Increasing sensitivity of photodiode amplifier

Question

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).

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 :)

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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.

Answer 1

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.

Answer 2

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.

Answer 3

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?