# Photodiode Dark Current Measurement

Discussion in 'Electronic Basics' started by GraemeC, Sep 28, 2007.

1. ### GraemeCGuest

I am trying to measure the dark current of a photodiode

I have Chathode of the PD connected to 1Meg ohm resistor which is
connected to +2V
The annode is connected to 0V

I obtain the following measurements
Voltage accross Resistor: 1.39V
Voltage accross Photodiode : 0.82V

I have assumed that the Dark Current (or indeed any current value) is
(2 - (1.39+0.82)) / 1Meg

=> I obtain a dark current of 210nA. This figrue however is less by
more than an order of magnitude than that I am expecting.

Is my understaning wrong?

Please don't ask why I have this set up am not using Op-Amp circuity!

Many Thanks

Graeme

2. ### Greg NeillGuest

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?

3. ### GraemeCGuest

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

4. ### John PopelishGuest

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

5. ### John PopelishGuest

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.

6. ### redbellyGuest

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).

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.

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

Regards,

Mark

7. ### JamieGuest

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.

8. ### GraemeCGuest

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

9. ### Guest

Hello 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

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
http://www.oldtemecula.com

10. ### redbellyGuest

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

11. ### neon

1,325
0
Oct 21, 2006
fisrt of all how do you see dark current in the dark. AFTER THAT YOUR READING DO NOT ADD UP 1.39 +82=2.21V HOW DO YOU GET HIGHER SOURCE FROM A 2 V. Furthermore is the photodiode connected to anything at all? like a base of a transistor or what?Torealy measure the diode current your meter must be in series with the 1mega there is going to be some inacuracy because of meter resistance. better yet add a 1k in series and measure viltage across then deternine the current.  