Photodiode Question

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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.

Here:

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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"> I'd appreciate that. We'll share whatever we learn. This is not very

Hmm I hadn't thought it would be that hard... but then I'm probably missing something. I was going to take the the 10nA current source and fed it into a TIA opamp circuit with 100Meg as feed back resistor, (giving me a volt of DC across the resistor) And then see how noisy it is. I can then compare it to 10 nA's from a photodiode. Do you care about high frequencies (above 10kHz-100kHz or so..) or very low frequencies? (There's 1/f noise in the FET opamp that starts to interfere at the low end.)

Our noise apparatus is on a truck somewhere between here and Portland, it may be back tomorrow though.

George H.

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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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Fine, as long as that 100M resistor has no excess noise of its own, and you account for its Johnson noise, and the TIA has very little input current noise. And all the supplies are very quiet. And everything is really well shielded.

We're going to make a voltage divider from two identical RUTs, and AC couple that into an opamp with a gain of +1000. The AC coupling RC will add shunt Johnson noise but no excess noise of its own. We're figuring on a polystyrene cap and a 1G resistor maybe. Opamp = ADA4817. I figure we'd measure from 1 KHz to 100K maybe.

John

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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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If you use two FET op amps going into two scope channels, and use the math functions to multiply the two inputs, FFT the results, and average over N traces, you'll get just the noise of the resistors, because everything else cancels out. How big N is depends on how far below the amplifier noise you want to go.

Quiet power supplies are of course vital.

Sounds like a really interesting measurement.

Cheers

Phil Hobbs

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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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Hi John, The current source is just a 10 Volt reference (REF102) feeding the big resistor with an opamp on the bottom driving the 'ground' pin of the reference. The references has a bunch of noise so I'm not sure how quite the current source will be. We'll find out though.

The ADA4817 is that the screaming 1GHz FET? I was going to use the much slower opa134.

George H.

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Sounds like a lot of math. That would be good for extremes, like cryogenics maybe.

Why FFT? Wouldn't the averaged product work?

The ADA4817 has 4 nv/rthz noise, and the Johnson noise from 50M will be almost a microvolt per, so we should see the noise pretty well. Input noise current is low enough to not make trouble. It's an amazing opamp.

If cranking up the DC voltage into the divider doesn't much increase the noise, we're done. If we jam 20 volts into the 100M:100M divider, full shot noise would make ... calculates furiously ... 13 uV/rthz, a huge signal.

John

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Yeah I'll have to check that the 100M is OK..you can always switch down to 10 and 1Meg.... You can measure changes in noise down below the amplfier noise...at 10% of amp noise it's still pretty consistant. I think the 100M will dominate the input amp noise.

I told my boss that the capacitor multiplier was my latest hammer. I've been hitting every noise nail that I find with it. (Within Teachspin this circuit fragment is called a Hobbs filter, because I first learned of it from Phil's book.)

I've got a five sided steel box with an aluminum front panel. I sorta gave up on magnetic shielding and we just try and keep all the magnetic crap away from the front end. The power supply is a brick on a rope that can live far away. Then keep the 'scope on the far side of the lab bench.

George H.

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Cool, the second opamp is looking at the voltage source feeding the ladder?

George H.

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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.