Photodiode series resistance -> dominant noise source!

So I'm doing another nanoamp photoreceiver.

Normally in this sort of circuit, the dominant noise source is the amplifier's input noise being impressed across the photodiode capacitance, resulting in a noise current density

i_N = e_N * 2 pi f C.

This "eNC noise" is insignificant at low frequency but at high frequency it rises to dominate the total noise.

I've measured the photodiode capacitance with a Boonton 72, and it matches the datasheet value for the part I'm using (Osram SFH206K), i.e.

10 pF @ 24V.

With an NP0 cap of that value substituted for the photodiode, the noise floor of the circuit is just where I calculate it to be, so no worries there--the amplifier is doing just what it's supposed to be doing.

However, with a photodiode, it's 3-4 dB higher than that, even in the dark. It's not leakage or shunt resistance, because it rises with frequency just like the other eNC contribution.

It looks like it's the series resistance of the photodiode, which appears in series with the photodiode capacitance and creates current noise just the same way as the amplifier's noise, except that it's only the PD capacitance that enters, not the total capacitance at the amplifier input.

Usually you can just ignore the series resistance unless you're running at zero bias, but not when you're down in the nanoamps!

Sooo, I'm spending time with the manufacturers' apps guys looking for information about the series resistance of 2-3-mm class PIN photodiodes in reverse bias.

It's always something.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs
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Isn't there a way to measure that along with the capacitance using a network impedance analyzer? Keeping the excitation really low, of course.

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Regards, Joerg 

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Reply to
Joerg

Probably. I don't have one, though, and anyway it's just as easy to cram it into the circuit and measure the noise, which is what I care about anyway.

Thanks

Phil Hobbs

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Phil Hobbs

"Gotcha" is the daily event :-( ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Reply to
Jim Thompson

I usually sim a pd with a current source and a capacitance, so now you have made my life more complex.

The fiberoptics detectors that we use are small so have pretty short RsCj time constants, 10s of ps maybe, so the noise will be way up in the spectrum. But even a 100 ps time constant could add jitter to a fast receiver circuit. But we launch milliwatts, which makes things a lot easier.

In a free-space app, might multiple small detectors and amps be better than one giant one?

What effect, if any, does bias voltage have on Rs?

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John Larkin         Highland Technology, Inc 
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John Larkin

Yup, me too.

A little matter of 100 dB or so better. ;)

Probably. Even just one amp with several parallelled detectors would be likely to reduce series resistance.

Using multiple amps works as long as the amps' input capacitance is sufficiently smaller than the individual PD capacitance--otherwise you just add more noise power from the intrinsic eNC noise of the amps.

It's mostly silicon resistance, which should drop as the bias increases. Many PDs speed up amazingly when you fully deplete the silicon, which is either a drop in Rs or eliminating diffusion delay, or probably both.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Yeah, more voltage speeds things up even after the capacitance levels out.

Incidentally, I have a lot (embarrasingly a lot) of First Sensor PS13-5B-TO5 UV-enhanced photodiodes.

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John Larkin         Highland Technology, Inc 
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John Larkin

Nice part if you have a bit of light--they're about the best UV PDs I know of, but their capacitance levels off at about 60 pF, which is about

10 times higher than the ones I'm using.

BTW by switching to Osram BPW34s (same area) and jacking the bias up to

30V through 100k gets me to within 1.6 dB of the result with the 10 pF cap. Its abs max is 32V, but its actual breakdown is over 100V. So it's probably soluble with the right PD.

I'll get some Vishay BPW34s and try them out. They're slower, so maybe they have thicker epi with lower resistance.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

That's interesting. Would it help if I order one and measure it with an HP 4194?

Reply to
JM

Sure, that would be interesting. Thanks!

So far an Osram BPW34 at 30V is down to 1.6 dB worse than the 10 pF cap, so maybe I can find the right PD for the job.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

OTOH, while a network analyzer is certainly nice, if you only need to measure C and ESR, an LCR meter might do. Maybe not one of the really cheap ones for dry electrolytics, but a halfway decent LCR meter that goes at least to 100 kHz should have a chance to "see" the ESR.

Dimitrij

Reply to
Dimitrij Klingbeil

? The thermal current noise of the series resistance? Does that mean the series resistance is higher than than the feedback R? That doesn't sound right... I'm probably not understanding.

George H.

Reply to
George Herold

Is it possible to calculate or estimate approximately how much resistance, are you seeing tens ohms, hundreds ohms or kilo-ohms etc?

piglet

Reply to
piglet

If you imagine that the TIA itself is noiseless, it puts exactly zero volts AC across the photodiode. But the PD isn't a pure capacitance, it's a seri es RC. Since there's zero volts across the series combination, the resistor 's Johnson noise voltage appears across the capacitor, leading to e_NC curr ent noise in the external circuit.

In my circuit this dominates when R_s is above about 60 ohms.

Cheers

Reply to
Phil Hobbs

OK - will do.

Reply to
JM

What is the physical mechanism of this resistance? Just the usual diode "bulk" resistance that all PN junctions have or is it something specific to photodiodes?

Is it at all frequency dependent?

Reply to
bitrex

You should be able to TDR it pretty easily.

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John Larkin         Highland Technology, Inc 

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John Larkin

McGuver method: Make an resonant circuit out of it with a coil that has very low losses. The resonant frequency will reveal the capacitance and the -3dB points will tell you the Q from which you can calculate what the resistance would be. This assumes that all other losses are insignificant compared to the contribution of the series resistance.

This could be done with an RF generator, a FET probe, and a scope. The exact capacitance of the FET probe needs to be known but it usually is. If not, run it sans PD and log the frequency (and then note it in the manual for next time). The Q in that case must be more than 2x better than with PD. If it isn't you need a better inductor. No kidding, I have polished coils in the past to get the Q up. Wenol paste is really great for that but don't get that on your skin or clothes.

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Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

ts AC across the photodiode. But the PD isn't a pure capacitance, it's a se ries RC. Since there's zero volts across the series combination, the resist or's Johnson noise voltage appears across the capacitor, leading to e_NC cu rrent noise in the external circuit.

Got it, Thanks.

GH

Reply to
George Herold

Would the series resistance be any different forward biased? Maybe just an I-V curve will show it?

George H.

Reply to
George Herold

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