# Question regarding the photodiode model

Discussion in 'Electronic Design' started by cuca, Aug 22, 2006.

1. ### cucaGuest

Question regarding the photodiode model. Just to clarify in my mind,
if a photodiode has no biasing voltage across it, i.e. it is operated
under short circuit condition using a transimpedance amplifier,
theoretically there is no leakage or dark current. Thus the shunt
resistance could be considered infinite or at least larger that that
quoted in datasheets where a 10mv reverse bias voltage. Am I write in
saying this?

2. ### Winfield HillGuest

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

3. ### Phil HobbsGuest

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

4. ### Phil HobbsGuest

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

Cheers,
Phil Hobbs

5. ### Guest

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).

6. ### Phil HobbsGuest

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

7. ### JoergGuest

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 ))

8. ### colinGuest

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

Colin =^.^=

9. ### Guest

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

10. ### cucaGuest

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

11. ### cucaGuest

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

12. ### Guest

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.

13. ### Sven WilhelmssonGuest

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.

14. ### Robert BaerGuest

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

15. ### cucaGuest

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

P.s. sorry for the "write" spelling

16. ### Phil HobbsGuest

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

17. ### Phil HobbsGuest

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

18. ### cucaGuest

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