A question about the floating bandgap reference in Razavi's book

[Hamid.Kiumarsi]

The schematic of this floating reference is at the end of chapter 11 of Razavi's CMOS book. (please refer to the attached file)
I am having a hard time to understand why Vgs9 = Vgs11. Based on this fact we Razavi has deduced that the voltage drop on R6 equals VBE4.
If Vgs9=Vgs11, then according to the fact that he assumes that M9 and M11 are identical, we should have ID9 = ID11
and since all emitter currents of Q1-Q4 are equal, then we would have ID9 = ID11 =====>>VBE4/R6 = (2Vt*ln)/R1
The problem is: then Vout = (R4/R6)*VBE4 + 2(R5/R1)*Vt*ln ====>> Vout = (R4+R5) * (2Vt*ln)/R1 or Vout = (R4+R5) * VBE4/R6

Where do I make a mistake?

Is ID9 = ID11?

 

[erikl]

Vout = (R4/R6)*VBE4 + 2(R5/R1)*Vt*ln ====>> Vout = (R4+R5) * (2Vt*ln)/R1 or Vout = (R4+R5) * VBE4/R6

Where do I make a mistake?

No mistake, I think, just a different presentation. But it doesn't contain the 2 counter-running temperature dependencies of VBE4 and Vt, which are necessary to calculate the correct resistance ratios (R4/R6) and 2(R5/R1) corresponding to the ratio of their respective temperature dependencies, in order to achieve a best-possible compensation at a selected temperature.

Is ID9 = ID11?

Yes. The right choice of resistors within the limits of the resistor ratios given above should make this possible.

 

[Hamid.Kiumarsi]

Then what mechanism in the circuit makes Vgs9 = Vgs11 possible?
If there was no opamp and node E and F were connected together, current mirrors would result in ID9 = ID11 which inturn leads to Vgs9 = Vgs11 (of course with the right choice of transistor's dimensions)
But now that node E and F are disconnected, the only information we have is VE = VF.
If VF remains the same before and after putting opamp in place, then we can conclude that opamp has no effect on ID9 or ID11 and they remain equal.

I seem to be a little bit lost in here.

 

[yuvan]

If VE=Vf & R4=R5, then the current through M10 & M9 are the same & So is M11 &M10 currents -->making Vgs9=vgs11. And lets assume Vgs9 >Vgs11. Then we will have node Ve not equal to Vf, which is not possible with the amp.

 

[Hamid.Kiumarsi]

If VE=Vf & R4=R5, then the current through M10 & M9 are the same & So is M11 &M10 currents -->making Vgs9=vgs11. And lets assume Vgs9 >Vgs11. Then we will have node Ve not equal to Vf, which is not possible with the amp.

Actually R4 is not equal to R5; see the original paper which proposed this topology: "A Low-Power Differential CMOS Bandgap Reference" by T. Brooks et al
Moreover, we have Vout = R5*ID10 + R4*ID9 + VEF (if VE and VF are different) which suggests that VEF being zero does not depend on the ratio of R4/R5.

I think the best way to describe that ID9 = ID10 is using the current law for the cut-set cutting through nodes E and F.
The algebraic sum of currents associated with any cut-sets is equal to zero. Then the current flowing into node E should flow out of node F.