Photodiode Question

People have been doing that sort of sampling for 25-30 years--first using coplanar transmission line on an electrooptic crystal like LiNbO3, where you look at the polarization shift in the light, and somewhat later using the plasma-optical effect, which can be either an optical-readout sampler or a switch. Dan Grischkowsky and Jean-Marc Halbout were doing it when I joined IBM, and before that, Dave Bloom was doing it at Stanford (he might have started it at Bell, I forget).

Using a femtosecond laser (a Hall-Haensch comb generator) will get you unmatched stability, but of course the experiment has to be synched to the laser rather than the other way round.

Telecom modulators are getting up above 50 GHz bandwidth, but only in the 1.5 um band, which is too long for GaAs.

Cheers

Phil Hobbs

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Hmm, Yeah this is a bit confusing. The photons make the charge carriers and reduce the resistance... and as Phil H. says there is generation and recombination noise. But this should only depend on the light level and not on how the device is biased. For a given light level if I double the bias voltage (and thus double the current) does the noise go up?

I'm using 100M and 1 G resistors from Ohmite to make a 10nA current source. I'll look at the noise next chance I get.

George H.

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sorry for the VERY naive question, but would some form of junction leakage have lower noise? Or are junction leakages at and above johnson?

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Thanks Phil, I must admit I don't quite 'see' how that works out. If I 'reuse' a photo generated charge carrier say ten times I get ten times the shot noise? (OK twenty times since I get twice the shot noise to begin with.) I'll have to think about that. The bias current divided by the photo current is something like the gain of the LDR.

Say (speaking of noise) I think I remember you mumbling sometime ago that you had a nice derivation of the Johnson noise formula. Something that didn't involve the counting of modes in an infinite transmission line as Nyquist first derived it. Have you written this up anywhere?

George H.

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Not naive at all! Phil is the expert here, but I'll answer anyway... (put my foot in mouth once again.) So, as I understand it, the thermal (Johnson) noise from a biased junction is 1/2 of the thermal noise from a resistor that has the same resistance as the junction. (R =3D kT/(eV * I). I beleive that Phil has actully used this 'trick' to make lower noise photodiode front ends... but it's a trick that is beyond my ability.

George H.

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What's amazing to me is that there are resistors that *don't* have shot noise.

Any time you have a current in which the electrons arrive randomly, there's shot noise. It's pure raindrops-on-the-roof statistics. That happens when the electrons are knocked loose by photons or when some thin barrier, like a P-N junction, doles out electrons across a surface. Metal wire is unique in having electron interactions that smooth out the flow. As far as I know, all semiconductor junction currents and leakages have shot noise. Tubes have shot noise.

The shot noise current depends only on the average current; it goes up as the square root of I. Of course, some devices have more noise than pure shot noise.

I'd appreciate that. We'll share whatever we learn. This is not very easy to measure.

John

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The electron arrival times are correlated by electron-electron scattering, which tends to smooth the flow out. Rolf Landauer (a late IBM Fellow whom I knew very slightly) showed that the shot noise in a metal resistor is reduced by a factor of Ls/L, where L is the length of the element and L_s is the mean free path for electron-electron scattering. In a typical device, that's 10 nm/1 mm, or 10**-5.

Cheers

Phil Hobbs

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It isn't original--it's just classical equipartition of energy plus the linearity of resistors and capacitors.

Consider a parallel RC circuit, isolated from everything else, and at a temperature T. Because it is a single classical degree of freedom, the energy stored in a capacitor has an RMS value of kT/2, which leaks away through the resistor with a time constant of t=RC.

In order for this to be statistically stationary (which thermal equilibrium always is), the rms power supplied by the resistor to the capacitor must be the same as the rms power dissipated in the resistor due to the voltage that's already on there.

Thus (1/2)*CV_n**2 = kT/2, so V_n**2 = kT/C.

Because the resistor is linear, we can consider the dissipation current (draining off the kT/C voltage) and the fluctuation current separately. (Key step.) The current in the resistor that is dissipating the capacitor's energy is

I_diss**2 = V_n**2/R**2 = (kT/C)/R**2

The bandwidth of this current is the noise bandwidth of the RC, which is

Since this is in thermal equilibrium, I_n**2 == I_diss**2, so the 1-Hz noise is

i_n = sqrt(kT/(R**2*C)*4RC) = sqrt(4kT/R), which is the classical Johnson noise formula.

Cheers

Phil Hobbs

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So, to paraphrase that: Shot noise is due to the carrier starting or stopping. The longer the device (greater transit time compared to recombination time) the more photons (or energy from an accelerating field as in a PMT) it takes to kick it the length of the soccer field (sorry about that analogy). But your current depends on the number of cariers that make it through the goal at the end.

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Yes, except that the recombination time can be much longer than the transit time--i.e. the same carriers can make the trip many times. Thus you can have quite large photoconductive gains, as in CdS, where it can be several thousand iirc.

There's a 1:1 tradeoff between gain and speed, because the current builds up and dies away on the scale of the carrier lifetime rather than the transit time. That's one reason CdS is so slow.

Cheers

Phil Hobbs

Cheers

Phil Hobbs

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Yup that's the argument that I know. I had several discussions about the lack of shot noise in resistors at the APS trade show. Someone said there was no shot noise in resistors because it would violate the Fluctuation-Dissipation Theorem. I don't know the F-D theorem well enough to dissupute the statment.

George H.

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Thank you for your reply.

Next naive question is Can the leakage be provided by a DC/DC converter circuit where the supply's noise is caused by caps and inductors, oh wait, no noise there. But I'm talking a physically realizable circuit.

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Excellent, Thanks Phil!!! All I need to convince myself of, is that the bandwidth of the RC is 1/4RC, but that should be easy.

George H.

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There was a big foofaraw 15 years or so back where some guy claimed to have figured out how to make an active circuit that functioned as a noiseless resistor. It was all over IEEE Spectrum and places like that.

There are various means to take quiet active things, e.g. the beta of a BJT or the g_m of a good JFET, and apply feedback so as to make the equivalent of a very quiet resistor. I recently designed a TIA that's shot noise limited down to the low tens of nanoamps in a 1-MHz bandwidth, using techniques like that. It's about 20 dB better than I thought I could do, which was a very pleasant surprise. You just have to get rid of the 300 kelvin resistors.

(John L. and I collaborated on it, along with one of his guys, Jonathan Dufour--you'll be able to buy them soon, if all goes well. Buy lots--I'll have two kids in college this fall, and John's ski place needs a new laboratory.) ;)

Cheers

Phil Hobbs

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It's pretty simple: the squared modulus of the transfer function is

|H(f)|^2 = 1/(1+(2*pi*f*R*C)**2), and the noise power bandwidth is

infinity BW = integral 2*|H(f))|**2 df 0

(The factor of 2 is for the analytic signal--otherwise you have to use the two-sided integral.)

Substituting tan u = 2 pi f R C, and using the identity 1+tan**2 u = sec**2 u, the integral becomes

pi/2 BW = 1/(pi*RC) integral cos**2(u) 0

Now cos**2(u) = 1/2 + cos(2u)/2, and the integral of cos(2u) from 0 to pi/2 is 0, so

BW = 1/(pi*RC) (pi/4) = 1/(4RC), i.e. pi/2 times the 3 dB bandwidth.

Cheers

Phil Hobbs

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Are you talking about active feedback? (Maybe those aren't the right words?) You send back some of the signal (from a system) to damp the response and this looks like a resistance to the system. But it=92s noiseless.

Did some physicists do this back in the 40's - 50's? Purcell? Do you have any references?

George H.

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That's great! The pi/2 really speaks to me since the two pole Butterworth filters I'm using have a 1.111 effective bandwidth.

George H.

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You can make a room-temperature amplifier that has very low noise temperature... I think some gaasfet-based microwave amps get down to something like 50K noise temps. And you can use such an amp to simulate a resistor well below room temp. It doesn't violate the laws of thermodynamics because the amp needs power to do its thing.

If you connect two resistors with wires, and they're at different temperatures, the Johnson noise difference becomes a thermal conductivity between them... the hotter one makes more noise, and pumps power into the colder one. I did the math once... it's something like 10 orders of magnitude weaker coupling than the thermal conductivity of any conceivable wires that you'd use to connect them.

John

Try looking for the patents on Faraday detector amplifier calibration sources. They are typically around 0.1nA with fairly low noise and very high reproducibility. Not sure how much of it is trade secret.

ISTR the resistors for this job are effectively single sourced and in the range 10^9 to 10^11 and some kind of doped conductive glass. They are low noise (at least for low frequencies) but also behave a little bit like they have distributed internal RC and battery behaviour which requires external compensation.

Regards, Martin Brown