Question regarding the photodiode model

No. The leakage current, V/R will be zero if V=0 (which it rarely does), but R will not change.

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 Thanks,
    - Win

No. If you want to increase the shunt resistance, run it at a small reverse bias. The I-V curve (current vs voltage) of a diode without photocurrent goes through (0,0) but not with zero slope. The shunt resistance is dV/dI, which is large but not zero at the origin.

If you're worried about the shut resistance of your diode, one of two things is probably true:

1. You're using an IR diode such as Ge or InAs, and it's leaky at room temperature.
2. You're trying to maximize your signal-to-noise ratio, and want to minimize the dark current shot noise.

Reason #1 is a good one, but reason #2 is almost always a bad one, unless your signals are in the nanoamps or below. Silicon and InGaAs photodiodes are so much better behaved with reverse bias, it'll blow your mind. Try 7x lower capacitance (which translates to 17 dB lower noise density at high frequencies), 10x better linearity, ... I could go on.

Cheers,

Phil Hobbs

My brain went into reciprocal space there for a moment. Should read, "large but not infinite".

Cheers, Phil Hobbs

The dark current won't be zero at zero bias - charge carriers promoted to the conduction band by thermal fluctuations, cosmic rays and local radioactivity look just like charge carriers promoted by photon absorbtion.

Leakage current will be zero. As has been mentioned in the other posts, the shunt resistance of the diode is not infinite, or even all that large, at zero bias - IIRR it is just kT -26mV at room temperature - divided by the leakage current at a volt or so of reverse bias (anything too low to give any significant avalanche multiplication - the reverse current at the rated reverse voltage is inflated by avalanching, which is why you don't want to exceed the rated reverse voltage).

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Bill Sloman, Nijmegen

So you don't believe that a wire with zero voltage across it carries zero current? It has conduction-band electrons all the time.

Cheers,

Phil Hobbs

Hello Phil,

I had one that was shorted but beginning to let off a stench. Then the insulation melted off and glued it to the table. The stench became unbearable.

Ok, it got helped along by the field of the tank coil of a 750W amp :-)))

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Regards, Joerg

http://www.analogconsultants.com

Interesting, so if I connect a very sensitive current meter accros a totaly dark diode will I see a current ?

Colin =^.^=

So you don't believe that you can detect photons with a zero-biassed diode?

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Bill Sloman, Nijmegen

Ok let me see if I get this write; if I have a photodiode operating under a short circuit current regime say for simplicity, there will always be a current or dark current generated simply because there will always be some mechanism that creates an electron-hole pair which gives rise to a current, in much the same way as that achieved via the photon radiation? In order to theoretically eliminate the dark current, the diode would have to be cooled and shielded from external radiation. Not that I am trying to eliminate this component I am just interested in clearing this up in my mind.

Then at this point we could define the shunt resistance will be the slope of the iv curve at 0v. Is this so?

Charles

Ok let me see if I get this write; if I have a photodiode operating under a short circuit current regime say for simplicity, there will always be a current or dark current generated simply because there will always be some mechanism that creates an electron-hole pair which gives rise to a current, in much the same way as that achieved via the photon radiation? In order to theoretically eliminate the dark current, the diode would have to be cooled and shielded from external radiation. Not that I am trying to eliminate this component I am just interested in clearing this up in my mind.

Then at this point we could define the shunt resistance will be the slope of the iv curve at 0v. Is this so?

Charles

That's my expectation. It does depend on where the dark current comes from and how energetic the corresponding charge carriers are when they get into the conduction band.

-- Bill Sloman, Nijmegen

Dark current from cosmic rays and radioactivity is very low. About one q per cm2 and second. In the absence of cosmic rays and at thermal equilibrium, the 'rectified noise current' will cancel the "charge carriers promoted to the conduction band by thermal fluctuations". If not we have a Maxwell's demon.

So in practice we will need a very sensitive current meter, I think. Another case if the diode and the current meter have different temperatures.

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Sven Wilhelmsson
http://home.swipnet.se/swi

1) You are *right*, NOT "write". 2) Furthermore, the theoretical capacitance of the photodiode in that case is zero (think about it).

This would seem logical to me but would appear to contradict what other people are saying!!

P.s. sorry for the "write" spelling

Capacitance is the derivative of charge with respect to voltage. Making the voltage zero does not make the derivative zero. It's like a boat launching ramp--when your wheels hit the sea water, the altitude is zero, but the slope is not.

For a silicon PIN photodiode, the capacitance will go down by a factor of about 7 as you increase the reverse bias from 0 to somewhere near the diode's max Vr.

Cheers,

Phil Hobbs

Carrier pairs in the depletion region, regardless of how they're generated, will get pulled apart by the junction E field, and appear at the terminals. But the quantities involved are sufficiently small (a few electrons/s at most) to be completely dwarfed by the diffusion currents and the Johnson noise current of the zero-bias resistance.

My beef is not so much with your physics (which is fine) but rather with the confusion this hair splitting is likely to produce in the OP--who doesn't seem to realize the difference between current and conductance.

Cheers,

Phil Hobbs

Thanks I realise that there are a number of other parasitic that will dwarf these issues As i am writing up my thesis, i was keen to get a firm grasp of the meanings or physics that lead to the model. Thanks for the great discussion

Thanks Charles

Phil Hobbs wrote: