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Help understanding Photodiode Amp

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dknguyen

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I'm a little confused about this photovoltaic photodiode amp that I have attached:

Doesn't the photodiode attempt to provide a voltage across across it that tries to pulls the inverting input below the negative supply rail? I know that the op-amp tries to hold the voltage difference across the photodiode to 0V. But that's not the case during startup or disturbances.

Doesn't this cause the op-amp to malfunction? How does an op-amp respond to something like that?

Also, just double checking:
1. This circuit will produce a positive output right? I've seen things all over the internet that have it going both ways for the photodiode in the same direction. I am a bit uncertain because...

2. I am conceptually having a difficult time seeing how the op-amp is able to drive the inverting input to 0V by driving current in the same direction as the photo-diode leakage currents which is causing the inverting input to go below 0V in the first place.
 

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For that circuit to work with a single positive supply, the op amp must work with the input at or slightly below 0V (which many rail-rail type op amps do).
So when the diode tries to pull the (-) input slightly negative, the output will go positive to generate a current through Rf sufficient to keep the input at 0V.
There's generally no problem during startup or disturbances. The op amp input may momentarily go above or below zero but that will cause the output to rapidly change so as to bring the input voltage back to 0V (as long as the output can stay within its linear range).
 
hi,
This Microchip PDF explains the PD OPA in detail.
E
 

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Look again, the photodiode does not have any leakage current because it has no bias voltage across it. Leakage current is used when the diode has a bias voltage which reduces its capacitance and allows it to switch on and off quickly. But the bias voltage causes "dark current' that you might not want.

The circuit you posted uses the photodiode as a tiny solar cell that generates a small current (the negative feedback cancels any voltage) when it has light. Then without a voltage across it does not produce a dark current but its higher capacitance slows it down a little.
 
Look again, the photodiode does not have any leakage current because it has no bias voltage across it. Leakage current is used when the diode has a bias voltage which reduces its capacitance and allows it to switch on and off quickly. But the bias voltage causes "dark current' that you might not want.

The circuit you posted uses the photodiode as a tiny solar cell that generates a small current (the negative feedback cancels any voltage) when it has light. Then without a voltage across it does not produce a dark current but its higher capacitance slows it down a little.
Not sure what that current is actually called. I just read someone else referring to it as a leakage current, probably because it flows in the opposite direction of a typical diode. Didn't have a better name for it since there isn't an actual current flowing through the photodiode during normal operation (I think) due to the op-amp drive.

It blows my mind that an op-amp can process inputs beyond its rails. Obviously, my knowledge of the internal workings of an op-amp is very weak.
 
The inverting input of the opamp has negative feedback so it is a dead short to the photodiode. Then the opamp amplifies the current (but there is no voltage) produced by the photodiode. The current is called Ish where sh is SHORT. The shorted current.
The other way of doing it is by biasing the photodiode with a reverse voltage then light causes it to have leakage current.
Look here:
 

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  • PhotodiodeAmplifers.pdf
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The op-amp might amplify the current but does it still not operate based on the voltages at its inputs? I'm just having a hard time visualizing how it works in the transient state when amplifying currents (before the inputs are driven to equal voltage).

In my head it's driving more current on top of a pre-existing current which is kinda backwards when the op-amp is meant to amplify voltage (where the output drives something opposite to what is on the inverting input to "cancel out" the voltage there so its equal to zero.
 
Your pic would confuse anybody. You have to draw the currents being positive toward a node. (e.g. the + input of the OP amp.)

It does, try to make the inverting input, the same value as the non-inverting input. To do so, it has to feedback the current Irf. Again, assuming that toward the op amp (-) input is positive.

The output, is therefore -Irf. You can put a voltage on the (+) inp and get Vout=Vb-Irf

Vb or the bias voltage is impressed across the device. Vos of the OP amp is too. So, in I-V converter designs, Vos is important.
 
In my head it's driving more current on top of a pre-existing current which is kinda backwards when the op-amp is meant to amplify voltage
1) Use a +/-10 volt supply for now so you don't have to think about voltages below the supply voltage.
2) The diode is making a negative current. It wants to pull the (-) input below ground. Lets say 3uA. The output of the amp will go up or positive. If the resistor is 1meg ohm then it will take 3 volts to make +3uA to counter act the -3uA from the diode.

So NO the the diode current is not on top of the resistor current. The diode current pulls down and the resistor current pulls up. They subtract to make zero.

upload_2015-11-28_18-12-17.png
 
At DC an opamp has a voltage gain of maybe 1 million or more! Then its voltage between the inputs is so small that it is almost impossible to measure. If the reference (+) input is at ground then the inverting input (-) is also practically at ground. That is why the inverting input is said to be a dead short when there is negative feedback to cancel most of the voltage. Many rail-to-rail opamps have inputs that work down to 0.3V less than the negative supply voltage that is usually ground. Here the input voltage is much less than that and is practically at ground.
 
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