Photodiode Dark Current Measurement

The photodiode and resistor are in series, so they share the same current. Given that you measured a

1.39V drop across the 1M resistor, what's the current in the resistor?

I don't understand that. Because the the value across the resistor varies as you adjust the bias voltage. The bias voltage shouln't effect the value of the photocurrent generated.

G

What device are you using to measure these voltages? I am concerned that it may be distorting these very high impedance source readings.

I think the dark current is just the current through the 1 meg resistor, so 1.39/10^6=1.39 uA is the dark current (while the diode drops what is left of the 2 volts, or

2-1.39=.61 volts), assuming the volt meter has an impedance many times higher than the 1 meg resistor. For a more accurate reading, you need to know the voltmeter's resistance and solve that formula for the parallel resistance of the 1 meg resistor and the meter's resistance.

It may be simpler to find out what the resistance of the voltmeter is, and just use it as the current sense resistor, with a higher applied voltage. It is probably a good idea to also take several results at several voltages and learn how the dark current varies as the applied voltage varies.

I think your formula is.

The reverse voltage across the photo diode certainly varies its dark current, which is a leakage current. Most photo diode data sheets show this voltage dependence.

Your measured dark current, if we neglect the effect of the voltmeter(s?) on your circuit, is 1.39 uA as John P. said. At least that's closer to the >2 uA you were expecting (*). Perhaps your >2 uA figure is a "max/worst case" value, or is the spec at a considerably higher bias voltage than the 0.6 to 0.8V you had.

You might redo the measurement using a 100k resistor, and see if you get close to the same current (i.e., 0.14V across the 100k resistor).

Also, I have some questions about your setup ...

1. What is the voltage of your "2V" supply, to the nearest 0.01V? The diode and resistor voltages should add up to the supply voltage.

1. Did you measure the resistor and diode voltages at different times with the same meter, or simultaneously with two separate meters?

Regards,

Mark

(*) GraemeC wrote:

I would not rely on measuring the voltage across the diode. the Load from the meter it self is going to influence the reading, especially in the case of a diode. I would how ever, use the measurements across the resistor and deduce the 10 meg shunt your meter is adding to it. The current in the 1 meg R will be the same of that in the PD.

You could use a meter like we have at work for special testing that places G/T ohm loads on the circuit. In the math you gave, you're actually subtracting more than you started with, which gives you a - number. This is what's giving you the problem.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
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To answer some of the questions.

I am using a HP 3401 DMM and a Fluke 115. I have taken the results from both sepertatly simulatneously.

When I measure the value across the resistor with a large amount light incident on the PD. I obtain a value in excess of 2V (2.6V is the maximum iv measued).

Graeme

I am not an engineer, rather a garage hacker since the early 1960's when all you needed was a Volkswagen van and you picked up love along the side of the road.

When I worked with "photo-diodes" over the years in the "many" mega-ohm range you must consider the effect of the magnetic fields that fill your work area, often heard as hum over audio amplifiers.

"I would shield the photo diode and ground everything."

If I used a cardboard columnating tube to narrow the field of the captured light (pin hole) I could detect passive light shifts/movement over a quarter mile away at high noon using a single photo-transistor and no lens. This is how we guarded our crops. LOL

• * * Christopher

Temecula CA.USA

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I would remove the meter that is across the photodiode (if that is how you are set up), and do the measurement with just a meter across the resistor.

Also: measure the actual source voltage, so you can infer the diode voltage. And, repeat the measurement with a smaller (say 100k) resistor to check for some consistency in what the diode current is.

Regards,

Mark