Delta-Sigma noise and bandwidth

The noise of the RC with a timeconst T is exactly equal to the noise of R averaged over time T. Is this the answer to your question?

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

Peter Dickers> I have a sensor that produces tiny currents which are converted to volts for

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Actually, the noise voltage of a parallel RC circuit (where the RC shapes the bandwidth) is: k*T/C. Yes, it does not depend on R. This may seem counter- intuitive at first, but consider: if R is increased, the noise voltage increases, but the bandwidth decreases.

-f

The bandwidth you mention for an rc filter is "correct". It is also the 3 db point. The 6 db point comes at twice that frequency, barring memory failure. The 9 db attentuation point comes at an even higher frequency and noise can be found at all of them. To calculate the noise "bandwidth" the transfer function need be summed, or integrated, over the entire frequency range not just to the 3 db point.

Hul

Peter Dickers> I have a sensor that produces tiny currents which are converted to volts for

Yes, except I don't think that it is so (I get a factor of two difference).

Peter

OK, perhaps I need to explain the calculations so that someone can put me right. I avoid putting equations into the original message because it puts people off.

Firstly, noise voltage doesn't sum because the components are random. The noise power (or volts^2) does. Noise power is 4kT per Hz of bandwidth i.e. the noise desity is constant with frequency. So for a perfect low-pass filter that has unity gain below some frequency f and 0 gain above then bandwidth is f and the noise power is

4kTf. The noise voltage is then V = sqrt(P.R) = sqrt(4kTfR).

Now, if we have a simple RC filter instead then the gain (in power terms) is

1/(1+(2pi.f.RC)^2) at frequency f (it would be the sqrt for votage gain). So the noise power is now 4kT integral 1/(1+(2pi.f.RC)^2) df (0 to infinity) which is 4kT.1/(2pi.RC) integral 1/(1+x^2) dx the final integral is a standard preschool one = pi/2 so we get 4kT. 1/(4RC). This tells me that the noise bandwidth is 1/(4RC), not 1/(2pi.RC) as I'd expect.

Then for a filter that averages over time interval t the gain (again in power terms) is (sin(pi.t.f)/(pi.t.f))^2. Note that that's pi not 2pi. So the noise power is now

4kT integral (sin(pi.t.f)/pi.t.f)^2 df (0 to infinity) which is 4kT.1/(pi.t) integral (sin(x)/x)^2 dx the final integral is slightly beyond preschool but is again pi/2 so we get 4kT.1/(2t). This says the noise bandwidth is 1/(2t), which is twice that for an RC!

This is totally counter intuative to me. So, clearly I have done something wrong.

Peter (the top posting is making it difficult to read, so I apologise for that).

A simple RC filter would have about 7.5 dB attenuation at 2f and about

12 dB at 4f and then continuing at -6 dB/octave.

Where did you get that "4" from. In my understanding the noise power would be simply kTB (not 4 kTB) or at room temperature -174 dBm/Hz.

In modern VHF/UHF RF-amplifier design, the noise figure would be well below 3 dB noise figure, i.e. at 3 dB NF the amplifier additional equivalent input noise power would be equivalent to the source (antenna pointing at 300 K ground) noise power of kTB.

At lower frequencies, the 1/f noise would be dominant.

Paul

Peter with "noise voltage doesn't sum" you have banned numerous decades of noise specifications. Look at the typical spec for input refered noise of op amps, for example.

Hul

Peter Dickers> > The bandwidth you mention for an rc filter is "correct". It is also the 3

I assume that you have started your calculations from originally calculating the noise voltage over an _open_circuit_ (unloaded) resistor.

In weak signal amplification system, you usually try to achieve a power matched condition, i.e. the amplifier input impedance is the same as the generator source impedance to transfer maximum of the signal power to the amplifier. The generator output signal voltage, as well as noise voltage is 1/2 of the open circuit values. Thus, the noise power transferred from the generator to the amplifier should be kTB, not 4kTB (B=bandwidth).

Paul

There are more than one flavor of bandwidth. The 1/(2 pi R C) bandwidth is the 1/2 power bandwidth; the bandwidth that you use for noise calculations is the noise bandwidth.

The noise bandwidth of a filter is the bandwidth that a perfect "boxcar" filter would have to result in the same response to white noise. It's equal to the integral of the square of the filter response over frequency; in the case of a simple 1st-order lowpass filter the noise bandwidth is equal to pi/2 times the 3dB bandwidth. So in your RC filter case it's your 1/(4RC).

If you're willing to do a lot of weird and questionable calculus you'll find that the noise bandwidth of a sinc filter is equal to frequency of the first null. Or if you use Parseval's theorem on the Fourier transform of the sinc filter you'll find out the same thing, but without having as much fun.

HTH

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Yes, that's a different problem that I have i.e. how come the noise voltage has sqrt(4) in it relative to the noise power. Ultimately I am only interested in the noise voltage into an ADC so the factor of 4 in the power doesn't change that.

Obviously the physics is the same for RF as for my stuff but the numbers are very different.

I'm looking at quiet readings at 1/sec and noisier up to 100/sec. The resistor values are typcally 200 Mohm but with a good source we're hoping to get down to 20 meg. If we are 'lucky' we'll have 100nA full scale. Ideally I need to read 4 decades down from there at better than 0.08% noise (actually two readings that track to better than 0.12%) - no its not many electrons.

Peter

Nope. This is one bit that I am sure I understand.

Peter

Obviously if the resistor is completely unloaded then this classical statistical mechanics calculation breaks down due to so-called ultraviolet divergences i.e. infinite noise power due to infinite bandwidth. In that case I have to think of the resistor as a black body radiator but I'm sure doesn't matter until IR frequencies.

Thanks, Paul, yes that explains the factor of 4 between power and volts. The transfered power is less than the 'radiated' power. In practice I am measuring current into an effective short (virtual earth) so power, here, is just a notational convenience to explain why I must sum squared noise volts or current.

Peter

Thaks. I feels good to be right for a change.

I don't see any questionable calculus involved. I even did a quick check numerically i.e. that integral sinc squared x is pi/2. The first null is at f=1/t. I got the noise bandwidth to be half of that even though 1/t seems more reasonable: integral [sin(pi.f.t)/(pi.f.t)]^2 df = 1/(pi.t)*integral [sin(x)/x]^2 dx =

1/(pi.t)*pi/2 = 1/(2t).

Yes this helps a lot.

Peter

Peter Dickerson escreveu: [snipped]

It is a lot of electrons, Peter. I wish the same number as dollars in my checking account!

Depends on you view of a lot. 100nA*1E-4*(0.08/100)/e is about 50000 electrons per second. So that's about 500 per reading. And that is assuming we get 100nA. Previous products suggest that 30nA is more realistic.

Peter

On Wed, 5 Nov 2008 08:01:55 -0800, Peter Dickerson wrote (in article ):

Just how low are the currents? Perhaps you need to look at the quantum mechanical limits of the electron statistics, not the Johnson noise.

I switch to QM and Ebers-Moll when I get down to pico and fempto amps.

-- Charlie Springer

I'm not sure what Ebers-Moll has to do with this. The noise is in a resistor, not a semiconductor junction. Of course Johnson noise is a QM phenomenon.

Peter

On Mon, 10 Nov 2008 00:29:10 -0800, Peter Dickerson wrote (in article ):

Just that about the same time I need to use the electron statistics instead of the average value from typical Johnson noise calculations, I also need a better junction model that is closer to real nonlinear behavior.

I have done jobs where sampling rates were impossible high, where in an IR diode temperature system statistical errors were 20 times greater than the specified error bands for the temperature output! In fact, this device is in all the AMAT supplied Intel Copy Exact RTP chambers. In a rapid temperature cycle the data is OK at the high end but miserable as it heats up or cools off. A cycle is only a couple of seconds and fortunately the accuracy of the curve at the cooler ends doesn't seem to make any difference -- or does it? The process control experts think the data is reliable so ???.

We are looking at a process that follows a temperature curve that is sometimes only 1.2 seconds long. How do you do it so fast? The lid of an RTP chamber is a honeycomb of gold platted chambers with halogen lamps in them that head a wafer from above. Temperature is read through a long sapphire rod from the bottom that looks at the underside of he wafer.

Here is the rub. In the lower half of the temperature curves the diode current is in the pico and fempto amp range. But sampling is at 100 to 1,000 Hz. Figure the variation.

-- Charlie Springer

Just replying to my original question to report the conclusions. Basically my calculations were sound but I didn't trust the answers. For an RC filter the relevent 'noise' bandwidth is 1/(4T), where T=RC. For a filter that simply integrates over intervals of time T the noise bandwidth is 1/(2T). So the natural assumption with 2*pi really is wrong. Some people said that the results should be the same, but its not so. A simple spreadsheet is enough to reassure me on that score. Some said that the noise bandwidth of a sinc filter is the frequency of the first null, but its twice that.

But lots of advice so thanks to everyone for help with the problem.

Peter