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Is this circuit stable?

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Regnum

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I found this current source in a Microchip's Application Note (AN687)

It looks as a positive feedback circuit... Is it really stable? Why?

Thanks
 

This circuitry is regulating current. For currrent, it is negative feedback.
 

stephen_ said:
This circuitry is regulating current. For currrent, it is negative feedback.
Click to expand...
I did not understand why?
If you assume Iref=1mA and change it for dI, then voltage at the output of the buffer A2 is increased, which results in increasing of the non-inverting amplfier A1 output .
Iref is increased and you have positive feedback. Also you can see, phase shift is 2pi around loop.
 

pixel said:
stephen_ said:
This circuitry is regulating current. For currrent, it is negative feedback.
Click to expand...
I did not understand why?
If you assume Iref=1mA and change it for dI, then voltage at the output of the buffer A2 is increased, which results in increasing of the non-inverting amplfier A1 output .
Iref is increased and you have positive feedback. Also you can see, phase shift is 2pi around loop.
Click to expand...

There is some mistake in your analysis. Let's analyze it again:
At first, we set the output voltage of amplifier A1 is V1, and the no-inverting voltage of amplifier A2 is V2.
If we assume Iref increase ΔI, because Iref=(V1-V2)/Rref, so V2 should decrease, and V1 will decrease too, and then Iref decrease..., the result is keeping Iref constantly.
The DC loop is a positive feedback, that just means V1 and V2 can't be regulated, they can be anything, which are only depend on the load.
 

I see now. I made error when I have assumed that dI increase and V2 increase in the same time...
This two opamps acts more like level shifter of constant amplitude, that always keep same voltage at the resistor Rref (small signal gain is 1)... Positive feedback is not dominant because of small loop gain (βA=Rrtd/(Rref+Rrtd)<1)...
 

    Regnum

    Regnum

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pixel said:
I see now. I made error when I have assumed that dI increase and V2 increase in the same time...
This two opamps acts more like level shifter of constant amplitude, that always keep same voltage at the resistor Rref (small signal gain is 1)... Positive feedback is not dominant because of small loop gain (βA=Rrtd/(Rref+Rrtd)<1)...
Click to expand...

You showed me the light! :)

A1 and its associated resistors (R1, R2, R3, R4) shape an unity gain differential amplifier stage, whose reference is shifted to (Iref * RTD) volts, through A2.
so that, if you apply a voltage at A1 inputs you get the same voltage on Rref, what determines the magnitud of Iref.
 

let assign V1 and V2 as stephen_ said:
At first, we set the output voltage of amplifier A1 is V1, and the no-inverting voltage of amplifier A2 is V2.

now we calculate Iout:

Iout = (V1-V2) / Rref
A2 output is V2 too (unity gain opamp).
assign V3 as non inverting input of A1
V3 = (V2-Vref)*R4 / (R4+R3) +Vref
As R1=R2=R3=R4 thus
V3 = (V2/2) - (Vref/2) + Vref = (V2/2) + (Vref/2)
and
V1 = 2 * V3
substitute
Iout = (V1-V2) / Rref = (2*V3 - V2 ) / Rref = (V2 + Vref - V2) / Rref
Iout = Vref / Rref
 

Davood, you can make it faster:
At the input of A1 amplifier is (V2+Vref)/2, and at the output of A1 is its double value V2+Vref. Then voltage difference on the Rref is simply V2+Vref-V2=Vref.
 

pixel said:
Positive feedback is not dominant because of small loop gain (βA=Rrtd/(Rref+Rrtd)<1)...
Click to expand...

Yes, you are correct!
 

That says Barthausen stability criterion:
|βA|<1 - stable
|βA|=1 - oscillates
|βA|>1 - unstable
 

use voltage to analyze the type of feedback is much easier.
 

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