The voltages across RA1 and RA2 are the same so they acts as a mirror. RB is not needed, but then RA1 and RA2 would have to be larger, so RB saves space at a cost of increased sensitivity to opamp offset. Together, the voltage across RA1 (or RA2) plus the voltage across RB make up the PTAT portion of the bandgap voltage.

M1 can be thought of as the output device of the opamp. One of the advantages of this topology is that it can source current to the load. Making the mirror using resistors instead of MOSFETs also results in less thermal noise.

student_of_analog

The Mosfet M1 current is the inversely proportional of the opamp voltage.   So negative feedback is achieved.

http://rfcooltools.com

Thank you RobG,

That was what was putting me off track...so is right to assume that the terminals are incorrect in the patent ?
or is there something else that is going on, that I have not been
able to see ?

rfcooltools.com wrote on Oct 10^th, 2011, 11:01pm:


I didn't see it at first. It usually takes me 3-4 times to get the polarity right on these things  :)

Student_of_analog - just remember the feedback from the delta-Vbe side to the output has to be negative for it to start-up. Typically, the PMOS is the 2nd stage of miller compensated amplifier, so the "opamp" in the figure is a single stage with the polarity opposite of what is shown.

loose-electron wrote on Oct 13^th, 2011, 1:25pm:


Well I farm out the work on the first two
It seems like they will patent anything. I used the same type of cap cancellation about 15 years ago... but I was using it so that the injected substrate noise was common mode.

Alexandar - The AC noise on the substrate will couple into the circuit via the caps on the bipolars. Since the left bipolar is larger than the right, the left AC current is larger than the right. The bipolars distort the current (partially rectify it). The distortion includes a DC component. Since the DC components are different on the right and left there will be a new DC operating point. The end result is that the DC value of the bandgap will change in the presence of AC noise.

On the other hand, if you balance the caps on each side the AC noise injected into the circuits will be the same. Since the AC noise is the same, the rectified current will be balanced and the operating point will remain the same.

No, in this case you are matching the depletion cap from substrate to base. The detailed description of the patent explains this better than I can, and is actually quite readable. I think there should actually be 38 caps on the right side for proper matching, but I assume that 32 was chosen because it gives a symmetric layout.

The base-emitter depletion caps should also be matched but they must be much smaller than the depletion caps of the base/collector or this would not have worked as I've explained. (Nor would it work if you weren't gaining up the voltage with R1 and R2. I don't remember doing matching Cbe, but the more I think back, the more I realize I forgot exactly what I did!

I'm not sure what you mean by tying an additional base resistor to ground - are you trying to match the base resistors? That's ok when the base is tied to ground, but that won't work at all in this circuit (R2 is already there). The base resistance in the figure is intentional: it nulls out the effect of base current drop in the R1/R2 network. Eq. 12 gives the value. Again, the patent talks about this in the explanation.

rg