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Old 8th April 2014, 04:01 PM   #1
AndrewT is offline AndrewT  Scotland
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Default Measuring Power Amp Output Impedance

I usually measure output impedance of a completed amp fed from the mains supply via the installed PSU.
I usually get around 40milli-ohms to 80milli-ohms.

Today I am measuring a 3886 chipamp while it is being fed from a dual polarity Lab Supply.

Results, all at 1kHz sinewave, for 3 different supply voltages:
Supply Vccee . output into 8r0 . into open circuit
. . 20V . . . . . . 2804.1mVac . . . 2806.2mVac
. . 40V . . . . . . 2822.9mVac . . . 2824.4mVac
. . 63V . . . . . . 2822.9mVac . . . 2824.4mVac

Output impedance = [Vopen / Vinto8r * 8r] - 8r

for 20V, 6.0milli-ohms
for 40V & 63V, 4.3milli-ohms

Q1.) is the output impedance really that low?
Q2.) is the formula correct?
Q3.) why does the low supply voltage impedance increase by 1.7milli-ohms?

The test (~1W into 8r0) output voltage and the output current are well below the limits for the chipamp output stage.
Yet the 20V result shows the amp gain has dropped from 10.25times to 10.18times.
The supply voltage is 10Vdc and the output voltage is 3.96Vpk

This result contradicts what I was being told in other Threads a few weeks ago.
I asserted that amplifier gain falls as the supply rail voltage moves down towards the output voltage.

Comments welcome
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Last edited by AndrewT; 8th April 2014 at 04:23 PM.
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Old 8th April 2014, 04:35 PM   #2
peufeu is offline peufeu  France
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Q1.) is the output impedance really that low?

Yes, and 6 mOhm is a bit high for an amp with lots of feedback (not very important, since the cable impedance will swamp it anyway).

Your way of doing it has the inconvenience of needing an extremely precise AC voltage measurement, which can be a problem.

It is much simpler to connect the amp output to GND and inject an AC current in the amplifier output, for example using another amp and a 8 ohm resistor. The voltage at the amp's output gives the impedance, an you can also visualize it on a scope, FFT it, whatever, to reveal all sorts of gremlins (crossover distortion has nowhere to hide). You can also inject, for example, a 50 Hz 1 amp current and a 1kHz 10mA current, with 2 extra amps. Then you can measure the intermodulation at the amplifier output. Etc.
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Old 9th April 2014, 04:58 AM   #3
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Quote:
Originally Posted by AndrewT View Post
......Output impedance = [Vopen / Vinto8r * 8r] - 8r.....
Q1.) is the output impedance really that low?
Q2.) is the formula correct?
Q3.) why does the low supply voltage impedance increase by 1.7milli-ohms?.....
It would seem from many sources that the expression is fine and about the only basis for practical low impedance measurement.
The method may vary among test authorities but this detail from a review tester at least clarifies how it may be done commercially.
The relevant section is the editorial in the conclusion.
Basic Amplifier Measurement Techniques | Audioholics
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Old 9th April 2014, 09:17 AM   #4
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Andrew, the small perturbations on the supply voltage should not change voltager gain, except when you get to clipping.

Remember that Zout is really a differential value. The correct way to measure it is as follows:

- measure Vout with a specific load, say 8 ohms, and calculate Iout.
Example: 8V out, at 8 ohms is 1 amp;

- change load to say 4 ohms, measure Vout again, and calculate Iout.
Example: now Vout drops to 7V, which gives Iout = 7 / 4 = 1.75 amps.

Now calculate Zout = change in Vout (8-7) / (change in Iout (1 - 1.75) = 1.3333 ohms.

Jan
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Old 9th April 2014, 09:32 AM   #5
forr is offline forr  France
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When I need to measure an amp output impedance, I first set the output voltage with a load, both having simple numbers.
For example, 8 Vrms on a 8 Ohm power resistor makes the amplifier delivering a current of 1 A.
Then I remove the load. The output voltage changes, usually only slightly.
The voltage difference is equal to the value of the output impedance.
No calculation.
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Old 9th April 2014, 11:05 AM   #6
Elvee is online now Elvee  Belgium
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Peufeu's method is the only valid way to measure the impedance, for a number of reasons.
One the most important is that for amplifiers with feedback, the output impedance is not real: it is complex, almost purely inductive.
When the small inductive drop is subtracted from the huge quadrature output voltage, it is completely overwhelmed, leading to enormous errors.

The differential load method is a remnant of the tube era, when the output impedance of an amplifier was of the same order as the load, or even much larger in some cases, but it is totally unapplicable to modern semiconductor amplifiers.
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Old 9th April 2014, 11:40 AM   #7
brig001 is offline brig001  United Kingdom
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Quote:
Originally Posted by peufeu View Post
It is much simpler to connect the amp output to GND...
I'm having difficulty visualising this... If the above was INPUT, I get it, but I'm not sure...

Brian
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Old 9th April 2014, 11:49 AM   #8
Elvee is online now Elvee  Belgium
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Quote:
Originally Posted by brig001 View Post
I'm having difficulty visualising this... If the above was INPUT, I get it, but I'm not sure...

Brian
He means the input (= no signal)
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Old 9th April 2014, 03:09 PM   #9
brig001 is offline brig001  United Kingdom
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Thanks for that - it makes sense now.

Brian
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Old 9th April 2014, 03:22 PM   #10
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Quote:
Originally Posted by brig001 View Post
Thanks for that - it makes sense now.

Brian
Mr. Pass has an article entitled "Mosfet Citation 12" at www.firstwatt.com. He described at the end of it the method as posted earlier by peufeu.
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