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[SOLVED] Bandgap Reference - OTA feedback connection

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bearhuhu

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Dear all,

I am trying to understand the feedback connection with the OTA in band-gap reference. The following is the circuit i am using now: shot.png

Can anyone please tell me why the branch in which BJTs in serial with the resistor is connected to the positive port of the OTA, not the negative one?

Thank you very much!

Regards,
Joya
 

In the architecture that you have shown, there are effectively two loops. One through just a single diode and other through a larger diode in series with the resistor R3. One of the loops will be in positive feedback and other will be in negative feedback. To obtain overall negative feedback for stability, you need the larger gain path to be in negative feedback. Now, the two branches carrying I1 and I2 currents can be thought of as common source amplifiers. If the load for the common source amplifier is larger in the I2 path, then gain of that common source will be higher and you have to keep it in negative feedback. Hence that branch will be connected to the non inverting terminal of the OTA as the common source path provides an inversion.
 
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    erikl

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Thank you so much Rakshitdatta!

I did some simulation to find out the effective resistance of the two loops. It turns out that the loop with a single diode has larger resistance before Va=Vb, after the intersection point, the loop with parallel diodes and a resistor has larger resistance. Now it makes sense to me :)
 

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