Capacitor-feedback for low noise

See Knoll's classic book, Radiation Detection and Measurement, for some ideas.

I think the ultimate discharge device is supposed to be a photosensitive jfet or something. Knoll has an enormous number of references at the end of each chapter. Some people just use a very high-value resistor across a big feedback cap, which is continuous (no dead time) and works fine in some situations.

APDs are high-voltage devices, probably not appropriate here. NEC has some fairly low-capacitance photomos SSRs that might be interesting.

John

We show in our book one way to solve the discharge-FET capacitance problem (how to avoid interfering with the 1.0pF feedback cap).

P 1039?

There's also the old trick of collapsing the opamp power supplies briefly, letting the esd diodes discharge the cap.

John

The idea is that, for very small zots of charge, like from a low-energy nuclear detector, you drive an opamp with just a cap for feedback; this gives you low-noise gain without Johnson noise. You differentiate the signal in later gain stages to get the spikes back. But eventually the input charge piles up on the cap and saturates the amp, so you have to time out data acquisition long enough to reset it, then run for a while longer. Many clever variants exist in the nuclear biz.

John

Hello Zigoteau,

I am not sure that I understand what you are doing and what DC level the cap will be at. Can you use FETs to discharge? You'll probably have to use a balanced scheme where the charge injection cancels out and any residual imbalance can be trimmed away.

If on a high DC potential a FET can be driven via a little ferrite toroid signal transformer.

Regards, Joerg

Dear John, Joerg and Winfield,

Many thanks for all your suggestions.

John and Joerg, no, I don't think I can use a JFET to discharge, because I will then be stuck with the reverse leakage current of the gate-channel junction, which is typically a few nA. The noise spectral density that I'm trying to reach is something like 1e-34 A^2/Hz, which is equal to the shot noise from a few tens of fA. That's why a photodiode has to be an avalanche type, because the usual run of PIN photodiodes have reverse leakage currents in the nA range.

John, how does the idea of putting the op amp supply to zero for a short period work? You've got to get the charge out of the end connected to the summing junction as well.

A resistor across the capacitor to discharge it continuously is no good because resistors are noisy. The noise level I'm trying to reach is equal to the Johnson noise from more than a teraohm. I am quite au fait with resistor feedback, and if you've got to have a resistor there, then an additional capacitor is a waste of time (in any case high value resistors have significant parasitic capacitance that you have to compensate for somewhere).

Winfield, could you give me details of your book? Any clues about the configuration you describe there?

Best regards,

Zigoteau.

That's wrong. Take some measurements on small parts. And even if it was right, which it isn't, there's a simple solution, as we show.

Using a fA opamp, reverse it, through 10k resistors. Uses the matched protection diodes to quickly do the job. A very nice elegant trick I've written about here on s.e.d. several times.

I don't see the problem. It's not necessary to end up with exactly 0V on the integrating capacitor, if that's what you're implying. It's only necessary to reset the differentiator capacitor to the same value, which is easy and automatic. But watch out for dielectric absorption. That's an issue.

Hi, Win,

Thanks for the tip, which I will check out.

I take it you are referring to Paul Horowitz and Winfield Hill, The Art of Electronics, Cambridge University Press (1989)?

Ah, are you talking about the protection diodes at the op amp input? OK, I'm with you. And I guess that the protection diodes of a fA op amp are guaranteed to have a reverse leakage of that order of magnitude.

Reversing the op amp power supply sounds dangerous, even with 10K resistors. Is that covered in your book? Do you use decoupling capacitors for the op amp supply?

Many thanks,

Ziggy.

That would be very low, e.g., e_n for 0.1 ohm is 0.04nV

Hi, Jeroen,

This sounds very interesting. Do you have a reference? Such an amplifier would not be DC coupled - would there be stability issues at either the upper or lower roll-off points? I will think about this.

The signal source is the current through a single protein molecule bound to a bilayer lipid membrane immersed in an electrolyte. The current flows through the protein molecule as ions, which are traded for electrons in a metallic conductor at two electrodes. Typical currents are of the order of pA, and switch between a number of different values as the protein switches between different conformations. The aim is to get as high a bandwidth as possible while still being able to see the molecular transitions.

Cheers,

Ziggy.

Hi, Jim,

Capacitors have no noise because there is no dissipation. Jeroen is right that inductors also have no noise.

If you want to calculate the noise you get from an arbitrary circuit, then you need a model for the noise behavior. The thermal noise of an impedance Z(f) can be modeled by a Thevenin equivalent circuit, where the voltage source in series with Z(f) is random with a spectral density of 4kTRe(Z(f)) V2/Hz. Equivalently, its thermal noise can be modeled by a Norton equivalent circuit, where the current source in parallel is random with a spectral density of 4kTRe(1/Z(f)) A2/Hz.

Cheers,

Zigoteau.

Nothing wrong with using 1 to 10T-ohms, easy, wire the sj end floating in air, or on a Teflon standoff. Sources? Willow has resistors to 100T,

Actually, a resistor makes plenty of sense. I often mention what I call the R-C-R trick to remove the effect of resistor self-capacitance. Alternately, I use a two-path scheme, with an RC to insure only a low-frequency path through the high- value resistor, and a precise capacitor feedback path for the high frequencies. Followed by an appropriate opamp network to bring everything back to flat response. The R-C crossover can be precisely set, say in the 10Hz region, with one adjustment. This approach avoids the problem of capacitance to ground with long high-value resistors, which I've written about here before.

Wild idea:

Run the front-end as an integrator, add downstream gain maybe, and digitize that, without any attempt to differentiate. Oversample a lot so you can do digital filtering tricks. When the system gets close to saturation, let the software kick in a weak current source to ramp back towards ground, or wherever you like the start point to be. Since the software controls everything and knows (or can calculate from observation) the down-slope, it can fix the data appropriately.

No timeouts. All we need now is a very low-noise switchable current source!

John

That's because most of us rarely have to process femtowatt signals at baseband. And because uncooled high-Q-tuned amps *can* have very low noise temps. And there's the 1/f thing, too.

John

There's still the sqrt(kT/C) reset voltage uncertainty to worry about--but correlated double sampling fixes that.

It's sort of interesting the habits most of us have--at low frequency, we're often tempted to look at the noise of a 300K resistor as an unavoidable cost of doing business at room temperature, whereas at RF, we'd be grousing about its indifferent 3 dB noise figure. The physics is no different in the two cases.

Cheers,

Phil Hobbs

Yes, the physics behind it is summarized in the fluctuation-dissipation theorem of statistical mechanics, which says that any mechanism that can dissipate energy has associated fluctuations at finite temperature. If this weren't so, you could make heat flow spontaneously from cold to hot.

The usual way to derive the Johnson noise formula for a resistor is to use classical equipartition of energy, which predicts that any single degree of freedom, e.g. the charge on a capacitor, has an RMS energy of kT/2. Classical equipartition is a very general consequence of statistical mechanics, and even in a quantum treatment, it can be shown to hold for frequencies

On 22 Aug 2005 12:52:17 -0700, "Zigoteau" wroth:

I've followed this thread with interest. Not because I think I may have something to add, but because I'm learning stuff.

That "throw away" line about caps not having thermal noise interests me. Can someone put some more words around that concept? What are the fundamental properties of caps that make them so different from every other component? Would there not be an equivalent thermal noise voltage developed across a cap's equivalent series resistance?

A successful explanation would also include transformers and inductors.

Jim

Did you consider transformer feedback? That doesn't generate any additional noise either and it affords a flat frequency response. It rids you of the problem of discharging the feedback cap and of the differentiator.

What's the application?

Jeroen Belleman

Thermal noise only comes about where there's an exchange between heat and electrical energy. So resistors have noise, but the AC impedance of coils and capacitors do not. The equivalent *loss resistance* of those components *does*, make noise, of course.

Also, it's possible to make room-temperature circuits with a real resistive impedance that nevertheless produce less thermal noise than a room temperature resistance of the same value.

Jeroen Belleman