I want to build a photodiode amplifier to get an AC signal at a frequency of 1 Mhz.
My photodiode has a parasitic capacitance of about 4pF.
I want my system to work in a dark room or with sun light beams on the photodiode. => The photocurrent would vary from about 0.1µA to 70µA. The AC signal will be about 0.3µA.
I tried to simulate different topologies (transimpedance amplifier, etc) without any good results.
It seems that I don't succeed to remove the DC component properly...
Any help will be appreciated !
Different topics I read:
Thank you !
Are you detecting the signal from an LED or laser? The first and easiest step in improving the ratio of AC signal to DC background light may come from optical filtering. A narrow wavelength filter can greatly reduce the background signal level.
The first link you provided in your message shows two possible two-op amp approaches. The first uses relatively low transimpedance gain to accommodate the high ambient light signal, AC-coupled into a second amplifier. The second approach uses a feedback op amp and transistor to counteract the ambient light level.
Can you provide some details on the circuits you have tried? This would give us some background on the op amp selections and power supply arrangements you have tried. We also might possibly spot simple problems that you have overlooked.
In reply to Bruce Trump:
Thanks for your answer.
The source signal il an IR LED. I plan to have IR filters and my photodiode has already an embedded filter :). I would like to have only a +5V power supply and if possible tiny packages like SC70 (prefered) or SOT-23 amplifiers.
More precision about my needs.
I have 8 photodiodes and a multiplexer. The final goal is to have this configuration:
Photodiode 1 |Photodiode 2 |... | MUX -> Amplifier -> Trigger -> µCPhotodiode 7 |Photodiode 8 |
I have one topology which seems to work in simulation: http://www.kirikoo.net/images/7force-pc2-2-20120415-221355.pngIt's a simple voltage follower -> multiplexer -> voltage amplifier. I can add a band pass filter centered on 1Mhz before or after the final amplifier.What do you think of thas topology ?
I tried to pu the multiplexer just after the photodiode but with no result. It seems all the photocurrent goes into the multiplexer parasitic capacitance...The schematic: http://www.kirikoo.net/images/7force-pc2-3-20120415-221355.png
Finally, with a transimpedance amplifier, I have to put a low gain for my system to work with sunlight (feedback resistor of about 15k-20k). Moreover, the feedback capacitor (for stability), generally founded on transimpedance configuration has to be very small for the system to work at 1Mhz. It's a problem to have capacitance of about 1p - 2pF...
I didn't found an efficient manner to compensate the effect of sun. I understand that it's better to have a transimpedance amplifier with a high gain but how to remove this DC component? That's the question :).
In reply to Vianney P:
Your two circuits operate the photodiode in photovoltaic mode which is requires that the load resistance be pretty low to achieve the desired frequency response. And, as you have seen, the additional capacitance of the multiplexer ahead of the first op amp makes this approach far too slow.
Does your first circuit appear to operate properly? It appears that the AC coupling should remove the ambient light signal. Is this where you are having a problem?
Yes my first circuit seems working in simulation. I have to prototype it. The only drawback with it is the need of a voltage follower after each photodiode.
I have to deal with a very small space on my PCB, a solution with the MUX directly after the photodiodes would be welcomed.
I have a transimpedance amplifier which works in simulation: http://www.kirikoo.net/images/7force-pc2-20120416-191114.pngDifferent points with this schematic:+ The MUX can be put just after the photodiodes- "Low gain" due to the high DC component (about 6mV pk pk in output).- "Little" bypass capacitor which could be hard to achieve.
Do you see a way to have a better design ?
I'm not sure you have simulated the multiplexing function. Keep in mind that the high-pass filter function that removes the ambient light level must stabilize to a new DC value when you switch channels. This will limit the speed at which you can scan the channels.
Any attempt to multiplex at the input of a transimpedance amp (TIA) is greatly hindered by the large multiplexer capacitance (approximately 100pF). You may be able to improve this with selection of a multiplexer with lower capacitance. In any case, it takes a wide bandwidth op amp to deal with this issue. With the combination of lower capacitance and a wideband op amp you might succeed with a TIA approach.
You have not commented on my suggestion of a two-op-amp loop for elimination of the ambient light level. This would allow use of a much larger feedback resistor in a TIA without overload. The current in T1 stabilizes to a value that balances the photodiode current due to ambient light. This does not, however, solve the problem of multiplexer capacitance. I've shown the schematic for this two-op-amp loop below for your consideration. This version is not designed for single supply operation and is just shown as an example of the approach.
Yes I noticed that the coupling capacitor will introduce a delay for the signal stabilization around the bias point of the op. amp.
I will look for some better multiplexers. By "wide bandwidth" you mean Gbpw / A >> 2Mhz ?
About your approach, I have some difficulties to make it working. What can you advise me to adjust it?
My first approach would be to adjust R1 and C3 to have a good gain + stability with no DC component. I'm not sure to understand the role of the integrator...
I should have provided a schematic that more closely resembled your situation. I've attached a new one with some annotations that may help.
R1 in this diagram could be set to 300k or so. With 0.5pF stray capacitance on R1, bandwidth would be limited to somewhat over 1MHz. Your ambient light signal of 70uA would create too much voltage on R1. R2/C1/U2 integrates the output voltage, compared with Vref (2.5V). It drives current source T1 to create a balancing current equal to the 70uA ambient light current. Thus, the net average DC current in R1 will be zero. This loop will hold Vout at an average voltage of 2.5V, even if the ambient light level varies.
This approach allow use of the highest possible transimpedance gain, R1 without overload from ambient light. The gain-bandwidth requirements of the circuit depends on the ratio of the impedance at the summing junction to the impedance of the feedback network. With 100pF at the summing junction and less than 1pF on the feedback, the circuit requires an op amp with hundreds of MHz GBWP to achieve 1MHz signal bandwidth. If you can get the capacitance at the summing junction much lower, you may have a chance of multiplexing the input.
Thank you for your schematic, it works great in simulation. I had to decrease R1 to about 330k but it's ok.
Moreover, I found a better multiplexer, a TI one, with 24pF input capacitance, 75ohms on resistance and 6pF out put capacitance. Multiplexing the anode of the photodiode seems to give good results. I have to prototype it now.
About the choice of the op. amp I think the LMH6601 or LMP7717 are good choice for the transimpedance amplifier, any suggestions on that ?
I have no personal experience with the LMH6601 or LMP7717 as I support the "OPA" branded products. Upon quick look, they look like reasonable possible selections. You may also want to look at OPA356 and OPA365 which resemble these "LM" products in some respects. Final choice could depend on factors I'm not delving into. I suggest some careful simulations with your choice to help optimize the circuit.
I have implemented your design idea into my photodiode application. My question is what is the purpose of the T1 and U2? Does it tottaly blocks the ambient light? I tryinging to detect modulated green led from the unit. and display the serial data on hyper terminal. Thank you for your help.
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