) look at real world voltages. No 500 ohm Rout then.

Dave, is bandwith a complication for Rout? It's a non linear amplifier property as wel, diminishing with output power.
 
There are not two values for Rout but an infinite set of values. The distinction between variables and constants is a fuzzy one. The only things in electronics which are really constant are things like the charge on an electron. However, for a well-designed circuit Rout will be roughly constant so we can calculate using it. It will, of course, vary with signal level and frequency and won't necessarily be a pure resistance.

The small signal model is fine provided you have small signals. Part of the art of engineering is knowing which approximation to use, and not complaining that 'theory is wrong/useless' when the wrong approximation has been used instead of the right approximation. It is not the fault of the approximation if an engineer naively uses it outside its domain of applicability.
 
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