feedback correction logic mechanism

Thread Starter

yef smith

Joined Aug 2, 2020
752
Hello , I know that feedback is a time domain phenomena.
in the circuit below is there a mathematical way I could see how exactly opamp in the middle sees some "error" and tried to fix it?
Thanks.


1709382669271.png
 

Irving

Joined Jan 30, 2016
3,885
The circuit shown is a precision voltage reference using two bandgap references Q1, Q2, the ratio between them is used to minimise4 temperature effects.; This article might help work out the relationship.
 

Thread Starter

yef smith

Joined Aug 2, 2020
752
Hello Irving, i want to look at it from different perspective.
Suppose the minus and plus input to the opamp is not equal V_minus=V_plus+delta_v.
How does the opamp will make them equal again?
Thanks.
 

Irving

Joined Jan 30, 2016
3,885
Hello Irving, i want to look at it from different perspective.
Suppose the minus and plus input to the opamp is not equal V_minus=V_plus+delta_v.
How does the opamp will make them equal again?
Thanks.
Did you read the paper I linked to? It gives all the formula you need to answer that question. Here's my 2p worth...

It doesn't have to make them equal, it just needs to find a point, for a given T where the feedback loop on the LH side is close to the feedback loop on the RH side to fix the drain current where the temperature/current curve on one side crosses the temperature/current curve on the other.

The output voltage Vg of the opamp gives rise to the drain current Id in all three MOSFETs as they are a matched triplet, Id = F(Vg), where F() is the gate to drain current transform of the MOSFET. The output voltage is Vo = Id * R4 = F(Vg) * R4
Also Vg = A(V_plus-V_minus) where A is the open loop gain of the opamp, so Id = F(A(V_plus - V_minus)

Taking the LH branch, the voltage at the -input of the opamp V_minus is a complex function G of temperature and the portion of Id that doesn't flow through R3, V_minus = G(T, id - V_minus/R3) [equation (7) in the paper cited above]
Similarly for the RH branch, V_plus = Id R1 + G(T, Id - V_plus/R2)

There is a unique point where, across a range of T, F(A(V_plus - V_minus) gives a constant Id and therefore a constant Vo.

But that's beyond my maths as it involves partial differential equations which I never really got my head around...
 
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