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
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 -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