Feedback loop gain higher than op-amp gain?

Hi guys,

I've been playing around with the design of a high-gain high-BW transimpedance amplifier (TIA).

I've read again and again the theory behind all this but I have a question that maybe is a bit stupid, but...

What happens if I try to set the TIA gain higher than the DC gain of the amp-op?

For instance, if I need a 50k ohms gain for the TIA, but my amplifier has a 10k DC gain with 10Mhz GBW?

The end result would be what, it would oscillate?

Thanks!

If you try to set the DC gain too high it will just limit as if you resistor isn't there.

TIA can be tricky if you want high gain and speed - it will probably oscillate anyway

Keith

Hey keith, thank-you for your reply, but are you sure this will happen?

I was doing some very simple math here, and what I got was:

Supposing a Op-Amp with gain A, infinite input impedance and zero output impedance, we would have that:

Vout = -Vin*A (negative feedback, positive input of Op-Amp grounded)

Vout - Vin = Rf*I (I is input current, Rf feedback resistor)

Then we have that Vout = Rf*I*(A/A+1) which would suggest that with a gain of 99, I would have 1% error on the output, but still a very high gain, supposing Rf in the tens of KOhms range...

I know that these situation would slow my response, since the low gain Op-Amp would take some time to re-estabilish the virtual ground, but depending on the frequency of operation this could be tolerable.

Am I wrong in my line of thought somewhere?

Thanks for the help!

Hi banvetor!

I think you mix two different "gain" expressions and this leads to some confusion.

* The opamp open loop gain A is a pure voltage gain (V/V) and can be as high as 120 dB.
* The "gain" of an opamp with feedback used as a TIA is expressed in kohms (V/I) and, thus, cannot be compared with A.

Under the ideal assumption that the opamp input resistance approaches infinite (no input current) you can get, of course, output voltage-to-input current ratios which correspond (theoretically!!!) with a figure in the order of 10exp5 ohms - even when the opamp dc gain is only, for example, Ao=10exp4.

Banvetor,

For a theoretical amplifier, I think you are right. I was thinking in terms of practical, high speed transimpedance amplifiers I have designed and used. There you have an input bias current and reach a limit where the output is based on the current gain of the amplifier. In an "idea"l opamp the current gain is infinite. I have attached a simulation using a "perfect" opamp with the feedback resistor stepped and it behaves as you say.



It is trickier for me to demonstrate what happens with a real amplifier because when you increase the feedback resistor too far the DC biasing fails before it limits by current gain alone. This is the best I could manage:



Keith.

Hi guys, once more, thank-you for your help!

It's been some time since you have replied, but it's not that I'm not interested in this, it is only that unfortunately my weekdays have been crazy ehehe...

Anyway,

I think I understood when LvW said that the infinite input impedance is allowing the infinite current gain... it makes perfect sense (for the moment, at least hehe). Anyway, since I will be working with CMOS amplifiers with gate input, I think it is safe to assume an "infinite input impedance".

Then, Keith, I am getting some very similar results here when I simulate with a real op-amp (with a voltage DC gain of 200). What voltage DC gain does the op-amp that you used for this simulation has?

What I would like to know is how the DC voltage gain influences the feedback transimpedance gain... I mean, if you increased your DC gain you could then make the 1Meg resistor work, but maybe not one of 100Meg?

Maybe you already have tried to explain that with the sentence "There you have an input bias current and reach a limit where the output is based on the current gain of the amplifier.", but I really did not understood this sentence very well... if you could elaborate a little bit more on this I would be very grateful!

Thanks again!

I will have to check, but off the top of my head, finite voltage gain results in a finite input impedance. A perfect transimpedance amp has zero input impedance. If the voltage gain isn't infinite then the input impedance will have a value. This limits speed due to photodiode capacitance.

Keith.

keith1200rs said:

I will have to check, but off the top of my head, finite voltage gain results in a finite input impedance. A perfect transimpedance amp has zero input impedance. If the voltage gain isn't infinite then the input impedance will have a value. This limits speed due to photodiode capacitance.
Keith.

Click to expand...

Keith, why do you think that voltage gain and input impedance are correlated?
Of course, neither the gain nor the input impedance can be infinite (ideal), but both are not dependent on each other.

LvW said:

keith1200rs said:

I will have to check, but off the top of my head, finite voltage gain results in a finite input impedance. A perfect transimpedance amp has zero input impedance. If the voltage gain isn't infinite then the input impedance will have a value. This limits speed due to photodiode capacitance.
Keith.

Click to expand...

Keith, why do you think that voltage gain and input impedance are correlated?
Of course, neither the gain nor the input impedance can be infinite (ideal), but both are not dependent on each other.

Click to expand...

The input impedance of a TIA depends on the DC gain. Ignoring the HF components, input impedance is the change in the voltage at the input of the TIA divided by the input current. With a low open loop DC gain the inputs of the amplifier need a significant difference in voltage to drive the output to the required voltage.

So, as an example, if you have a 1k transimpedance and an amplifier with an open loop gain of 1000, a 1mA input will require 1mV on the input to produce the required 1V output, resulting in a TIA input resistance of 1 ohm. If the open loop gain is increased to 10,000, then only 0.1V will be across the inputs at 1mA so the resistance is 0.1ohms.

It is also important when you look at the dynamic performance. Amusing it is a photodiode with finite capacitance then you ideally want zero input impedance for maximum bandwidth.

Keith.

Hi Keith,

thanks for your reply. There was a misunderstanding between us.
I was of the opinion that the subject of discussion was a current feedback amp (CFA) which sometimes also is called TIA. Sorry.

No problem. I hope we were discussing the sort of TIA used for photodiode amplifiers otherwise I don't think I am helping!

Keith.

keith1200rs said:

No problem. I hope we were discussing the sort of TIA used for photodiode amplifiers otherwise I don't think I am helping!

Keith.

Click to expand...

Hi Keith, my application indeed is for photodiode amplifiers...

I've just started reading this book:

Photodiode amplifiers : op amp solutions / Jerald G. Graeme. - New York [u.a.] : McGraw Hill, 1996

which will hopefully help me understand all the issues in the design of the TIA.

Currently I'm not seeing (in the book) much of mathematical explanation of the conclusions, but maybe it is further down the book...

This is quite old now, but still relevant I think. It is mainly about the noise analysis but still useful:

focus.tij.co.jp/jp/lit/an/sboa060/sboa060.pdf

This also has some useful information on bootstrapping:

**broken link removed**

Generally my interest in in high speed, low noise designs for laser rangefinders, although I have designed very slow ones for fluoroscopy.

Keith.