Amp output Z

The 70 ohm reactance isn't in *parallel* with anything ! It's in series.

Vacross600R = 10V * ( 600 / modulus ( 610-71j )) I think. Sorry, don't fancy doing the complex math right now, it's late

Graham

Oops. That is treating ro as if it is real, which it isn't. Changing the external load resistance will rotate the phasor, so I over simplified the problem.

Hey Rex, Love your equation but it yields the dreaded 13 ohms. E1=14.81Vac, E2=14.50Vac, R1=600 ohms, R2=300 ohms. Is that the correct answer at 400 Hz? By inspection I say it's about 70 ohms but cannot prove it. The 71 ohm reactance is in parallel with the 600 ohm load resistor. Regards, Harry

[snip]

Harry, when you change your resistive load you are changing the voltage divider of it and the series capacitor mentioned. So you have changing phase angles. Except you claim it's not in series.

It would have been better if you'd have included a diagram of how you are hooked up and where your measurement point(s) are instead of other people guessing from your verbal description.

Robert

See

Why don't you have any real test gear ? Life's so much simpler.......

Graham

Humourless s**ts !

Graham

Ummmm - as a follow-up. I prefer +0 -0.5 dB. That's what I design to anyway. On a good day, -0.3dB even !

Graham

Ok, forget all that. As I mentioned, my first approach was wrong.

Here's my attempt at one approach to a proper attempt to find what I think you are trying to get...

Here is your circuit as I see it.

----Ri-------Xc------- | | Es Rm Vm | | ----------------------

Es is the source voltage and Rm is the load you are applying.

I think you are changing the value of Rm and measuring the ac voltage Vm across that resistance.

As I understand it, you are trying to calculate Ri from the measured values. You said you measured (Vm, Rm) as (14.81, 600) and (14.50, 300)

The total impedance that Es sees is Z = Rm + Ri -jXc

Xc = 1/(2 pi f C) = 1/(6.283 * 400 * 5.6E-6) = 71.05 ; lets call it 71

The current flowing: i = Es / |Z|

or from the measurement: i = Vm / Rm

Equating the i equations and solving for Es:

Es = Vm/Rm * |Rm + Ri - jXc| = Vm/Rm (sqrt((Ri + Rm)^2 + 71^2))

Es is constant for both measurements so, substituting your measurements:

14.81/600 (sqrt((Ri + 600)^2 + 71^2)) = 14.50/300 (sqrt((Ri + 300)^2 + 71^2))

when I solve that for Ri (with the help of math software), I get

Ri = 0.567 ohm

Just for fun, I calculated Es at about 14.9 V

Hopefully, I didn't completely screw this up.

So Spehro is right. There is not much there to measure with your setup and Xc dominates.

Just as a sanity check, I removed Ri assuming the circuit is only Xc and Rm. I let Es be 14.9 V and calculated Vm for Rm=600 and Rm=300. I got

14.80 and 14.50, which is almost exactly what you measured. So we can assume Ri is low enough in your circuit to be in the noise with your measurement.

Hello Harry. I think you are missing something.

I Ro X +->-----/\\/\\----||----+

Vm1/Vm2 = (Ro-jXc+Rm2)/(Ro-jXc+Rm1)*Rm1/Rm2

Since your DMM doesn't give you phase, you can't solve for Ro.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

I read in sci.electronics.design that Harry Dellamano wrote (in ) about 'Amp output Z', on Mon, 11 Apr 2005:

I am astonished at all the discussion on this. It actually isn't helped by your loose terminology. You've been using 'output impedance' to mean both the output source impedance and the load impedance. You also seem confused between series and parallel (or perhaps you are mixed up between Thévenin and Norton equivalents).

You DON'T need complex numbers as such, but you DO need to take the reactance of the capacitor into account.

Assuming that the output source resistance is 10 ohms, as shown in the various circuits that have been posted, The voltages you should measure with your DMM are related by:

600/{sqrt(610^2 + 71^2)} and 300/(sqrt{310^2 + 71^2)}

Their ratio is 0.9655. How does that compare with the ratio of the voltages you measure?

--
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

Exactly. Harry does some rather cool and important work, and has a nice lab. Just a glitch.

Probably why I didn't feel like posting multiplication by conjugates and stuff well after midnight and resorted to a bit of hand-waving to describe the results... ;-)

Best regards, Spehro Pefhany

--
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Embedded software/hardware/analog  Info for designers:  http://www.speff.com

I read in sci.electronics.design that Harry Dellamano wrote (in ) about 'Amp output Z', on Tue, 12 Apr 2005:

To get 14.81 V out, you need supplies exceeding +/- 21 V, otherwise your output waveform is clipped.

--
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

You could try measuring the voltage across the cap for starters. Won't be in phase with the load - but it'll tell you the load current !

I'm sorry but you have to be the biggest example I've seen to date of believing simulations over real practice. If you'd learnt Electronics 101 you wouldn't have to ask this kind of stuff.

Graham

OMG - you've had to resort to teaching him basic *electricity* !

Why, Why, Why do ppl post questions here illustrating their total absence of knowledge of the art / science of electronics ?

Is it that the education sytem is f***ed ? ( most likely IMHO actually ) or do ppl just get degrees by attending class ? ( also quite probable IMHO )

Graham

[snip]

Harry. I see now that you gave in another post, E1= 14.81Vac, E2= 14.5Vac for R1= 600 and R2= 300.

Plugging those numbers into the above sum gives X= 35.8 ohms.

That's with an assumed Ro of 10 ohms.

If Ro= 0 the sum gives X= 72.5 ohms.

--
Tony Williams.

Yep, I should have reduced it to what the meter actually shows. However I'm pretty sure you can't get the answer you want.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

I suspect that Harry has just had a temporary brain phart, that's all..... I recognise the symptoms. :(

--
Tony Williams.