Amp output Z

But you're missing the phase angle difference at the two differing loads.

...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.

Harry, Few things are suspect. Your equation has some sign and parentheses problems. It might be Ro=(E1-E2)/((E2/300)-(E1/600))

Your equation is for pure resistance. This is a complex number problem. Use Spice to measure the output amplitude like you would with a DMM. I'll bet you end up with the same wrong answer. You need to deal with complex numbers since your series capacitor introduces a significant phase angle at 400 Hz. At 400 Hz your output impedance is 10-j71 Ohms.

Mark

--- Here's your circuit:

+---[10R]--> >----+ | | | [5.6µF] | | [GEN] + | | | [600R] | | +----------> >----+ | | GND GND

What you know for sure is that you have an AC voltage source (the generator) in series with 10 ohms and that before the generator's connected to the load, {5.6µF in series with 600 ohms) you're getting

10V at 400Hz between the uncommitted end of the resistor and ground.

What you want to know, you say, is what to measure so that the "output impedance" will be 71 ohms.

I'm confused. There is no 71 ohm output impedance anywhere in the circuit. There _is_ a 71 ohm reactance which the capacitor will exhibit at 400Hz, and which will influence the impedance of the circuit, but all you need to know to determine that is the output frequency of the generator and the capacitance of the capacitor.

Likewise, you don't need to know anything to determine the impedance of the load except the resistance of the resistor and the reactance of the cap. Knowing that, you can write:

Z = sqrt (R² + X²)

and plugging in what we have so far, determine the load impedance:

Z = sqrt (600² + 71²) ~ 602 ohms

So I don't understand what you're trying to do.

-- John Fields Professional Circuit Designer

--
Dunno, but here\'s are your circuits:
 

    +---[10R]---------+------E1A
    |                 |
    |              [5.6µF] 
    |                 |
  [GEN]               +------E2A
    |                 |
    |              [600R]
    |                 |  
    +-----------------+
    |                 | 
   GND               GND



    +---[10R]---------+------E1B
    |                 |
    |              [5.6µF] 
    |                 |
  [GEN]               +------E2B
    |                 |
    |              [300R]
    |                 |  
    +-----------------+
    |                 | 
   GND               GND

How about replacing E1A, E1B, E2A, and E2B with the voltages you
actually measured at those points?

Zee math ;-) I'll write it down and scan it in the morning.

...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.

Mmm.. I think more like 69.5 ohms..

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Ok, so I got this amplifier with an internal output resistance of 10 ohms and we are outside of all feedback loops. It is coupled to the 600 ohm resistive load thru a 5.6uF capacitor. The output voltage of the amp is

10.0Vrms at 400 Hz with no loads connected. 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)). I need to measure the Zo using a DVM and not Ro. By observation it must be 71 ohms. What must I do?? Harry

Oops, I simulate with SPICE using the normal 1 amp AC current source and get 71 ohms at 400Hz which sounds correct. I measure the actual circuit with a good DMM and using the above formula calculate about 13 ohms. So how do I measure the actual circuit to get the proper 71 ohms output impedance? thanks, Harry

Hello Harry,

Just a brief comment. I never had much luck with DVMs in such situations. In our lab I use a true RMS meter for that, in this case a Rhode&Schwarz UVM which reads exact levels from 10Hz to 15MHz.

For comparative measurements in the audio frequency range it is possible that even the sound card in your PC could be superior to a regular DVM.

Regards, Joerg

Hi Joerg, thanks for your response. But I am only dealing with pure sine waves, THD

*Measure* with SPICE ? You can only *simulate* !

You sound confused. I think you now confused me too.

*Exactly* how are you measuring Zout ? I.e. method, technique etc - detail please.

Graham

Assuming my guess is right on your setup, I think you'll have problems with this way of measuring Zo. Xc will dominate (95% of the output change) the *magnitude* of output change, and you're not changing the loading a huge amount, so there's not a lot left to get any kind of usable result, even with fairly bold assumptions. But I'll wait for the equation from Jim to plug the numbers in, it's late. ;-)

Maybe you can get rid of the cap or increase it by several orders of magnitude for this test?

I'll refrain from mentioning "boil"..

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

I didn't check your formula in detail. Are you sure it's right ? Maybe the change in load from 600 to 300 R is too small to get a meaningful result ? Don't forget that the output Z will be 10R-71j.

Maybe the complex part is causing the trouble ? Is the load purely resistive ?

Graham

Hello John,

I believe Harry wants to verify the output impedance of the amp. Considering that it's in the 10ohm range and the load is a few hundred the required accuracy is a bit of a stretch for the AC circuitry of many DVMs.

Regards, Joerg

We are trying to measure the output Z across the 600 ohm resistive load at

400HZ. We measure about 13 ohms using the DVM method and the formula given previously. No complex math is necessary because we are looking across a stepped resistive load. We step the 600 ohm load to 300 ohms to obtain delta Eo. By inspection the output Z is 600 ohms resistance in parallel with a 70 ohm reactance. Is that not about 60 ohms impedance? What am I missing? Harry

Hello Harry,

Assuming the 13ohms are correct these would add to the Z of the cap. I don't see anything missing. But all that is only good at 400Hz.

Regards, Joerg

Hey Spehro, If it is 69.5 ohms how do I measure/calculate this given only the load to vary and measure with a DMM. I get 13 ohms when I measure/calculate. Do I need other equipment and/or the node on the other side of the coupling cap? I do like your 69.5 ohms but can't measure it. How is your cooking show doing? Harry

Harry, If I understand correctly, you are measuring E2 and your ro is everything back from there including the Xc.

I get this equation:

ro = r1r2(e2-e1)/(e1r2-e2r1)

r1 and r2 are 300 and 600 for your test and e1 and e2 are the voltage across it. Hope I got the math right. I don't understand why you have Xc in parallel with something in your explanation.

Hello Spehro,

Once I pulled a little prank when the financials were discussed. Asked about what my confidence level would be for the margin I replied "plus minus 3dB". Some jaws dropped so I quickly explained that this was a joke and promised to myself never to do that again.

Regards, Joerg