Amp output Z

Can you explain that again using proper science and analytical methods ?

Graham

WooHoo ! John came to the same conclusions as me ! OK I used 'mod' as shorthand.

I bet it's spot on !

Graham

"Brain Phart. I recognise the symptoms." :(

I read in sci.electronics.design that Pooh Bear wrote (in ) about 'Amp output Z', on Tue, 12 Apr

Would you like to explain what you think is wrong with it?

I think the op-amp may be clipping; 14.8 V r.m.s. is awfully big.

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

When you rationalise Rm1/(Ro+Rm1+jXc) to Rm1/sqrt{(Ro+Rm1)^2+Xc^2)} you are automatically including the effect of phase. Likewise for the similar but inverted expression containing Rm2.

Hey Poop Bear, I originally stated that the voltage was 10 volts to keep everything in simple terms but when someone wanted real data, it was taken with a 15.0Vrms voltage source. The amplifier is not clipping, has +/-50Vpk compliance and as stated all pure sine waves with

It gave the wrong answer for Es

Sure, but the OP said that the source voltage was *10* volts not 14.8 !

Graham, puzzled

Did I measure the output voltage at the open 600 ohm load? No, this unit is not stable with open loads so that measurement is not possible. I do have the node on the other side of the 5U6 cap if that is helpful but no readings at present. Your twin sure did give me a lot of POOP about my technical abilities, are you the same bear? You don't seem smarter than the average bear. Harry

You can measure all you want, but you can't accurately compute Zo from those measurements.

...Jim Thompson

Oi ! wash your mouth out !

OK I just did a quick simulation with 15V open circuit volts @ 400Hz,10 ohms amplifier output resistance, 5.6 uF coupling cap and a choice of 600 or 300 ohm loads.

I get 14.66 V across the 600R load

I get 14.16 V across the 300R load.

Did you measure the voltage in the no load situation with the DVM ?

Graham

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

I posted the results of using an assumed 10 ohms for Ro earlier. I'm a bit surprised at the problems people have with rather simple impedance calculations. However, the calculation of Ro from the measurements is messy and, under the stated conditions, not very reliable. Rather than seek an explicit expression (Ro = something huge and horrible), it may be much easier to put numbers into the equations and work from there. You can measure the open-circuit output voltage and put that value into the equations.

A similar problem occurs when attempting to measure the impedance looking back into a mains supply (which is important in a number of situations, including the theoretical bases of IEC/EN 61000-3-2, -3, -11 and -12), although in that case the reactance is inductive. I had a big argument with someone on a newsgroup a while back about whether a solution is possible using only resistive loads. There are two equations and two unknowns, E and Ro, so a solution must be possible unless one equation is a multiple of the other (so you really only have one equation) or they are inconsistent (representing parallel lines on a graph, such as y = x + 3 and y = x + 4.892). I consulted my colleague who is a professor of electric power engineering in Germany, and he confirmed that a solution is possible, but messy. One problem is that if you do the algebra one way, you find that the whole source impedance (Ro

• jX) cancels from the equations. Talk about babies and bath water!

I just asked Mathcad to solve it symbolically, and the expression for Ro has eight terms in the numerator.

I read in sci.electronics.design that Pooh Bear wrote (in ) about 'Amp output Z', on Tue, 12 Apr

I calculated the ratios, and based on 15 V open-circuit, and the results are:

V600/Voc Mine: 0.9770 Pooh Bear: 0.9773

V300/Voc Mine: 0.9433 Pooh Bear: 0.9440

V300/V600 Mine: 0.9655 Pooh Bear 0.9658

I guess we agree a bit.

Nice work, but I think the capacitance is a component and therefore assumed to be known. It is the 10 ohms resistive component that is in question (the effective impedance of the amplifier) and is to be solved for. The trouble is that a value so much lower than the lowest load (300 ohms), if it really is that low, it produces little real, measurable voltage change.

But John, C1 is there. Are you saying the 71 ohm reactance of C1 has no effect?? Harry

John, What do your numbers mean? Is Zo around 10 ohms or 70 ohms, and if 70 how do I measure it. Regards, Harry

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. Harry

I still think that my calculation that arrived at Ro of about 0.5 ohms is correct for the measurements you made. This is not Zo because of the largish Xc. Zo is complex, Ro is the real part.

John's point is that an accurate calculation of Ro is pretty much impossible this way. The slightest error in measurement will change the answer. There is only a tiny difference in voltage values if Ro is .5 or

Hey Jim, your killing me! Are you saying that we cannot measure Zo because there is a coupling cap there? Given that the unit is not stable with an open load but I can measure the voltage on the either side of the coupling cap and vary the load. Instead of varying a resistive load, why don't we vary a capacitive load? That should cause a voltage divider with the coupling cap and yield readings to prove the Zo is about 70 ohms, not 12 ohms. Regards, Harry