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I'm studying the following Vbe-based current reference:

enter image description here

The author of my book says:

Observe that the circuit has two feedback paths around the gain stage. One positive, the other negative. Assuming transistors M1 and M2 matched, equal currents are injected into R and into the diode connected transistor Q1. The voltage across a diode changes as the logarithm of the current; therefore, the feedback on the resistor side is typically stronger than the one on the bipolar transistor side. Transistor M2 and resistance R form an inverting amplifier. Therefore, in order to ensure stability, node B must be connected to the positive terminal of the differential gain stage.

I've understood that the loop which contains M2 is the negative feedback loop and the loop which contains M1 is the positive feedback loop (to check it, it is sufficient to break the wire, apply a small voltage increase and note that M1 and M2 are common-source configurations, which thus invert the signal). But what does it mean, from a mathematical/electrical point of view, that "the feedback on the resistor side is typically stronger than the one on the bipolar transistor side"? What would happen, from a stability point of view, if inverting and non-inverting terminals of the op-amp would be exchanged? How to work in general when we have two (or more) feedback loops?

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    \$\begingroup\$ how will you start up this circuit? \$\endgroup\$
    – analogsystemsrf
    Aug 29, 2019 at 3:29

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The change in voltage with current is greater on the resistor side than on the transistor side when the current is in the normal operating range for this circuit.

The derivative of voltage with respect to current is R on the resistor side and the dynamic resistance of the diode on the other side. At, say, 1uA the dynamic resistance will be around 26K ohms. With around 0.5-0.6V across it (to match the voltage drop of the diode-connected transistor at 1uA), the resistor will have to be about 500-600K so it will be much higher than the dynamic resistance of the diode.

Note there are two different things going on here- the bias point is set by the resistor relative to the (more-or-less fixed) diode drop. The stability depends on negative feedback so the derivative of the voltage with respect to current.

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    \$\begingroup\$ Thank you for your answer. So: the currents which flow in the two branches are equal (say 1 uA), and since the small signal resistance looking into the emitter is very small with respect to R, as a consequence of Ohm's law we get a larger voltage drop on R than on 1/gm. Thus the negative feedback is stronger than the positive one: if I break each loop and apply a small positive voltage increments at each loop, the negative loop will return back (in amplitude) a larger increment than the positive loop does. Is it correct? \$\endgroup\$
    – Stefanino
    Aug 29, 2019 at 8:13
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    \$\begingroup\$ Yes, that's correct. \$\endgroup\$
    – Spehro Pefhany
    Aug 29, 2019 at 12:36

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