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Design >> Analog Design >> feedback loop of PTAT
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Message started by Go,TB on Oct 11^th, 2012, 2:24pm

Title: feedback loop of PTAT
Post by Go,TB on Oct 11^th, 2012, 2:24pm

Hello, all

Please check the schematic in attachment. It’s one way to generate PTAT current. For example, Q2’s area is N times of that of Q1. The two Vbe difference will dump onto R to produce the PTAT current. Obviously, the quality of the PATA current has a lot to do with how well i1 can match i2.

It looks like that the negative feedback loop (Q1, Q0, M0, M2) can help to force vx equal to vy (i.e., i1 = i2). But I just wonder how the feedback loop works on this kind of structure. This feedback loop is quite different than the conventional structure (e.g., two explicit input terminals and one output terminals of an op-amp. output signal somehow sensed and returned to negative input terminal to be compared with positive input). In my feeling:

1. It’s very hard to identify where the negative/positive inputs and outputs of the op-amp are in the ckt of this kind.

2. Even if the current of Q1 and current of M1 can be regarded as positive and negative inputs to an op-amp, the vx=vy (or i1 = i2) can’t be guaranteed.

Please comment on this topic and feel free to correct me. Thanks a lot!

Title: Re: feedback loop of PTAT
Post by Vladislav D on Oct 12^th, 2012, 2:09am

In this circuit, the error amplifier is a transistor Q0. The inverting input is the base, and the non-inverting input is emitter. The feedback sets voltage at Vx=Vdd-vbe(Q0). vbe(Q0) is defined as a dvbe(Q1-Q2)/R = Vt x ln(N)/R

Title: Re: feedback loop of PTAT
Post by raja.cedt on Oct 12^th, 2012, 8:29am

Hello,
this has -ve and +ve feedback, they will make sure that Vx and Vy are set at rite common mode, vx=vy is wrong. Since this circuit is single ended it wont make two potential equal rather it makes two potential difference constant. coming to the fig, -ve fb get you vy=vdd-vbe(Q1) and +ve fb get you vx=vdd-vbe(Q6) .

Thanks,
Raj.

Title: Re: feedback loop of PTAT
Post by Go,TB on Oct 15^th, 2012, 10:42am

raja.cedt wrote on Oct 12^th, 2012, 8:29am:

Well. if Vx is not equal to Vy, the ckt can't work well at all. As I pointed already, the negative fdbk does make the ckt to have the same currents in the 2 branches.