need help with a photodiode & opamp circuit

[cryptopsii]

I have a circuit that has to receive a low-duty pulse, emitted from a IR LED 1-2 meters away. The wave does not use a carrier or any form of modulation so the receiver is also pretty simple.

The receiver use an IR photodiode of a matching wavelength. Here the schematic:

(or https://www.freeimagehosting.net/uploads/1e4bb80d94.png whatever)

This circuit work ok, but it has ZERO noise protection, so if there is sun, the diode produce a constant amount of dc. the same is true for indoor lights. (except it produce 60hz instead of dc)

So I thought about simply using a high-pass filter since the signal i receiver is around 1kHz but i just can't figure where to put the capacitor & resistor...

the opamp is a obscure NMJ062 and is dual package, so if your solution imply a voltage follwer or any dual-op amp configuration it's fine. Also, the opamp isn't
rail-to-rail (It's like -.5V on each rail)

So any help/suggestion / better design idea would be greatly appreciated.

 

[keith1200rs]

Is the first amplifier saturating in bright light? If not then all you need to do is add another stage with AC coupling (or a 2 pole high pass filter if you like).

Remove the resistor to ground anyway - it just increases the noise.

Who makes the NMJ062? I cannot find any details of it.

Keith.

 

[FvM]

Apart from the question, if the 1M resistor to ground is required or not, the circuit doesn't work, because the photodiode is reverse connected.

P.S.: I guess, you mean NJM062, a TL062 clone from NJR.

 

[cryptopsii]

@keith1200rs: thx for the answer,
I will give a try with the filter and see what happen.

@FvM: yes the op amp is an njm062
here a link https://www.futurlec.com/Sanyo/NJM062.shtml
but the photodiode is the right way. It is used in reverse, It normally block
voltage, giving no current and if it receive light, it leak more current
(dark =~ nano Amp, light = uAmps)

 

[FvM]

but the photodiode is the right way

I'm not talking about a different photodiode circuit. You're using a standard circuit, where the photodiode is biased at Ud of 0V. That's O.K. But the voltage at the OP output voltage would be negative, which of course can't work with single supply. You can provide a bipolar supply and see, what happens. (Or it's only a drawing error and the real circuit is different). If you are in doubt about the photo diode current polarity, there are previous threads at edaboard related to the same topic. See e.g.

Regarding the connection of +ve OP input. The input voltage range requires a bias above GND. But you should better use a voltage divider than a high ohmic resistance.

 

[iamdink]

You should use differential opamp setup to remove CM noise. These are very common and used many places.

https://en.wikipedia.org/wiki/Operational_amplifier_applications#Differential_amplifier

I often provide gain in a first stage then used the differential amplifier stage to remove the CM.

 

[tinska.h]

Sorry to high jack a thread...

I'm trying to filter out a DC-offset from photodiode output as well. In my configuration the photodiode is biased to virtual ground and op amp is configured as transimpedance amplifier with non-inverting input connected to virtual ground. I'm only adding a capacitor between op amp output stage and adc input. However it seems that signal becomes slowly bipolar where Vmean is at VCC/2-Voffset. Where Voffset is from 0.2V to 0.4V (it drifts). So my question is what is the best solution to high pass a unipolar signal from a photodiode? Only purpose here is to remove DC-offset caused by sunlight.