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I would like to have your opinion on this circuit :

enter image description here

The source voltage V12 is a feedback of an other loop hidden. So the operating point of the system is changing.

When the 47 Ohm resistor R1 is connected to the NPN, the system seems to be stable according to LTspice and it makes sense according to the theory :

enter image description here

Nevertheless, if the operating point is making working the PNP and not the NPN, the system is unstable as the resistor of 47 Ohm is "removed" from the open loop gain.

enter image description here

What I am asking is the following : if the circuit is working with the NPN, i.e with the stable part of the system, can the system be lightly unstable? I ask this question because I have a circuit which looks like to be unstable. Sometimes the output op amp seems to change completely of state and then go back to its value. It is particurlarly the case when the 47k5 resistor is not placed. But it has a low effect on the open loop gain and so on the stability ...

Thank you,

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Instability occurs when the PNP is active because the feedback path to the op-amp uses a 47 Ω resistor (R1) and, when that happens, the influence of C6 is very significant. This could add significant phase shift to the feedback and cause instability.

Compare this when the NPN is active. The NPN's emitter is "strong enough" to over-power the effect of C6 and this means the system is more stable. I'm assuming the the inductor is shorted out by the red line you appear to have added.

if the circuit is working with the NPN, i.e with the stable part of the system, can the system be lightly unstable?

It can be closer to the borderline of instability (C6 being the big factor to worry about).

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  • \$\begingroup\$ One could say that when PNP is active as well, C6 will see an even lower BE impedance (// of 2 bjts) so the pole will shift further down. \$\endgroup\$
    – edmz
    Commented Mar 25, 2023 at 14:59
  • \$\begingroup\$ @edmz either the NPN or the PNP are active but, not both at the same time --> look how the bases are wired. You are thinking of an impossible scenario. \$\endgroup\$
    – Andy aka
    Commented Mar 25, 2023 at 15:00

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