How to know the resistor value of Brokaw bandgap reference?

[mpig09]

Hi all:

The attached pic is Brokaw bandgap reference circuit,
I try to know the "RX formula", but I didn't understand
how to get this formula?

Does anyone know how to get this formula?

Thanks for your reply.

mpig

 

[dick_freebird]

Rx is bias current cancellation, and depends on the BJTs' base
current (in turn, on hFE and operating point and their tempcos).
It's a dirty trick which is difficult to tune for production when
your foundry doesn't run product-optimized flows (like AD does).
If your BJT process/temp attributes and your available resistor
sheets/TCs are sloppy, bias current cancellation may go over
the top (cubic residue) and leave you worse off than without
(certainly, less predictable lot-to-lot).

Cut-and-try is all I've ever managed, and Ive pretty much
given up on this kind of thing for power management grade
references; data acq, 16 bits+, you're going to have to do
something pretty special (and you'd prefer a foundry with
at least as much interest in a proper outcome, meaning
proper models, really proper).

 

[mpig09]

Hi dick_freebird:

Thanks for you share your experiences.
The RX in my design will get poor result, so I remove it.

But in my step, this action is try and error, I wish to know
how to get the formula, but I can't find it.

If you have tiime, could you share more design experience that about "depends on the BJTs' base current (in turn, on hFE and operating point and their tempcos)." ?

I know your experience will help me more understand RX purpose.

Thanks for your reply.
mpig

 

[erikl]

I wish to know how to get the formula, but I can't find it.

You can find its derivation in P. Brokaw's original paper A Simple Three-Terminal IC Bandgap Reference, IEEE JSSC, Vol. SC-9, No.6, December 1974, pp. 388-393, item C. Base Current Error Correction (p. 390).

 

[mpig09]

Hi erikl :

Thanks for your suggestion.

I have studied the paper, but I try to derivative the Formula,
I am not smart, so I can't get the same result.

Could any help me?

mpig

- - - Updated - - -

upload the paper for everyone.

mpig

 

[erikl]

I think Brokaw's result for R₃ in his Fig. 3. is correct - not the Rx equation in your first posting.

 

[mpig09]

Hi erikl :

After I double check,
they are the same equation.

Does anyone know how to get this equation?

Thanks.
mpig

 

[erikl]

Does anyone know how to get this equation?

I think Paul Brokaw has explained it quite well in his a.m. item C. Just study it thoroughly!

 

[mpig09]

Hi erikl :

Thanks for your suggestion.

I have studied the paper, but I am not smart that I can't get the
Formula (3).

Hi all:

If you know how to derive the Formula (3) that in Paul Brokaw' paper, could you share it?

mpig

 

[erikl]

If you know how to derive the Formula (3) that in Paul Brokaw' paper, could you share it?

The following text and equations (Brokaw, p. 390) relate to his Fig. 3. on his previous page.

If E is taken to be the circuit output voltage (Vout) in the absence of base current for Q1 and Q2, then E’ (still Vout, but changed due to the base currents of Q1 and Q2) resulting from considering R3 and the two base currents is given by

E’ = E + R4(ib1 + ib2) – ib2R3(2R1/R2)(1 + R4/R5) . (3)

This relation contains a term due to the base currents through R4 and an offsetting term due to reduction of ΔV_BE by base current through R3. If E’ is set equal to E, (3) can be reduced to a constraint on R3.

So, with E' = E (no change of Vout by the base currents) :

R4(ib1 + ib2) = ib2R3(2R1/R2)(1 + R4/R5)
R4(ib1 + ib2) = ib2R3(2R1/R2)((R4+R5)/R5)

(ib1 + ib2) = ib2R3(2R1(R4+R5)/R2R4R5

R3 = (ib1 + ib2)/ib2 * R2R4R5/2R1(R4+R5)
R3 = (ib1/ib2 + 1) * R2R4R5/2R1(R4+R5)

Expressing the relationship between the base currents in terms of a parameter P1 = ib1/ib2 permits (3) to be reduced to

R3 = (P1 + 1) * R2R4R5/2R1(R4+R5). (4)

In the case shown in Fig. 3 the collector currents and hence the base currents are assumed to match, making P1 equal to 1 and resulting in the reduced expression shown in the figure.

R3 = 2 * R2R4R5/2R1(R4+R5)
R3 = R2R4R5/R1(R4+R5)
R3 = (R2/R1)*(R4R5/(R4+R5))

HTH!

 

[mpig09]

Hi erikl :

Thanks for your kindly help.

mpig