discoveries on output impedance

[unclejed613]

i have found out through investigation into amplifier behavior that the output impedance of a solid state amp is (or is pretty darn close to) Av/Aol. however, even though this seems to be a very close approximation, i fail to see mathematically how one can derive ohms from a gain ratio (one type of quantity being derived from another type of quantity), unless i'm somehow skipping over some math steps because the impedance just happens to numerically match the feedback ratio. anybody got any thoughts on this????

[glennb]

The output impedance of a typical SS amp depends on several factors. The average open-loop o/p imp is about the same as the OPS emitter resistors. When the loop is closed using NFB, the VAS driver o/p imp comes into play, which varies the o/p imp with frequency and the available NFB factor to reduce that impedance. The output damping inductor can easily double the o/p imp. Speaker cables add more...

[Eva]

The closed loop output impedance is the open loop output impedance divided by the feedback ratio... The open loop output impedance of your circuit may happen to be just 1 ohm at the frequency you are testing...

[unclejed613]

i am attaching the simulation i'm using. as i change the output resistance (equivalent to open loop output impedance) R3, the measured output impedance remains at about 20 microohms until i get above .02 ohms, then the output impedance changes linearly, but still remains in the milliohm range (even with output resistors of 10 and 20 ohms)

[andy_c]

Originally posted by unclejed613
i am attaching the simulation i'm using. as i change the output resistance (equivalent to open loop output impedance) R3, the measured output impedance remains at about 20 microohms until i get above .02 ohms, then the output impedance changes linearly, but still remains in the milliohm range (even with output resistors of 10 and 20 ohms)

Hi,

Have a look at the document that Rodolfo posted here. It has the derivation of the output impedance of a feedback amplifier with resistive open-loop output impedance and single pole compensation.

[Conrad Hoffman]

Not surprising, since that op-amp model probably has huge gain that you've reduced to one. The ratio is so high that any smallish output resistor is negated. Were that real life was so simple!

[Tim__x]

There's a better answer. The open loop DC output impedance of that "ideal" opamp is... drumroll... 1 ohm.

Yet again Eva's advice is quite reliable.

[unclejed613]

not quite..... if that were the case, then the minimum output impedance would be 1 ohm..... not happening..... i am getting results down to 2 microohms.

i looked at rodolfo's document, but it's difficult to do math when a lot of terms in the equations are undefined......

[Tim__x]

No, the opamp's 1 ohm open loop output impedance is just that, open loop. Closed loop it is brought down to (1ohm+R3)/((Aol+1)/Av).

With R3 in your schematic set to zero, DC output impedance will in fact equal 1/((Aol+1)/Av), which is also equal to the equation you came to: Av/Aol.

Try making R3 9 ohms, then zero (1p is fine). The difference in DC output impedance should be a factor of ten.

Edit: I've attached a simulation file for you to take a look at, it runs once with Rs at 1 pico ohm (~0) and then once at 9 ohms. I also corrected a small omission in my equation

[unclejed613]

tnx, i also got a better simulation from the SWCad yahoo group that seems to work as well. part of my confusion came from my running the sim at only 1khz. the sim i got from the yahoo group runs a .ac sweep, and clarifies a lot of my questions, as well as showing the changes in Zout with frequency.

the reason i originally drew my sim the way i did, was because when i actually build an amp, and test it, i will be using an amp with a very high damping factor (such as a crown) in place of the voltage source. that will give me a real world "direct reading" output impedance indication. one tech i know thinks that's a scary way to test an amp, but it IS through a load, and not much different than bridging an amp. actually it is less stressful, since all the amp under test has to do is hold a ground potential, rather than swinging the opposite direction, and it is only 10 volts.