regarding bandgap reference output

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darksteez

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I've been trying to design a brokaw bandgap reference using 65nm technology, like I read the paper on this and they presented a simple schematic like this
and like i copied this circuit into my schematic and some did some hand calculations to calculate the ratio of R2/R1 with n=8. I found out that R2/R1= 5.87kOhm(approx). However, when I took R1 =1k and R2 =5.87k, the BGR output was still of CTAT nature so I kept increasing the PTAT contribution by increasing R2 to 27k-28kohms. How does this make sense? this much deviation from hand calculation?
And secondly, like I've not provided any amplification to the amplifier by adding another set of resistors to the output of the amplifier like this
so shouldn't the BGR out be near 1.2V while doing DC sweep across temperature? but my simulation shows that its already on around 2.37V. Why is it so? Please help me out with this atleast. This has been eating my head for weeks.
I'm attaching my schematic and output down below:

 

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* Maybe your op amp is a rail-to-rail type, but can you confirm it is? Or is it a generic type (eg. 741) which needs a bipolar supply because output never gets within 2 volts of ground?

* Alternately, what happens if you delete the op amp? Can you apply 1.2V bias to the transistors and obtain expected results?

* Many bandgap schematics use mosfets rather than bjt's. Did the author stipulate that the difference could be important? Did you try mosfets in your topology?
 

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The op amp can be a "real" op amp or a much simpler (sloppier) FET loop. Its only job is to find the happy place.

The load resistors should be to the rail, not current source fed. Maybe that takes amp inputs someplace not-good.

CTAT vs PTAT may be in the source resistor TCs. But an insane-o operating point hides much
 

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BradtheRad said:
Alternately, what happens if you delete the op amp? Can you apply 1.2V bias to the transistors and obtain expected results?
Click to expand...
sorry but if i delete the opamp, how would I make sure that equal currents flow through both the branches?
BradtheRad said:
* Many bandgap schematics use mosfets rather than bjt's. Did the author stipulate that the difference could be important? Did you try mosfets in your topology?
Click to expand...
the author indeed said, that the variation of Vbe across BJTs act as a CTAT, wouldn't that change if I use a MOS instead?
 

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regarding the opamp thing, I modelled this basic opamp using verilogA, with it having a 10^6 amplification factor and Vdd and VSS supply limiting values. I can share the code as well if you want.
secondly. if I make the load resistors to be raily fed, the the V(be) drop across the BJT reduces significantly and the current through them also reduces to picoamperes.
 

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One thing your circuit lacks is a startup bias, which is sufficient to keep the amp front end correctly biased to "want to snap in" but not so strong as to bend normal operation.

If you want to shift common mode voltage I recommend using a lower-Z drop like a diode connected FET, not something "soft".
 

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if i delete the opamp, how would I make sure that equal currents flow through both the branches?
Click to expand...

By disconnecting the op amp and by applying bias 1.2 V artificially you're testing to find out if the circuit behaves as you expect it ought to behave. That is, whether resistor values are correct, whether supply rails are correct, whether you should replace 0V ground with a negative supply,...etc.

A principle in op amp operation is that the output adjusts itself so that both inputs adopt a volt level so they are equal to each other. You state a gain of 10^6 (which is reasonable) however op amps also normally use a feedback signal of some sort, for the purpose of adjusting gain.
 

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