open-loop gain vs. closed-loop gain in TIA

Hello and thanks for the long detailed answer.

I poseted taht question in another forum, and there someone suggested that the used OpAmp has to be unity gain stable since it is somehow working with gain = 1 for the transimpedance case, can you say something about that.

He said it is because the voltage gain is supposed to be 1, I could quite understand his thoughts since the input "sees" only a current and no voltage, but I would be glad to hear your thoughts about it.

Hello Seb,

An ideal current source can be modeled as an infinite impedance. When connected to the input of a transimpedance amplifier having a large feedback resistor the noise gain of the amplifier is A_V = 1 + Rf/Ri. If Ri is infinite, or even an extremely high resistance, the gain is very close to +1 V/V. Therefore, a unity-gain stable amplifier is required for stability, but that is an oversimplification. The feedback resistance, in conjunction with the input capacitance of the op-amp and shunt capacitance of the current source - commonly a photodiode, can easily create enough delay in the feedback loop to cause the unity-gain stable amplifier to become unstable. The instability is due to diminished phase margin.

Rather than recreating much of what has been written about transimpedance amplifiers I will refer readers to an EDN article, penned by Bob Pease, on the subject. Bob provides a good discussion about their stability, noise, and bandwidth concerns and how they are intertwined:

http://electronicdesign.com/analog/whats-all-transimpedance-amplifier-stuff-anyhow-part-1

Regards, Thomas

PA - Linear Applications Engineering

Hello guys,

I just read the post to refresh some TIA concepts, and I'd like to add another question.

Is the assumption of the closed-loop gain:

A_CL = Vo/I_PD = R_F

just a simplification? I mean, it is true that most of the current will flow through RF, but if there is a Ci (OpAmp + C_PD) and a R_SHUNT from the PD... Shouldn't we take that into account in the current calculations at the summing point?

EDIT: The question was better explained and answered below

I read quite a whole lot about that, I think the most to be distuingisht in your simplification is the frequency, since a capacitor is essentially a short for high frequencies and open for low frequencies.

So the feedback capacitance, wheather soldered in your TIA or not ( then there will always be parasitic capacitance), will effect your closed loop gain, since it takes over feedback impedance at a certain frequency upwards.

If you really want to understand all of this you have to study some notes, book, app notes and so on. Former TIs Ron Macini is always a good starting point with his "OpAmps for everyone".

I think I was not clear enough, sorry.

It's not about the effect of the feedback components (parasitic or not) on the close loop transfer function. When I said the following formula was a simplification:

A_CL = V_O/I_PD = R_F

I meant

A_CL = V_O/I_PD = Z_F  (including any kind of parasitic components)

But I still think that's a simplification, since we don't take into account the part of the current from I_PD that goes directly to ground via the input capacitance of the OpAmp or C_PD.

As you said, we don't need to care about those elements when we are at DC, but shouldn't they be taken into account when we increase the frequency? (We do take them into account for the feedback factor Beta)

Thanks for the feedback!

EDIT: The question was better explained and answered below

In my opinion this is taken care of in diagrams, of course not in a formula, so if you consider a diagram, frquency over closed loop gain, the closed loop gain will not be stable but fall at a certain point and this is due to capacitance.

From Greame ( for theoratical viewpoint :) ) page 39: This has not yet something to do with resiatnces but it shows how every formula can be made harder and harder to follow and to understand, which is basically why there are those assumptions, since every engineer wants to keep it simple as possible until problems hit in, like oscillations offset noise etc.

Vout = iphoto * Rfeedback / ( 1 + 1 / Aopenloop*beta)

explenations

loop-gain: Aopenloop * beta

since normally the loop gain is really high , like 120dB the equation can be told much easier with the ideal equation which is again your postet equation.

But with higher frequencies matters change. And so on

There are more complicated formulas also int hat book,, so if you want it the hard way get it it is well worth a read, I consider it the basis of every work surrounding TIAs for photodiode applications, all other app notes basically derive more or less from this book until today :)

I finally spent some time making the math myself and now I see it more clear.

You are right, the common expression for the closed loop gain in a TIA is, in all the bibliography:

A_CL = V_out/I_IN = -Z_F/(1+1/(beta*A_OL))

And that was exactly the expression that puzzled me. For some reason I thought that, in order to obtain it, the standard feedback system transfer function:

A_{CL_standard} = V_out/V_IN = A_OL/(1+beta*A_OL)

Was used and then V_IN was replaced by I_IN*Z_F

But I was wrong. To obtain A_{CL_TIA} I just used simple circuit analysis starting with:

V_OUT=-V_{IN_neg}*A_OL (since V_{IN_pos} is connected to ground)

V_OUT = I_F*Z_F+V_{IN_neg} (with I_F being the current flowing from V_OUT to V_{IN_neg} through Z_F)

These two lead to:

V_OUT = I_F*Z_F-V_OUT/A_OL   [Expression A]

And then we check the currents in the summing node:

I_F=V_{IN_neg}/Z_C - I_IN (with Z_C being the total impedance between the OpAmp terminals: photodiode, OpAmp input, parasitics to ground)

Using again the first expression, the sum of currents looks then:

I_F = - V_OUT/(A_OL*Z_C)-I_IN

And if we use that in [Expression A] we get:

V_OUT = - (V_OUT*Z_F)/(A_OL*Z_C)-I_IN*Z_F-V_OUT/A_OL)

From which we can finally derive:

A_{CL_TIA} = V_OUT/I_IN = - Z_F / (1 + (Z_F/Z_C+1)/A_OL)

Which turns out to be -Z_F/(1+1/beta*A_OL)

Since beta, the fraction of the output that goes again into the loop (V_{IN_neg}/V_OUT) is Z_C/(Z_F+Z_C), a simple voltage divider cause the impedance of the Input Current source is infinite (or was already included in Z_C).

I hope this mathematical elaboration helps others in the same way it has helped me clearing concepts about the real closed loop transfer function of a transimpedance amplifier without simplifications.

Thanks seb for posting, and I will definitely take a look to the book you suggested.

Have fun catching photons ;)

Gefeba said:

a simple voltage divider cause the impedance of the Input Current source is infinite

Actually in this term there is something interesting too: for TIAs where the Rfeedback can get very high, in value comparable to the resistance of the current course, the offset value at the ouput will raise once this happens, since i tis also building something like a voltage dividor, or actually a voltage divider not something like.

But glad I could help a bit.

Thomas, would you min explain how to determine the open loop gain for transcendence amplifiers? I need to know this to determine stability. I've posted a separate question here: e2e.ti.com/.../504360