photodiode current to voltage conversion

[e1000]

I am trying to quantify light for a homebrew spectrophotometer. I am using a fairly sensitive photodiode along with a low voltage op amp (I got both on digikey) I took a collegiate physics class, and I am technically minded, but electronics is not my field so I'm a little lost here.

When I use the circuit on the left and measure the current between point A and B, I get a voltage response that seems to be linearly proportional to the incident light. If I measure between A and C, i get a voltage that is about 15% lower than the supply voltage and is unresponsive to light.

My circuit seems opposite from the current to voltage converters I have seen online, shouldn't the circuit on the right work better? Also, shouldn't I get a voltage by probing between the output and the ground (A to D)? When I built that circuit with the same (1 Mohm) resistor and I probe point A and D, I get constant voltage that it a little lower than the supply voltage, and it is completely unresponsive to light, when I substituted a 4.7 Kohm resistor, it is almost the same story (also unresponsive to light).

 

[karesz]

Hi e1000,
let me know your type of PD (Photo Diode) pls.
K

 

[betwixt]

One circuit is measuring the diode current, the other is measuring the voltage it produces. I suspect however, that your problem lies with the power supply. Does the amplifier have built-in biasing and as the input voltage/current is ground referenced, can the amplifier handle this. You may need to use a split power supply to keep the amplifier happy.

Brian.

 

[FvM]

Assuming a single power supply with the "ground" symbol representing the most negative supply voltage, only the left
circuit can work, because it has a positive signal output. The right circuit would need a negative respective split supply
in any case to allow a negative output voltage.

Both circuits are however operating the photodiode with zero bias, as a current source.

As mentioned by betwixt, single supply also requires a common mode range including ground.

 

[Braski]

IMHO, you have to use the left circuit, but you have to add a voltage source on the "+" IN of the OP-AMP. Thanks to virtual short circuit, you will bias the photodiode in its inverse region, obtaining correct operations. I think 1 or 2 V will work. Then check the offset performances of you amp, especially current offset.
When you use a photodiode, in fact, you measure the increasing in the inverse saturation current induced by the photogeneration.

The circuit on the right is wrong for me.

 

[snafflekid]

the left circuit is correct. technically either circuit works.

It helps to understand what is happening inside the diode when light strikes it. A photon hits the depletion region which generates a hole-electron pair. This gets swept across the depletion region towards the terminals by the built-in 0.7V of the junction and generates a current. The way the diode is connected on the left side causes positive current to flow through the resistor toward the diode, which generates a positive voltage at the opamp output (remember that if this is a single supply opamp it cannot output negative voltages which is why the right circuit does not work).

The opamp forces zero volts across the diode, if you recall how virtual ground works in the negative feedback system. This is important because if the voltage across the photodiode is not kept constant, the current generated by the light will generate a voltage, then you do not have a current proportional to photons. You could set up the system to create an even larger negative bias on the diode if you wanted and that would make the carrier generation even more efficient, but should not be necessary if this is in fact a photodiode.

I do not know why you are not seeing a voltage proportional to light on the output. The diode may be generating so much current that when it is applied to the 1Mohm resistor it would be higher than the power supply. In that case the output would just sit near the power supply level. Try a smaller resistor value or higher power supply. Also remember that eyes react to light logarithmically but your circuit will react linearly and that might be important to you.

 

[FvM]

but you have to add a voltage source on the "+" IN of the OP-AMP

The circuit on the right is wrong for me.

I wonder, why these simple photodiode cicrcuits are often causing confusion?

You can, of course, reverse bias the photodiode. This way, you reduce the junction capacitance and slightly increase
the speed, but also get increased reverse leakage current. For a DC or low frequency application (e.g. chopper modulated
photometer), zero bias is the preferable operation mode in my opinion.

By reversing the photodiode polarity, you get an inversed output signal. As said, it only works with dual supply. There
are however possible reasons to use a dual supply with a high performance I/V converter anyway. E.g. the fact, that the
best available OPs are probably needing a positive common mode voltage. Or the requirement of an output signal scale
down to true zero.

As another point, some popular photodiodes for photometric applications have the cathode connected to the case. They
would be preferably operated in the right circuit, with dual supply and a negative signal output.

 

[e1000]

Here is my photodiode;
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=475-2579-1-ND

The responses so far are encouraging, but one issue that is still getting me is why I am seeing the light response voltage between points A and B, I always thought that I should see a voltage between the output and the ground? Is this possibly related to the lack of a virtual ground?