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[Pease Porridge]
What's All This Transimpedance Amplifier Stuff, Anyhow? (Part 1)

Bob Pease  |   ED Online ID #4346  |   January 8, 2001

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One of the first things you learn about operational amplifiers (op amps) is that the op amp's gain is very high. Now, let's connect a feedback resistor across it, from the output to the −input. When you put some input current into the −input (also known as the summing point), the gain is so high that all of the current must go through the feedback resistor. So, the output will be VOUT = −(IIN × RF). That's neat (Fig. 1). While we used to call this a "current-to-voltage converter," which it is indeed, it's also sometimes referred to as a "transimpedance amplifier," where the "gain" or "transimpedance" is equal to RF.

There's a whole class of applications in which this configuration is quite useful and important. An important case is when you need an op amp to amplify the signal from a sensor, such as a photodiode. Photodiodes put out current at high impedance (high at dc), but often they have a lot of capacitance. If you just let the photo diode dump its current out into a resistor, there are two problems (Fig. 2). If the sense resistor is large, then the gain can be fairly large, but the response will be slow and the time-constant will be large: τ = RL × CS. But if you choose a small sense resistor to get a small τ, the gain will be low. The signal-to-noise ratio (SNR) may also be unacceptable. How can you avoid poor gain and/or poor response? Kay garney? (That's Nepali for "What to do?")

To avoid this terrible compromise, it's a good idea to feed the photodiode's output current directly into the summing point of a transimpedance amplifier (Fig. 3). Here, the response time is not RF × CS, but considerably faster. Plus, the gain can be considerably larger, because now you can use a larger RF. This helps improve the signal-to-noise ratio too!

When you connect up the diode like this, the first thing you realize is that the darned thing is oscillating! Why? Well, it's well known that the input capacitance of an op amp (and its circuitry) can cause instability when the op amp is used with a feedback resistor. You usually need to add a feedback capacitor across RF to make it stable. In the old days, it was stated that:


So if you have a unity-gain inverter with RIN = RF = 1 MΩ, and the input capacitance of the op amp is 10 pF, then you're supposed to install a feedback capacitor of 10 pF. That's what people said for years. The LF156 data sheet stated this, and it still does. But that's not exactly true. A complete explanation is a bit beyond the scope of this column, but in practice you can usually get away with a much smaller feedback capacitor. In many cases, you can get a response that's improved by a factor of five or 10, and still not get excessive (more than 5% or 10%) overshoot. In practice, you have to tweak and optimize the feedback capacitance as you observe the response.

The formula for the optimized amount of CF is, if:


but if:

the feedback capacitor CF should be:

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    Reader Comments

    Hi Bob, could you please illustrate how the formulas for Cf was derived? I could not understand how the GBW comes in, please enlighten. Thanks a million! :)Frances.

    frances -January 07, 2004

    Hello, Francis,

    If the Cf is too small, the Noise Gain will keep on rising up and will EXCEED the gain-bandwidth product. The loop closure will be faster than 180 degrees, or 12 dB per octave. The loop will oscillate.

    If the Cf is JUST right, the noise gain will rise up and just meet the gain-bandwidth product as it flattens out. Thus, if the gain-bandwidth product were different, the Cf would also be different, at a sort of square root rate. Just look at it in a geometric way, so that noise gain at a given frequency is the same as the amplifier's gain.

    If the Cf is bigger than that, the reponse will be stable, but slower than necessary. THAT is a matter of compromise.

    If you want the complete analysis, it will take several pages of fine print, and I can mail it out; we gave this on our recent seminar tour.

    Tell me your snail-mail address. Best regards. / rap

    Robert A. Pease -June 01, 2004

    Note: the ISBN number for "Photodiode Amplifiers - Op Amp Solutions" is incorrect. The correct number is 0-07-024247-X.

    Dean Ter Haar -July 13, 2004

    To complete the article, I would suggest that the author: 1) add the photodiode bias in the pictures (now it seems like photodiode is self-biased), 2) and say a few words about possible issues related to coupling of the circuitry with external 50-Ohm impedance environment (secondary amplifier stages and connection with the bias); it can be important for high-speed applications.

    Aleksandr Verevkin -July 21, 2004

    Great article -- just what I was looking for. But I can't find "Part 2"! Does it exist? *** Not yet. /rap

    Thanks for the reference to the book by Jerald Graeme. *** If all the tricks in my column and in Jerry's book ain't enough, (A) write me your question or (B) Get the book by Philip Hobbs. It's about $130 and worth it. You can just look it up on Amazon.

    Can you provide more information on using a series resistor chain for the feedback resistor? My application will probably need about 60 megohms of feedback. *** You can assemble six 10-meg resistors in a zig-zag -- won't take much space.

    More resistors means that the "diameter" of the feedback loop is increased. How is noise pickup (and ultimately, noise gain) affected by using multiple resistors in the feedback loop? *** If you put six 10-meg resistors in series, you may be able to use a superior grade of resistor, such as metal film, with better tempco. The capacitance can be decreased by 2x to 4x, if that's important.

    *** Usually the loop size (inductance) is a negligible matter, unless your L acts as a big loop receiving antenna. And in case there are large magnetic fields around. In that case, be sure to arrange the R's so the inductive pickup cancels out. Best regards. /rap

    Dean Ter Haar -August 02, 2004

    Great article but what does GWB stand for? One of my downfalls is never being able to remember what letters in formulas stand for. Folks don't include that usually.

    Steve -October 27, 2004

    I would like to get a copy of the detailed article mentoned by Dean Ter Haar. How could I get one?

    Tim -February 14, 2005

    Why does adding the second pair of FETs improve the noise? Thanks.


    Abbas -April 19, 2005


    Can I have the complete analysis? You may email it to me. Thanks.


    Abbas -April 19, 2005

    are you still planning on writing the followup(s) to this article? i know it would be warmly received

    james -May 03, 2005   (Article Rating: )

    Nice article. Can you tell me more about the PCB material, layout and practical implementation of an amplifier with high (>10 MOhm) FB resistor?

    Thank you Mihail

    Mihail -May 06, 2005   (Article Rating: )

    The gain for my trans-amp. is very high at very low frequencies (10-100 Hz) , but drops off VERY quickly as I increase the frequency. What can I do to allow for a higher gain at higher frequencies (1-5 kHz)?

    Stephen A. Long -June 13, 2005   (Article Rating: )

    awesome, you've enlightened me into the new world

    Anonymous -July 08, 2005   (Article Rating: )


    Anonymous -August 16, 2005

    "Nikai Ramro article" in Nepali, meaning "very good article" It helped me understand transimpedence amplifier and the application

    Yogendra -October 26, 2005

    GBW is for Gain Band Width

    ameya -November 16, 2005

    well,it is a nice article especially for a beginner like me.Please I still more courage on doing exercises.Thanks

    ralph -November 28, 2005

    Can u explain thouroughly how to design a current to voltage converter?? What should consider when choosing the right amplifier for this purpose?

    Lam -March 02, 2006   (Article Rating: )

    very helpful, thanks!

    Anonymous -March 13, 2006   (Article Rating: )


    where can I get the comparison of all different topologies of the Transimpedance amplifiers? I looked in IEEE explorer but coudn't find....

    Anonymous -March 15, 2006

    In regards the Op Amp, are current feedback Op Amps better for such a current to voltage amp. I read this is due to the lower impedance on the inverting input. Will this be enough to avoid the need for neutralisation, or will it still need to be applied somehow?

    Jeff -April 11, 2006

    good one !

    Anonymous -May 23, 2006   (Article Rating: )

    Actually I'm doing my project,title Otical to electrical converter.In the project, I just need to design an amplifier.In addition, I have design my circuit. But the problem is, the gain for low frequency is very high, but the gain is going to drop when the frequency get higher. Can you tell me why is this happened and how am i going to solve it? Thanks.. :)

    Mei Wan -August 02, 2006   (Article Rating: )

    Does (part 2) exist? if yes will you please mail me that.

    Najib Lugar -March 12, 2007   (Article Rating: )

    Hi Mei Wan: I'm writing my final project in engineering college and the tittle is similar as yours (Electrical-optical mediaconverter). Is it possible to send me a mail? keep in touch

    \\Najib Lugar

    Najib Lugar -March 12, 2007   (Article Rating: )

    excellent article! thank you bob!

    Anonymous -March 21, 2007   (Article Rating: )

    Although the article is great, it does not show the basic idea behind the circuit in the introductory part; as a result, the readers "cannot see the forest for the trees". In order to really understand the circuit, they need to know what the op-amp really does there.

    Actually, the idea is extremely simple and well-known from our human routine (maybe, because of that I needed years more to reveal it:). We frequently use this technique to "neutralize" some negative quantities by equivalent positive quantities and v.v.

    In the circuit of a transimpedance amplifier, the op-amp does exactly the same - it compensates the "harmful" voltage drop VRf across the resistor by an equivalent "antivoltage" Vout = -VRf; as a result, a virtual ground appears. For this purpose, it adds so much voltage in the circuit as it loses across the resistor R; in this way, it actually "helps" the input excitation voltage source, which creates the input current (note that the two voltage sources are connected in series, in one and the same direction - "- +, - +" - so that their voltages are added).

    Finally, we use the compensating (op-amp) voltage as a "mirror" output (just like in life when we estimate indirectly some positive quantity by an equivalent negative "antiquantity" and v.v.). The advantages: first, the load is connected to the common ground; second, it consumes energy from the "helping" voltage source instead from the input source (so, it may be low-resistive enough). Maybe, some disadvantage is the inverted output.

    All the circuits with parallel negative feedback (i.e., all the inverting op-amp circuits) exploit this idea. By the same way, the op-amp compensates the voltage drop across the capacitor in a charge amplifier and in an op-amp integrator, across a diode - in an op-amp logarithmic converter, etc.

    If you want to know more about this topic and the basic ideas behind circuits, visit my site of www.circuit-fantasia.com and the Wikipedia pages about transimpedance amplifier, voltage-to-current converter, virtual ground and negative resistance... or just email me.


    Cyril Mechkov -April 15, 2007


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