# Correct use of op-amp to amplify photodiode in reverse bias

So I have a photodiode in reverse bias, from which I intend to measure my pulse.

Currently, my schematic looks as follows:

simulate this circuit – Schematic created using CircuitLab

Weirdly, this produces a perfect curve with a good amplitude and a response around 380mA (Systole) to 420mA (Diastole).

From my schematic, I understand that I'm not using the operational amplifier correctly, hence why I have tried doing something like this, as it would make better sense:

simulate this circuit

The following schematic still produces a pulse, but the amplitude is not good and I can't see the dicrotic notch as in the previous. (Don't mind the motion artifact)

What am I doing wrong here? How do I correctly use the operational amplifier? If I attach my analog read directly to the output of the Op-amp I get around 0.5V, i.e. close to nothing.

Lastly, when I drive my non-inverting input high, I actually do get some response at my output that looks okay.

• What are the power supply voltages for the op-amp in your first schematic? Commented Dec 15, 2018 at 17:14
• 1) Answer The Photon's question. 2) Read from the output of the op amp. I suspect that you are using a single +5 supply, Then the op amp output is always zero (since it is trying to drive below ground, and the op amp is having no effect at all. I suspect you've done something wrong on your second circuit - double-check your connections. Commented Dec 16, 2018 at 6:36
• sorry, the voltage across the rails is for + 5v, and - ground Commented Dec 16, 2018 at 12:06

There are a couple of things in here that don't really make sense. It would be good if you provided a little more context for the application, the time series you showed doesn't have a time axis label but I'm assuming from the terminology that you are measuring heart beats or something similar.

I'll break down the issues and suggest an improved circuit:

1. The point of the 1 MΩ + 1 µF resistor capacitor pair is to low pass filter. I'm guessing the point of this was to reduce noise from the 5 V supply. It kind of doesn't make sense to put this where it is on the op amp. This is filtering the output signal of the op amp and isn't really helping smooth out anything in the biasing voltage of your photodiode. I would suggest moving it to before the photodiode and lower the resistor a little bit. Maybe a 1 kΩ + 22 µF pair would do it (this would have a low pass filter of 7 Hz). Maybe that isn't enough but you don't really want to raise resistance too much and very large capacitors are more expensive.
2. You're not really using the op amp in the best way here. You want to read our signal out after the op amp with an active transimpedance circuit (for best linearity). It would put the 4.6 kΩ + 47 nF pair in feed back of the op amp. If you need more gain just up the resistance value and lower the capacitor. This RC pair acts like a low pass filter f_c = 1/(2*piRC). So for more gain maybe select 220 kΩ + 1e-9 nF could also work if you need really high gain. Just play around with values to get the bandwidth and gain you want at the output. See schematic for what I mean.
1. This op amp is usually operated single sided. That means you apply 3 -32 V on the +V pin and ground on another supply pin. If you just hook the circuit up with no light on the sensor you will be 'railing' or at the edge of its voltage range. If the sensor is alway run with some light on it then it might always be near the center of the range. Otherwise you will need make some changes to make the dark voltage is always offset a little from either your 5 V or 0V (GND) supply.

simulate this circuit – Schematic created using CircuitLab

I think you should try something like this:

simulate this circuit – Schematic created using CircuitLab

You will probably have to adjust the resistor value to get a strong signal.

Your opamp provides a dc-nulling function, with 1 second tau.

Your output is the error input to the opamp. Is this useful to you?