Whats the difference between those two bandgap schematics?

circuit analysis

I am attaching two bandgap schematic here.
The schematic in fig(1) is not working fine. but fig(2) works.
Whats the reason behind this. what exactly the difference between these two schematics.
thanks in advance.

Re: circuit analysis

The two circuit is the same.
Or you can add startup circuit.

Re: circuit analysis

Two circuit are not the same. Only fig2 can work. You can try to analysis the circuit by first image there is a +ve voltage change across Vs of Q2 and Q1, this action will increase the biasing current on both branches, the Vs of Q1 and Q2 rise accordingly, as there is a resistor on the right branch ,Vsq1 will rise in larger magnitude. Which with reduce the voltage across Vsq2 and Vsq1. Negative action is formed, hence Vsq2 is forced to be the same as Vsq1.

Re: circuit analysis

u think, there is a positive feedback in figure1.

Re: circuit analysis

The circuits look equivalent to me. I think that the negative feedback action described by hylas also happens in the other circuit. However both these circuits NEED a startup circuit. If you don't have a startup circuit it may happen that your simulation indicates that one of the circuits is not working (if this is the case you should have all currents equal to zero).

Re: circuit analysis

This action will not happen in figure 1. For the same analysis, +ve voltage change accross Vsq2 and Vsq1 leads to smaller biasing current, as the voltage change on the left branch due to the current change is much smaller than the right branch, voltage across Vsq2 and Vsq1 will increase further. +ve feedback action occurs, hence figure 1 doesn't work.

Re: circuit analysis

Dear Hylas, I was unable to understand your explanation. Maybe if you explain better what are the voltages you call Vsq1/Vsq2 and +ve, it is easier...

circuit analysis

hey max - actually the tough part of simulating startup is finding a circuit that WONT start in the sim.. usually you get false operation in a circuit prone to startup problems, and only the silicon shows you that the startup glitch is real! hehe

anyway, about your homework. a good rule of thumb is to NOT diode connect above a ptat resistor as it will mostly give you positive feedback.

Anyway, let me try to explain hylas' point.

first, consider the quiescent point of both of these circuits equal. how it looks at first glance...

now in figure 1, if the current in the right branch increases a tiny bit, diode connected q1 mirrors this current increase over to the left branch. this causes the q4/q3 mirror to again increase the current in the right branch - this cycle immediately goes into runaway - positive feedback!

in figure 2, if the current thru the resistor increases, the mirror holds the branch constant. the voltage sags a little but does not cause runaway.

the only consistant solution for fig 2 is when the voltage across the left diode is equal to the voltage across the right diode + resistor combo, AND when the currents are equal.

the startup problem occurrs beacuse Vleft=Vright=0 and I=0 is also a self consistant solution. but I think you get the startup problem.. Hope this helps!

Re: circuit analysis

ya...u need a startup circuit
for simulation purpose, u can ramp ya power from 0ns-1ns, 0ns-100ns, and 0ns-10us. This will help double check ya startup circuit.

Re: circuit analysis

a good rule of thumb is to NOT diode connect above a ptat resistor as it will mostly give you positive feedback

hi,electronrancher

I think you are are right. Can you explain why "a good rule of thumb is to NOT diode connect above a ptat resistor as it will mostly give you positive feedback". I have no idea about that. Can you give me some reference?

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