I have build a 2-stage amplifier circuit for amplifying the current from an InGaAs photodiode. It uses 2 OPA4350 op-amps. The first stage is a TIA with a feedback resistor = 1MΩ. The 2nd stage is a voltage amp with a feedback of 10kΩ. It works very well with little DC offset and very good sensitivity (it saturates to 2V with a 5mW uncollimated IR laser located 3m away). enter image description here

However, the device that we are wanting to measure the light output from is VERY weak. We require a 20ms exposure time on our InGaAs camera to see a pixel illumated to ~150ADC counts (I'm trying to find out what this means in terms of incoming photons).

I know the "correct" answer is "Improve the light output of your device". But we have no easy way of achieving this. We're already using the most powerful laser at our disposal and require a very fine linewidth.

I have considered modifying my circuit to create a charge integration amplifier with a capacitor reset switch to mimic the exposure time of the camera.

Are there any other ideas I could consider to sum up the photocurrent before reading out? If I'm not seeing any instantaneous photocurrent on the output of my 1st stage (or even my 2nd stage), is there any point in modifying the circuit?

I'm open to any suggestions :)

EDIT: Based on some of the great responses (special thankyou to @Matt), I'm going to try the following circuit. Ignore the component values. I've just been playing around. Until I know the photocurrent, I'll experiment with the feedback capacitor and reset timing. For the switch, I'm considering a TS12A4516.

Does anyone see anything blatantly incorrect with this, or have any ideas for improvements. I'll also pressure my optics colleagues to find a lens solution.

Thank you all. enter image description here

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    \$\begingroup\$ The common way to use a photodiode is using it in reverse bias, I do not see any bias voltage applied here. The first stage is a Trans Impedance Amplifier (TIA), as the output of a photo diode is a current that makes sense as a TIA has a current input and outputs a voltage. Your 2nd stage is another TIA. That makes no sense as the output of the 1st stage is a voltage, not a current. So basically the 2nd stage does nothing. \$\endgroup\$ Commented Jan 14, 2021 at 11:47
  • \$\begingroup\$ Is the light modulated or is it just a slow varying intensity? \$\endgroup\$
    – Andy aka
    Commented Jan 14, 2021 at 12:12
  • \$\begingroup\$ "The 2nd stage is a voltage amp with a feedback of 10kΩ." As @Bimpelrekkie says, your second stage is another TIA. Also 10k\$\Omega\$ is not a quantity of feedback for a voltage amplifier. \$\endgroup\$ Commented Jan 14, 2021 at 13:56
  • 1
    \$\begingroup\$ 4 fA per root hz current noise on your opamp isn't terrible, but you could also do better, which would allow you to have more gain in the first stage without adding additional noise. To pick a random opamp, the OPA657 is 1.3 fs per root hz, so 3x better. There may be even better choices. \$\endgroup\$ Commented Jan 14, 2021 at 15:18
  • 1
    \$\begingroup\$ Why would you have to get a lens made? Just buy one from a catalog. \$\endgroup\$ Commented Jan 14, 2021 at 21:23

3 Answers 3


For detecting very low radiance sources an integrating TIA is the way to go. Just remove the feedback resistor and size the capacitor to set the gain of the first stage. Add a switch in parallel with the capacitor to reset. Not seeing much instantaneous current isn't surprising for small signals and is exactly the problem that an integrating amplifier can solve.

You should put some time into calculating the expected irradiance on your detector and use this to determine how much current you get. This will allow you to properly size the feedback capacitor and better determine how long of an integration you need. It sounds like you are feeding this into an ADC. You will want to select the capacitance and integration time such that your capacitor fills up to somewhere near the maximum input voltage of the ADC for the highest irradiance you expect. Doing so usually improves SNR and otherwise you will be wasting some of your ADC bits and in the extreme case can add significant quantization noise, further reducing SNR.

Once assembled, you should be able to watch the output voltage ramp up during an integration with a higher slope under illumination than in the dark. If your signal source is comparable to the dark current of your diode you will need to subtract the dark current contribution after you complete an integration.

As was pointed out in the comments, your second stage amplifier is another TIA, and shouldn't work as drawn. You probably can do what you want with just one stage, which is usually best for noise performance. You also should be able to set the reverse bias of your photodiode as 0 V can work, but isn't going to be as good as a bit of reverse bias.

  • \$\begingroup\$ @Bimpelrekkie. I have the option of reverse biasing the PD. I chose to run in photovoltaic mode to keep dark noise to a minimum. From what I understand, Photoconductive mode increases the speed, but not the sensitivity, at the cost of dark noise.user1850479 \$\endgroup\$
    – Miakatt
    Commented Jan 14, 2021 at 20:36
  • \$\begingroup\$ @Miakatt Do you know that the dark current is that much of a concern? How much photocurrent and how much dark current would you expect in reverse bias? Operating at 0 V introduce nonlinearity in your measurement. Is this a problem? What are you trying to do with this system? \$\endgroup\$
    – Matt
    Commented Jan 14, 2021 at 20:38
  • \$\begingroup\$ Mmm. Non-linearity may be an issue in the future. Right now, getting light out of the system is the goal. It's a silicon photonics system with an array of InGaAs PDs. We're at the prototype stage and just need to get some signals out to proceed \$\endgroup\$
    – Miakatt
    Commented Jan 14, 2021 at 20:49
  • \$\begingroup\$ I tried this using a N-type MOSFET in series with the feedback resistor (reduced to 1K), but it didn't increase the voltage. It just shorted the output at each reset pulse. If you think a CMOS switch will work, I'll try implementing that. I was just trying to experiment with the parts I had at hand. Any suggestions are very welcome. Cheers. \$\endgroup\$
    – Miakatt
    Commented Jan 14, 2021 at 20:52
  • \$\begingroup\$ @Miakatt Putting your reset switch in series with a resistor seems like a bad idea as it will greatly increase the amount of time the reset takes. If wired backward you would have forward biased the body diode limiting the max integration voltage. You can try reducing your lasers power and see how low of a signal you can detect, then compare this to what you expect from your scene. What is the background like? If you have too much background IR then you are going to have a lot of extra noise. Make sure it is shielded if that is a concern. You also want the lowest noise op-amp possible. \$\endgroup\$
    – Matt
    Commented Jan 14, 2021 at 21:00

I know I'm a bit late in the day but....

With the circuit as you've drawn it you'll need a dual power supply (positive and negative voltages) for the op amps because when light shines on the photodiode the output of the first stage will try to go negative and if there is no negative supply there will be no output from either the first or second stages.

If you rotate the photodiode (anode to ground) then when light shines on it, the output of the first stage will try to go positive but the output of the second stage will try to go negative and so still no output of the circuit without a negative voltage supply. Even with a negative supply things would be no good because presumably you need a positive going output from the circuit to be useful to a ground referenced ADC.

So the solution, rotate the photodiode (anode to ground) and replace the inverting second stage with a non-inverting amplifier. With this configuration the outputs of both stages would go positive when light shines on the photodiode and there is no need for a negative voltage supply for the op amps.

  • \$\begingroup\$ @Miakatt See recent answer \$\endgroup\$
    – user173271
    Commented Jul 1, 2021 at 23:48
  • \$\begingroup\$ Sorry for the very late reply. You are right about the need for dual power. I had to design a negative supply too. I should have done it the way you said, with non inverting amps. \$\endgroup\$
    – Miakatt
    Commented Aug 16, 2022 at 3:57

The simplest answer is to put a collecting lens on your photodiode. This will collect more light and produce a higher output. Note that this assumes a rather small PD area.

I presume that your schematic is missing a resistor between the two op amps. As shown, you should have an output which is always saturated one way or the other. If you're using single-supply op amps, then you need to configure the second op amp as a non-inverting amp.

"I know the "correct" answer is "Improve the light output of your device". But we have no easy way of achieving this."

Sure there is. Get a beam expander on eBay and use it. You don't say what photodiode size is, but a 1 mr beam with a 1 mm beam diameter will be 4 mm across at 3 meters. If the active area of your PD is smaller than the beam diameter falling on it, you're going to lose energy, and a beam expander will help with that.

If you can, modulate your laser and make a phase-sensitive detector for your receiver. This will allow you to put another stage of amplification without worrying about stray light or offsets.

With generalities out of the way, let's talk about your setup. 5 mW on a "normal" PD (sensitivity about 0.6 A/W) should give a PD current of about 3 mA, which will give 3000 volts on the TIA. Obviously you're getting a lot of scattering/absorption. Tell us more about exactly what medium you're looking through. It is really reasonable that you're getting as little current as you say?

And what is this camera you speak of?

  • \$\begingroup\$ I sounds like the laser is used for test only. It sounds like the actual source to be measured is something else with a much weaker signal. But yes, OP needs to give more details. \$\endgroup\$
    – Matt
    Commented Jan 14, 2021 at 23:58
  • \$\begingroup\$ Sure. The system is a silicon photonics chip with the light exiting 200nm x 45nm waveguides (Yes, it will be diffracting!). We place the PD sensor at the end of the waveguide, hoping to capture the output light. In the full system, these waveguides will be 250um pitch, matching the Hamamatsu array we will use (G8909-01). For now, I'm using an FCI-InGaAs-75 from OSI. This is in a TO-46 case (sensor ~1mm deep). Yes, a lens is the way to go, but in lieu of having that, I'm trying to pull off an electronic miracle! :-) The camera is a GoldEye G008 InGaAs camera from Allied Vision. \$\endgroup\$
    – Miakatt
    Commented Jan 15, 2021 at 10:19
  • \$\begingroup\$ @Miakatt Your waveguide is 200 nm × 45 nm and your PDs are 80 um? It looks like you only have one waveguide per PD? Thats a pretty big difference difference in size. You will have a lot of extra (unnecessary) dark current with such large diodes, which will really hurt SNR and might make what you are trying to do impossible. You need to calculate what your dark current noise is compared to your expected signal to determine if this is even possible. If your waveguides feed directly into the diodes I dont think a lens will help at all in this design. \$\endgroup\$
    – Matt
    Commented Jan 15, 2021 at 13:41
  • \$\begingroup\$ Hi @Matt. Sorry! It's 200nm x 450nm, not 45nm. But even still, yes, there's a big difference in size. 75µm ~ 80µm are the smallest OTS devices we can get. I'll have to wait to hear back from Allied Vision (camera manufacturers) to get an idea of the photocurrent based on the camera measurements. \$\endgroup\$
    – Miakatt
    Commented Jan 15, 2021 at 14:38
  • \$\begingroup\$ @Miakatt You can calculate it yourself. Take the incident power on the diode, divide by the energy of a photon of that wavelength and multiply by the elementary charge. This gives you an upper bound assuming all photons get through. To fix this you can look up or estimate a quantum efficiency, its probably between 0.1 and 0.8, and multiply by that to get a more reasonable estimate. I would expect the manufacturer to have provided this information. I don't think the manufacturer will be able to give you a better idea than you can calculate yourself since they don't know what your source is. \$\endgroup\$
    – Matt
    Commented Jan 15, 2021 at 14:50

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