Which point is right? for stability simulation!!!

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Hello All DiyAudio member
I have one question of the open-loop gain simulation/Test
in the previous discussing, I know how to simulate open-loop stability,
it should add an AC source just before a high-impedance input, like OPAMP negative input side
but this has one point that make me feel stranger
why not add the AC source just before the B circuit (whole feedback loop)
an amplifier feedback circuit (B circuit) always has divider resistor, error amp (OPAMP negative side)
if we add the AC source just before the high-impedance input,
it seem we ignore the divide resistor and capacitor
why not add the AC source before the divide resistor, and don't break the "B" circuit
This confuse me very much, I have simulate those two situation, if I put AC source before
the divide resistor (not the high impedance node), also can call I put the AC source between the output and feedback side
the result is very strange, like a mess
but if I put AC source just before a high-impedance input, the result is perfect
Could someone teach me what the root cause is??
in the simulation, that say "add an AC source just before a high-impedance input" is right
but what the reason is??

thanks a lot, this will be a great help for me
have a nice day!!

Best Regard!!!
 
Read up on Tain and Middlebrook, join the LtSpice users group on Yahoo

LtSpice example file "LoopGainProbe2":

"This example is based on posts contributed by Frank Wiedmann to the independent users' group at http://groups.yahoo.com/group/LTspice

[1] Michael Tian, V. Visvanathan, Jeffrey Hantgan, and Kenneth Kundert,
"Striving for Small-Signal Stability", IEEE Circuits and Devices Magazine,
vol. 17, no. 1, pp. 31-41, January 2001.
"

also the Intusoft Spice website has articles, examples of loop gain measurement
 
mclarenpingu said:
Hello All DiyAudio member
I have one question of the open-loop gain simulation/Test
in the previous discussing, I know how to simulate open-loop stability,
it should add an AC source just before a high-impedance input, like OPAMP negative input side
but this has one point that make me feel stranger
why not add the AC source just before the B circuit (whole feedback loop)
an amplifier feedback circuit (B circuit) always has divider resistor, error amp (OPAMP negative side)
if we add the AC source just before the high-impedance input,
it seem we ignore the divide resistor and capacitor
why not add the AC source before the divide resistor, and don't break the "B" circuit
This confuse me very much, I have simulate those two situation, if I put AC source before
the divide resistor (not the high impedance node), also can call I put the AC source between the output and feedback side
the result is very strange, like a mess
but if I put AC source just before a high-impedance input, the result is perfect
Could someone teach me what the root cause is??
in the simulation, that say "add an AC source just before a high-impedance input" is right
but what the reason is??

thanks a lot, this will be a great help for me
have a nice day!!

Best Regard!!!

Remember that inserting the AC source is only a way to **approximate** actually breaking the loop, to try to examine the OPEN-loop response (even though the loop is actually still closed), as is needed for this type of stability analysis.

The approximation is 'better' if the AC source is inserted just before a very high impedance point in the loop. The approximation is 'worse' if the AC source is inserted before a lower-impedance. Q.E.D.

The "loop probe" method that has now been mentioned again, in this new thread, gives a much better approximation of the open-loop response while still running the closed-loop system. But its approximation still might not be valid above xx MHz, IIRC.
 
Thanks a lot!!!!

according the paper of Tian

he said

To analyze the stability of the feedback loop, it does not matter
whether the designer views voltage or current as the signal of
interest; both produce the same answer, as long as the dc impedances
are not disturbed when the loop is broken. In practice, this
requires careful selection of the break-point location. For the
voltage driving case, the break point should be located where the
impedance Yf looking backward from the break point is sufficiently
smaller than the impedance Ye looking forward from the
break point; the opposite condition,Yf <<Ye is necessary for the
current driving case to give a correct result. Generally, for an actual network, it may not be possible to find a break point that satisfies either of these extreme conditions.


Thanks Gootee and JCX and DiyAudio member help me to solve this problem, I think I find the answer!!!!!

Thanks you very much!!
 
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