can anyone explain to me why moving to configuration (b) overcame the problems I was having with the photodiode saturating?
Many thanks,
John
Electronic – Photodiode in photovoltaic mode, overcoming saturation
photodiodesaturation
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The circuit on the right side makes use of an LM 108 and a capacitor on pin 8 ( due to the datasheet pin 8 of the op amp stands for Comp). Not having this pin on L358 what can I do for that?
The LM108 has pins for external compensation. As you've already figured out you will not find these pins on the LM358 because the LM358 is internally compensated (e.g. the capacitor is within the chip). So if you use the LM358 you don't need that capacitor.
Morever the arduino analog pin normally indicates a value btw 0-1023. Now using the op amp I get a maximum value of 767 ( this means approx. 3/4 of 1023). Is there any particular reason for that or am I doing any mistake?
The output of your amplifier does not generate a voltage that the arduino would interpret as full scale. You can increase the amplification by choosing a bigger resistor between the output and the negative input. If you want your amplification 25% higher just make the resistor 25% bigger.
Note that the voltage at the LM358 output can't go as high as the supply voltage. It will always have a maximum output voltage that is 1.5V to 2V below the supply. That means, if your arduino analog input would give you a full scale at - lets say - 5V, then you need to supply your OpAmp with 7V at least. Otherwise the output will clamp and never reach the full scale.
You'll also likely don't want to connect your OpAmp output directly to the analog input. Analog inputs have a maximum input voltage, and you should never exceed this. If you do current will flow through a protection diode. If the current is to large that could blow up your arduino. Adding a resistor between the OpAmp output and the analog input will limit the current without influencing the reading much. 2.2k is what I usually use as a rule of thumb value.
Well...
Checking the datasheet of your "photodiode", it appears to be a photocell, and not a photodiode. Also, there are no specs concerning how much current it can provide, which means we're in a bit of trouble to select an appropriate opamp...
I have also experimented with the transimpedance photodiode circuits but without any success
They're all inverting, and current-input... This is not a problem for your single-supply circuit provided you wire the sensor in the correct way, but note that your sensor is specified for a voltage output. Its light-to-current specs are not in the datasheet. So, using a voltage-input circuit (as posted) seems wiser...
Since we don't know how much current the sensor can put out, you should measure its short circuit current and stay well below that. Check leakage current on C1 in your schematic.
Now, back to the non-inverting amp. LTC1050 has:
- "Low Noise 1.6µVP-P (0.1Hz to 10Hz)" which might actually be a bit on the high side unless you use lots of averaging, since you're measuring in this range....
- Input common mode goes to GND (good)
- Output swings to GND (good)
- Open loop gain is high (130dB) so good
- Input offset +/- 5µV: this will be a problem if the input offset is -5µV then you will lose the first 5µV of input range, since output cannot go below ground
- Input bias current is very low: good
- Output drive current is also very low, about 1mA with 5V supply, which means your 3.3k feedback resistor is too low, 330k would be better.
Try increasing the feedback resistors, and if that doesn't work, post your layout.
--- EDIT
About feedback resistors: compare resistor noise with noise from the opamp, which isn't that small... Increasing feedback resistor should help, as the output current capability of this opamp is tiny.
Now, your data:
- The voltage output outputs 10uV/Wm-2
- The current output mode outputs 46.68 nA/Wm-2
So, opamp input bias current (max 30pA at 25°C) is alright.
However, if the sensor can only output this extremely tiny current, you're in trouble... Any humidity on the board or capacitor leakage will screw things up. A transimpedance amp would be better, as it maintains the input voltage at 0V, which neutralizes leakage currents.
Best Answer
Current flows across the photodiode from cathode to anode: potential is built up against the forward conduction direction of the diode, this is what actually gives you saturation. The answer from @krufra is wrong.
In this 'photovoltaic' mode charge builds up across the diode like a capacitor and is dissipated across your 50 Ω resistor (case A). The responsively drops as more light is incident. You can think of it as the electrons having to do more and more work charge is built up, the diode responds less as the voltage builds.
In case B you have simply lowered the resistor value making it easier for charge to flow away from the anode. With less of a potential build up across the diode then you would expect to see less of a non-linear drop off of the diodes responsively to light. I'd imaging that your second circuit might product a lot less signal.
The best way of getting linear (voltage proportional to amount of light) that doesn't saturate is by using a transimpedance amplifier. This can be relatively cheap with modern op amps. Olin's answer is mostly right but photovoltaic mode is when the diode is hooked up so that it develops a voltage across some resistive element where as photoconductive mode is when it is configured (with a transimpedance amplifier) to see 0 Ω resistance across its output.