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RobG

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Bozeman, MT

Re: Trans-impedance amplifier
Reply #15 - Feb 25^th, 2013, 10:30am

raja.cedt wrote on Feb 24^th, 2013, 3:22am:

Dear all,
it is a trans-impedance amplifier. I found from prof Bertan Bakkaloglu student thesis (page 11, fig 6).

http://repository.asu.edu/attachments/56952/content/LaFevre_asu_0010N_10920.pdf

@Vladislav D:yes i agree that +ve fb circuits works with <1 gain, but i don't think it has PVT insensitive property like -ve feedback. Can you point out me any design with +ve feedback, if you know.

Thanks,
Raj. �

That is a quality team on the thesis but that circuit polarity just seems wrong. I can't see it doing anything except seeking one rail or the other, but I've been wrong before. On the other hand, writing is very hard work and it is easy to overlook mistakes like this.

Have you tried it?

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Lex Senior Member Offline Posts: 201 Eindhoven, Holland	Re: Trans-impedance amplifier Reply #16 - Feb 26^th, 2013, 1:46am RobG wrote on Feb 23^rd, 2013, 1:25pm: ... The best use for this circuit is to send it to a competitor that you don't like. Brilliantly commented. =D
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Trans-impedance amplifier Reply #17 - Feb 26^th, 2013, 7:44am hello, @robg: i forgo to ask, can you please explain how it will be useful for sensing power supply variations? Thanks, Raj.
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RobG

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Bozeman, MT

Re: Trans-impedance amplifier
Reply #18 - Feb 26^th, 2013, 7:52am

raja.cedt wrote on Feb 26^th, 2013, 7:44am:

hello,
@robg: i forgo to ask, can you please explain how it will be useful for sensing power supply variations?

Thanks,
Raj.

Oh, I was joking, which maybe I shouldn't do since I could be taken seriously. As far as I can tell the positive feedback will force the output to the positive rail or the negative rail so the output will Vss or Vdd-Vgs. In other words, it is a measure of the power supply.

Maybe Rd is small enough that it will find an operating point, but in that case the M1 bias current will be (Vdd-(Vo+Vgs3)) so it will be dependent on Vdd.

I could be wrong, but I can't see how it would work.

Did you get the email I sent you?

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raja.cedt

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Re: Trans-impedance amplifier
Reply #19 - Feb 26^th, 2013, 12:11pm

Dear robg,
fine, i understood what you are saying. Before posting this thread i am pretty sure this was mistake while drawing fig. i just want to make sure once so posted . recently saw few loops working with +ve feedback with less than 1 loop gain (please find the attached fig) and some are both with +ve and -ve feedback.

Do you think +ve fb with <1 gain have all good things as -ve fb had?
can you tell me how this attached unity gain buffer works?
BTW..i didn't get your email...
Thanks,
Raj.

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RobG Community Fellow Offline Posts: 570 Bozeman, MT	Re: Trans-impedance amplifier Reply #20 - Feb 26^th, 2013, 3:07pm The NMOS in that circuit acts more like a cascode for the PMOS that is bootstrapped to the input. I don't think of that as a feedback circuit but I suppose you could - the loop gain would be essentially nil from the PMOS drain to the input.
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RobG

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Bozeman, MT

Re: Trans-impedance amplifier
Reply #21 - Feb 28^th, 2013, 7:50pm

OK, heard through the grapevine that this transimpedance amplifier is supposed to be run open loop. Now it makes more sense. Start thinking about it by removing Rf. Then you will notice that the small signal voltage at the gate of M3 is I_IN*Rd. M3 then just serves as a voltage follower. The gain is simply �Vo/Iin = Rd.

When you include Rf it adds positive feedback which will enhance the gain. My initial scratchings (which are often wrong) indicate the gain will be Vo/Iin = Rd/(1-Rd/Rf).

However, I'm not sure that is how Rf was intended to be used because the description in the thesis does not match the figure. It says that the gain is �Vo/Iin =Rf and describes the operation as if it were a negative feedback transimpedance amplifier. (Especially one where the current from M1 is mirrored and then buffered by M3, which would work fine.) Therefore I think the figure is in error although it will function if Rf>Rd (also note you will have 0dB PSRR if you measure relative to ground, maybe worse when Rf is included).

Assuming the figure is in error, having written a few papers I know how easy it is to miss something like this in the final edit especially when it isn't the main focus of the paper. Every once in a while I have to remind myself not to be too critical of other people's work.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Trans-impedance amplifier Reply #22 - Mar 1^st, 2013, 10:33am Hello all, i am extremely sorry for mis understanding this feedback. It's my mistake, it is -ve feedback. But still i don't have solid reason to say -ve feedback but some how i will try to figure out the reason. Thanks, Raj.
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RobG Community Fellow Offline Posts: 570 Bozeman, MT	Re: Trans-impedance amplifier Reply #23 - Mar 1^st, 2013, 10:34am raja.cedt wrote on Mar 1^st, 2013, 10:33am: Hello all, i am extremely sorry �for mis understanding this feedback. It's my mistake, it is -ve feedback. �But still i don't have solid reason to say -ve feedback but some how i will try to figure out the reason. Thanks, Raj. ? I think it is positive feedback.
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raja.cedt

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Re: Trans-impedance amplifier
Reply #24 - Mar 1^st, 2013, 10:44am

No rogb,
I am sure it is -ve feedback. However i don't have solid ans...One simple test in small signal gain expression if you get all coefficients +ve then it's -ve feedback. Please find the attached doc.

As you might have already noticed circuits looks like +ve fb, but they have some where inherent strong -ve feedback. One simple example gm=1/R circuit.

I will update if i have some good answer.

Good night,
raj.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Trans-impedance amplifier Reply #25 - Mar 3^rd, 2013, 2:48am Dear Robg, one more example, according to razaavi the following fb is -ve but i though it is +ve. do you know how it is -ve fb.. Thanks, Raj.
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Lex Senior Member Offline Posts: 201 Eindhoven, Holland	Re: Trans-impedance amplifier Reply #26 - Mar 4^th, 2013, 1:33am raja.cedt wrote on Mar 3^rd, 2013, 2:48am: Dear Robg, one more example, according to razaavi the following fb is -ve but i though it is +ve. do you know how it is -ve fb.. Thanks, Raj. Go through M1 (no inversion), then M2 (inversion). So, one inversion -> negative feedback.
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Trans-impedance amplifier Reply #27 - Mar 4^th, 2013, 2:07am dear Lex, M2 non-inverting.So +ve fb. Thanks, Raj.
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Lex

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Eindhoven, Holland

Re: Trans-impedance amplifier
Reply #28 - Mar 4^th, 2013, 2:16am

raja.cedt wrote on Mar 4^th, 2013, 2:07am:

dear Lex,
M2 non-inverting.So +ve fb.

Thanks,
Raj.

Oops. Sorry. You're right. I mistakenly saw M2 for a PMOS. I guess it is time for a coffee.

/edit
I think you're right about the positive feedback part, but if the loopgain is smaller than 1 it can still operate. Please provide the figure number next time (8.57). It helps to look up the context.

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RobG Community Fellow Offline Posts: 570 Bozeman, MT	Re: Trans-impedance amplifier Reply #29 - Mar 4^th, 2013, 7:43am That looks like positive fb to me. I found it in his book but I couldn't find where he said it was negative fb.
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