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In this circuit, if a resistor of 1 M\$\Omega\$ is connected between A and B, what is the effect on the stability of the quiescent point?

The analysis I made led me to believe it would provide positive feedback and thus won't stabilize the point, but I dont feel safe enough as to assure it. Which is a good method for verifying this? I couldn't think of a way to check it in a circuit simulation

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  • \$\begingroup\$ This simulator works tinyurl.com/y6ybgh77 drag 1M into position and see gain ratio \$\endgroup\$ – Sunnyskyguy EE75 Feb 23 at 22:50
  • \$\begingroup\$ That's a really useful simulation, but what I'm interested about is the stability of the operation point rather than gain \$\endgroup\$ – MPA95 Feb 23 at 22:57
  • \$\begingroup\$ Change one of the gain bias resistors and see output offset without feedback and with reduce due to H gain \$\endgroup\$ – Sunnyskyguy EE75 Feb 23 at 23:03
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The analysis I made led me to believe it would provide positive feedback and thus won't stabilize the point, but I dont feel safe enough as to assure it

Your analysis is correct. If point A rises slightly, T2's collector drops slightly. When T2's collector drops slightly it turns on Q9 more and its drain rises in voltage and, if that drain (point B) is attached to point A it will raise point A a bit more and rapidly the circuit hits the rails due to positive feedback.

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  • \$\begingroup\$ I made the same reasoning, but I don't understand why turning T9 more directly implies increasing its drain voltage. For what I understand, that increase would increase T9’s drain current, and intuitively I’m tempted to say that that current increase increases the current over RL, increasing the voltage of the drain node, but I’m not sure if that's the mechanism by which the node voltage is increased \$\endgroup\$ – MPA95 Feb 23 at 22:55
  • \$\begingroup\$ If T9 turns on more, the resistance between point B and the pos rail drops hence, point B voltage rises. \$\endgroup\$ – Andy aka Feb 23 at 23:15
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For positive-feedback to cause oscillation, you'll need a total loop gain (Forward Gain * Feedback Gain) greater than ONE.

If you want extremely precise settling, even a small bit of positive feedback can cause over-shoot, or under-shoot, and thus invalidate your settling predictions.

I suggest your install the 1MegOhm feedback, and then apply 1 millivolt squarewave to Vs (left hand back of the diffpair). If that looks well-behaved, then apply 1 volt just for fun.

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