How to decide the polarity of the op amp in bandgap reference?

[juice9]

Hi guys,
Greetings from Juice!

I am new here and seeking for some help on the op-amp based bandgap reference.
I tried to analyze the polarity of the op-amp used in a bandgap reference.

My understanding is the function of op-amp is to force the voltages at the two input to be the same. So in the graph, when voltage at point A increase, I1 increases, and in order to make point B voltage follow point A voltage to increase, the op-amp output need to increase. In this way, point A should be positive input. But this can't explain why op-amp output need to drop if point B voltage increases.

So can you kindly help to explain how the polarity of the op-amp is decided?

Thanks!

 

[erikl]

For stability, the negative feedback loop must have higher (absolute) gain than the positive feedback loop.

 

[LvW]

For stability, the negative feedback loop must have higher (absolute) gain than the positive feedback loop.

Yes - negative feedback must be dominant over positive feedback.

 

[prestonee]

They are referring to there is a feedback loop for A and a feedback loop for B, typically I1 and I2 do not equal, and as they mentioned you want the system to have an overall negative feedback, since the same feedback drive the pfets at the top, and 1 branch will have a factor higher current then the other, its voltage impact will be larger and therefore should be the neg input.

 

[dick_freebird]

I usually decide by hooking it up, and then if it latches to
one rail or the other instead of closing in on the bandgap
voltage I conclude it wants flipping. Call me lazy.

 

[juice9]

They are referring to there is a feedback loop for A and a feedback loop for B, typically I1 and I2 do not equal, and as they mentioned you want the system to have an overall negative feedback, since the same feedback drive the pfets at the top, and 1 branch will have a factor higher current then the other, its voltage impact will be larger and therefore should be the neg input.

Thanks for your explanation. So I should consider the two branch separately. With the same voltage change at the top, because Q2 area is larger than Q1 area, the I2 change will be larger than I1 change. That means the gain at point B is larger than point A. That's why point B should be the negative input. (Gain B > Gain A).
Am I right?

 

[LvW]

Feedback factor HB must be larger than HA. However, to achive this, the resistor values must be known.

 

[FvM]

Thanks for your explanation. So I should consider the two branch separately. With the same voltage change at the top, because Q2 area is larger than Q1 area, the I2 change will be larger than I1 change. That means the gain at point B is larger than point A. That's why point B should be the negative input. (Gain B > Gain A).
Am I right?

As a general hint, this standard bandgap circuit is well explained in text books like Razavi, Design of Ananlog CMOS IC. You may want to review your favourite text book in a first order.

The usual implementation uses equal "top" resistors (R1 and R3 in your schematic) and thus I1=I2 in balanced state. Vbe voltage follows an exponential law, as a result the Vbe change is equal for both transistors, independent of transistor areas and also I1/I2 ratio. It's the series resistor R2 that sets up the gain difference and overall feedback polarity. Feedback factor |HB| is larger than |HA|due to R2 working and thus you get overall negative feedback.