Amp output Z

Well- that is all well and good but what does it accomplish? Starting from the basic scalar equation:I1^2*((Ro+R1)^2 + Xc^2)=I2^2*( (Ro+R2)^2

• Xc^2), this rearranges to: |Zout|^2=Ro^2+Xc^2=(I2^2*(2*Ro*R2+R2^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I2^2) so that if a third measurement was taken at load R3 with resulting I3, then you would have a second equation of form: (I2^2*(2*Ro*R2+R2^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I2^2)= (I3^2*(2*Ro*R3+R3^2)-I1^2*(2*Ro*R1+R1^2))/(I1^2-I3^2), a messy but linear equation in Ro which is then solved and result substituted back into the |Zout|^2 equation.

You don't have enough measurements to fully determine the output inpedance.

It looks to me as though the locus of possible output impedances is a circle with center at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the solution is an output inpedance consisting of a 28.27 millihenry inductor in series with a 13.1078 ohm resistor. The reactance of that inductor just cancels the reactance of the 5.6 uF capacitor, and the

13.1078 ohm resistor provides the drop you see when you apply 300 and 600 ohm loads (the open circuit voltage would be 15.1335 volts).

If there is phase shift between E2A and E2B, then the solution lies somewhere else on the circle diagram.

I read in sci.electronics.design that Harry Dellamano wrote (in ) about 'Amp output Z', on Wed, 13 Apr 2005:

My results are the calculated voltages which your 600/300 ohm test would produce if the resistive component of the output source impedance is 10 ohms and the (capacitive) reactive component is 71 ohms. The magnitude is sqrt(71^2 + 10^2) = 71.7 ohms

If you want to determine the resistive and reactive components separately, and you can't measure phase difference, don't use two different resistive loads, use two different frequencies, and measure the open-circuit output voltage and the output voltage with a single resistive load at both frequencies. Of course, the open-circuit voltage should be (nearly) the same at the two frequencies.

In this case, too, the explicit equations are Ro = huge mess and Co = another huge mess, I think. So write the equations you get from Ohm's Law, which have Ro and Co implicit, plug the known numbers into the equations and go from there.

Note 'Co' in the above, because with two different frequencies you have two Xo values, giving you three unknowns, Ro, Xo1 and Xo2. But Xo1 and Xo2 are related by f2/f1, because Co is constant. That is the third equation you need to reduce back to two variables, either Ro and Xo1 or Ro and Co.

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Have you considered measuring the voltage - not at the load - but at the amp output - before the coupling cap ?

What do you get ? That result will teel you how much effect is due purely to the amp's output resistance instead of having to guestimate it.

Wish you'd said all that in the first place.

Graham

I read in sci.electronics.design that Harry Dellamano wrote (in ) about 'Amp output Z', on Wed, 13 Apr 2005:

The mere presence of the 5.6 uF capacitor tells you that it is at least

71 ohms at 400 Hz. Of course, it varies with frequency. The *resistive* part may be only 10 ohms, which give 71.7 ohms for the magnitude of Zo at 400 Hz.

To get less than 25 ohms, you must increase the capacitance value. There isn't much point in calculating an exact value, because you can only get

22 uF, 33 uF and 47 uF, and even 33 uF may be difficult. 22 uF is 18 ohms at 400 Hz, so Zo = sqrt(18^2 + 10^2) = 20.6 ohms. I'd go for 47 uF if 400 Hz is the lowest frequency of interest and you have room for 47 uF.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
\'What is a Moebius strip?\'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

--
No, but first let me repair the snippage above:

"Harry Dellamano" a écrit dans le message de news:NrC6e.8223$jd6.3292@trnddc07...

ohms

which

I

measure

must

Harry,

I've not seen someone suggest the obvious: why don't you directly measure the output impedance?

Terminate your amp to 600R to make it stable. Inject some current into the output through a 1K or more resistor or a current source.

Measure the voltage and phase and then correct for the output load (600R resistor + current source impedance).

Any measurement error will have much less impact than with your method.

--
Thanks,
Fred.

--
I\'ll wait... ;)

Nothing to it!

Just measure it at 13 GHz. ;-)

Cheers! Rich

The OP said: " If I measure the Zo with SPICE using a swept 1 amp current source at the output, I get about 71 ohms which is the Xc of the coupling cap. If I use a DVM on the real circuit and measure the output voltage at two different loads, 600 ohms and 300 ohms, I get about 13.0 ohms using Ro=((E1-E2)/(E1/600)-(E2/300))."

There is obviously a discrepancy between the real circuit and his model.

Then somebody else said:

Then Harry responded:

What I have done is to determine what the output impedance of the *real* circuit

*must be* to account for the *measured* E2A and E2B. (replace the 10 ohm resistor with the impedance I have described.)

Harry said: "..so I got this amplifier with an internal output resistance of 10 ohms"

In the real amp it must be something other a pure resistance of 10 ohms to account for the voltages he measures.

The E2A and E2B he gives can only happen if the effective output impedance is substantially less than the 71 ohms attributable to the 5.6 uF capacitor alone. Harry himself calculated it to be about 13 ohms. So there must effectively be some inductive reactance in there somewhere. If he could tell us the relative phases of the E2A and E2B voltages, there would be only one solution instead a circle diagram of possible impedances.

This problem is not well described or measured. I can only do calculations on what Harry has provided. I have determined the locus of output impedances that must exist to account for the voltages Harry said he measured. If that leads to an impossible situation, one might take that as an indicator that better measurements are called for, or a better Spice model. I'm not sure we can figure out the source of the discrepancy based on what Harry has told us.

Harry, If you can change the output capacitor to a large value, like

10,000uF, then you can treat this like a pure resistance voltage divider. Or, connect up a network analyzer and measure directly.

I'm in process of building up a dual AMD Opteron computer. I'll pass on details if you want. It will be a week or so until I get my benchmarks tested out. I'm interested to see what happens with the better memory access speed.

Mark

Run our "benchmark" and see how it compares ;-)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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I read in sci.electronics.design that John Fields wrote (in ) about 'Amp output Z', on Wed, 13 Apr 2005:

There IS an error; your result violates Ohm's Law, Thévenin's Theorem and the Superposition Theorem. You can't change 10 - j71 into 8 + j0 be adding 600 ohms externally. But I can't spare time at present to delve into what you did, in order to locate the error precisely.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
\'What is a Moebius strip?\'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that The Phantom wrote (in ) about 'Amp output Z', on Wed, 13 Apr 2005:

The mind boggles! How on earth do you imagine you can generate an inductor from a circuit that has two resistors and a capacitor in series?

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
\'What is a Moebius strip?\'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

As I said "The output voltage of the amp is

10.0Vrms at 400 Hz with no loads c "Hi John, I am looking for Zo not Ro. I cannot measure the open circuit output voltage because the amp is unstable with no load. This is a customers design that I have to redesign. He believes that the output Z is about 10 ohms which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but cannot prove it. The spec requirement is less than 25 ohms."

So, which is it? Is the output voltage of the amp with no loads connected equal to 10.0 Vrms, or is the amp unstable with no loads connected? My best guess is that in the OP he was talking about the simulated amp, and in the "Hi, John" post he was talking about the

*real* amp. It would help avoid confusion if he gave more detail and tell us whether he's talking about the simulation or the real amp.

If that 10 ohm resistor is in place, then you can't have the voltages he gives us for E2A and E2B without some inductance in the circuit.

But if you allow the 10 ohms to become .54874 ohms, and the open circuit voltage to be

14.927 volts, then you can get the 14.81 volts with a 600 ohm load, and 14.5 volts with a 300 ohm load. And in this case, it's actually 14.81 V @ 6.747 degrees, and 14.5 V @ 13.301 degrees. But, he's not seeing the phase angle. Too bad; if he were, that would help figure out what's going on.

I th>If we look at the total circuit once again, and insert the value of

So you subtracted 14.8V from 15V and got .2V. That is treating the voltages like they are in phase, but they aren't. You have to subtract them as phasors which will give you a different scalar answer than 0.2V.

Those readings would pretty much give an instant first order approximation to Rout.

Well I know how to measure and calculate the output imdeance of amplifiers for sure which is more than you seem capable of ( never mind even explaining in your original post what you were trying to do ).

I'm still not sure if you want to know Rout or Zout of your amplifier.

Since 5.6uF measures -70j @ 400 Hz you clearly have a problem if it's included.

Graham

I have no difficulty understanding that John. ;-)

Graham