photodiode voltage

Find another diode? Call them and ask what the spec means?

The response goes to 1 nm! That's like an X-ray, I think.

George H.

That might be a starting point. The data sheet makes no sense.

The Hamamatsu uv photodiode that I've used has similar contradictory voltage specs. It seems to be a tradition. The killer is the c-v curve. I don't want to run these things at 5 volts.

And it's rising, off the graph!

Maybe it's over the whole temperature range? Perhaps 'our' Russian x-ray PD guy (I've forgotten his name...) will know.

Yeah... IDK what the x-ray absorption mechanism is.... just blasting electrons out of their orbitals at a guess. I wonder if the peak at ~15 nm is some deep electron state in Si?

(The inner most electron should bind with an energy proportional to Z (Z is the nuclear charge) Si is Z=14 , 14*13eV ~ 180 eV ~7 nm.

(13 eV is the rydberg.. hydrogen binding energy) George H.

Yes, of course. The graph is possibly misleading. As the energy goes up (wavelength goes down) the probability of an interaction (efficiency) goes down, but the number of charge carriers produced goes up. So, it actually gets LESS sensitive, statistically, but the output pulse will be bigger.

Lots of people use Silicon detectors to detect X-rays (the modern digital medical/dental X-ray systems, for instance) and charged particles (nuclear physics). Lower energy photons and particles are stopped in the Si, higher energy ones deposit some energy there and punch through.

Jon

The graph is amps per watt, and the energy per photon covers a 1000:1 range, so the 1 nm photons are relatively infrequent at a given power, but sure pack a punch.

'Breakdown' means different things in photodiode datasheets. For MPPCs/SiPMs, the 'breakdown voltage' is sort of like the Early voltage for a BJT: it's defined as the point where the extrapolated gain vs. bias voltage line predicts a gain of 1.0, and you're expected to run the device a few volts above there.

That datasheet arbitrarily defines 'breakdown' as the voltage where the reverse leakage is 1 uA, but doesn't quote an abs max voltage. The leakage rises very gently with bias, so the avalanche voltage is obviously quite a bit above the 25V range of the graphs.

Cheers

Phil Hobbs

I wouldn't mind running that one at maybe 20 volts (I have 24 handy) to keep the speed up, but my customer wouldn't let me. The data sheet could have recommended operating conditions, but they usually don't.

Leakage doesn't bother me because we can AC-couple and capture short bright light pulses, and auto-zero between pulses when we know it's dark.

This is not a 15 nm transmission peak, but the absorption peak (edge 100 eV) of a thin "dead" insensitive layer of the detector.

Huh, OK thanks. I guess one needs to know the expected background rate to know if it's (the spectral feature) a bump or a dip.

George H.

I wonder if this sort of windowless photodiode can detect single photons or particles. It must be close.

I could try the similar Hamamatsu part if I can get a source.

You mean like a single x-ray? Let's guess... a 1 keV Xray would make ~1,000 electrons in Si if the QE was near perfect. 1k electrons in 1 ns (Response time of PD) is... ~1.6 x 10^-7 amps... into 50 ohms that will be hard to see. maybe you can build a 1k ohm FB TIA?

You're still groveling for 100 uV at best.

There may be big mistakes in the following story. When I was at FEL a group was trying to make a tunable X-ray source. (don't ask for any details.) They had an impossible time seeing any signal because of the background electron flux. A break through came when Marcus figured out that you could block the electrons and see the x-rays* by putting the right thickness of lead in front of the photodiode detector.

George H.

At modest light levels, in the infrared, we see shot noise in the photodiode current, so I'd guess we might see single hits at 1000x the photon or particle energy. That's my best guess without doing math.

I could whack one with a glow-in-the-dark alpha source or a Coleman mantle or something.

I'd have thought something like beryllium for the shield--higher Z gives faster X-ray attenuation in general, doesn't it?

Re: seeing 1000 electrons in 1 ns

It really depends on the photodiode capacitance. A pHEMT bootstrap will have an output impedance of a few ohms and a noise floor of about 0.3 nV/sqrt(Hz) in the flatband.

To see a 160 nA signal in a 400 MHz bandwidth with 20 dB SNR (which is about what you want to keep the false count rate reasonable), the total noise current has to be less than 16 nA in that BW.

For a pure e_N*C situation, the noise in a bandwidth B is

I_N(B) = 2 * pi * e_N * C * sqrt(B**3 / 3).

Solving for the allowable C for a given I_N and e_N, this is

C < I_N /(2 *pi * e_N * sqrt(B**3/3))

If we allow I_N to be 16 nA, then with e_N ~ 0.3 nV/sqrt(Hz),

C < 21 pF.

I've built biochip preamps that can detect 60 electrons in 2 ns, but their input capacitances were much lower, below 1 pF.

Cheers

Phil Hobbs

I'd have thought something like beryllium for the shield--higher Z gives faster X-ray attenuation in general, doesn't it?

Re: seeing 1000 electrons in 1 ns

It really depends on the photodiode capacitance. A pHEMT bootstrap will have an output impedance of a few ohms and a noise floor of about 0.3 nV/sqrt(Hz) in the flatband.

To see a 160 nA signal in a 400 MHz bandwidth with 20 dB SNR (which is about what you want to keep the false count rate reasonable), the total noise current has to be less than 16 nA in that BW.

For a pure e_N*C situation, the noise in a bandwidth B is

I_N(B) = 2 * pi * e_N * C * sqrt(B**3 / 3).

Solving for the allowable C for a given I_N and e_N, this is

C < I_N /(2 *pi * e_N * sqrt(B**3/3))

If we allow I_N to be 16 nA, then with e_N ~ 0.3 nV/sqrt(Hz),

C < 21 pF.

I've built biochip preamps with noise floors of 10 electrons in 5 ns, but their input capacitances were much lower, below 1 pF.

Cheers

Phil Hobbs

Probably the inner electrons, not the valence band, will do the absorptioin at that energy. See here:

that spike on the left, around 2 keV, is the 1S shell; similar spikes at lower energy tell the story almost all the X-ray absorption is due to the inner electrons. One rule of thumb is that you get some charge multiplication, like 1keV/30eV or so, before the decay cascade is complete, but that's on a per-photon basis, so 1 keV in gets closer to 30 electrons than 1000.

Yeah Beryllium is great for letting electrons through. In this case I think they wanted to let the X-rays through, and stop the electrons. But I could have it backwards.

I'll have to look that over.

George H.

Higher energies are easy to detect with a cheap silicon photodiode:

Fun, thanks. GH