Ferrite bead model and real behaviour

Oh boy, can't wait to hear the answers. If the loss is linear (in amplitude) all the way down to zero, then I think there has to be noise. (mumbles something incoherent about the fluctuation/ dissipation theory) Otherwise I don't know.

Maybe you could measure it? How big is the "resistance" at 10 Mhz or something?

George H.

The resistive part of the impedance exhibits Johnson (thermal) noise, but the reactive part doesn't. That's on account of the fluctuation-dissipation theorem, as you say. (Dead useful, it is.)

Cheers

Phil Hobbs

Thank you for answer. It is not good for me, but I believe it is true.

I am looking for a way to reduce the "dark" noise of the photo-detector fro nt-end. The photo-sensor C together with L of leads and PCB traces makes a series resonant circuit. In order to damp it, we use a simple R today. The Johnson noise of this R, flowing at high frequencies through C, dominates t he total "dark" noise of the circuit across the whole bandwidth. I had a sm all hope that "may be" a bead will do a job without noise. So, there are st ill no "free lunch" in electronics, at least in this universe. Could you propose another way to damp it with less noise? Thank you

A better layout and some reverse bias would be my first choices. I can give more detailed help if you tell me:

What photodiode? How is it biased? What's the bandwidth? What's the minimum expected photocurrent?

Cheers

Phil Hobbs

At any frequancy, the Johnson noise must be the same as for a resistor that has the same ohmic value at that frequency. Conservation of Energy forces that.

The reactive component doesn't make noise.

What happens if the loss is amplitude dependent? I only ask 'cause I made a resonance LC circuit with a pot core and ceramic cap. At very low amplitude the Q was kinda amazing.. (up near 1000) but at higher amplitude it dropped to ~300-400. (a while ago I don't have the exact details.)

George H.

How long are the leads? What size resistor is it?

George H.

Depends on the physics. For instance, a BJT's emitter current has shot noise but no Johnson noise--it behaves like a resistor of value 1/g_M

The Johnson noise formula assumes thermal equilibrium, which means zero bias among other things. If the resistance is linear, it produces the same noise at all bias currents, but not otherwise, in general.

The BJT's noise is half what you'd expect because the forward bias suppresses the reverse diffusion current. At zero bias, the two diffusion currents are equal, so you get the full Johnson noise. (The derivation of that is pretty slick, and only takes a couple of lines of algebra.)

Cheers

Phil Hobbs

An FDR isn't a very good model, because it implies no phase shift, and imaginary properties (like non-causality). Of course, any real, physical, two-terminal "FDR" has to have reactance, so it's very easily approximated with a finite string of RLC units.

Most SPICE models you will find, are a single RLC, or RLC + DCR. This works okay over a few decades, but not usually to great accuracy over the entire specified range of the FB. Tweaking a model to match the published curves isn't hard to do.

Anyway, even such a simple model answers your question: yes, it adds noise, and only at the frequencies where the resistance component is significant. If the FB is chosen so the resistance peak occurs at frequencies outside your passband, then system noise is unaffected.

Another way to think of it: if you can replace your series resistor with an R || L, such that the F = R / (2*pi*L) cutoff frequency is above the top passband edge, while still achieving the desired damping, then you're set. Or to put it another way, you've constructed your own ferrite bead, but with more accurate components.

This can only work, if the series resonance you're trying to dampen, falls at a frequency above the passband. Which means the lead lengths must be kept short enough for this to be true.

Mind that real ferrite beads also saturate under DC current bias, and have interesting AC bias properties (the inductance rises for small (nonzero) amplitudes, then falls as amplitude approaches saturation). This isn't a problem for small signal applications, but precludes ferrite beads for high current filtering.

Tim

A bead is modeled similar to this...

...Jim Thompson

Phil,

Is the purpose of reverse bias to reduce the capacitance of the PD?

Clifford Heath.

Thanks. Does it reduce the sensitivity or change the response curve of the PD, since a photon now has to overcome a bigger barrier to get detected?

Clifford Heath.

Yup. In a PIN diode the reduction can be as much as 7-8x. Since the shot noise of the leakage is virtually never the limiting noise source, that huge factor comes pretty well free.

The main thing to remember is that the noise of the bias supply is as important as the noise of the TIA--a nanovolt on the supply side of the capacitor does exactly the same damage as a nanovolt on the TIA side. So careful filtering, capacitance multipliers, and so on are very much in order.

Cheers

Phil Hobbs

Oooh ooh! I can answer that one!

No. If anything, it'll increase the sensitivity (probably by a totally insignificant amount -- for that you'll have to ask Phil again). The photocurrent always goes "backwards" through the diode, so the bias actually helps speed it along.

I don't think it changes the response, but for that I'll have to defer to Phil.

The usual PD use case is in photo-conductive mode. It's rare to see a published circuit for one in photo-voltaic mode.

(I know there's at least one circuit in photo-voltaic mode in production, and moreover it's measuring the diode voltage and not its current. It's for a line-following robot toy, and the designer wanted it to work in a wide variety of illuminations. The response of a PD as a voltage source is roughly logarithmic, so the difference between two of them does a great job of measuring the proportional difference in the light striking each one -- which is exactly what we wanted. It worked over decades of ambient illumination. The guy who asked me to do it was very pleased that I not only solved his problem (he was trying to use photoconductive mode), but did it by taking out components. Muntz to the rescue!)

Nice, thanks! I guess if one photon dislodges one electron, and the capacitance is 1/7th, the voltage step will be 7x...?

Grumble.. 1) As close to the PD as possible, use damping bead(s) as a common mode choke; use as much variety of damping characteristics as possible. This (theoretically) should add no noise to the differential PD signal. 2) If necessary, use a ferrite type that has the highest Z starting at the lowest frequency of NON-interest; one on the signal line and one on the return line (to balance the differential loss); put a shunt capacitor between these lines both before and after those beads. Like a pi filter.

It is a start...

Err...PROVE there is such a thing as a "photon".

Or, better yet, the speed will be 7 times faster.

Hi, Everybody The conversation is going to be wild a bit. Thank you for almost immediate and informative replies. Pavel