Noise and gain in transimpedance amplifiers

In cases where you can make an ensemble of multiple instances of the same random function, sure. The classical example is using a bunch of noisy MOSFETs to measure a quiet resistor, by (as you say) cross-correlating their outputs pairwise and averaging the result. MOSFETs don't have any low frequency current noise to speak of, so the measurements are independent. Given enough time, you can make a good measurement measure maybe 20 dB below the FET noise floor.

However, the case under discussion is a single instance, where there's no way to distinguish signal from noise once they're mixed together. (Of course if I'm wrong about that, and you find a good general method, you can make your fortune.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs
Loading thread data ...

It *is* inconsistent. the "e_N*C_d" noise predicts that e_out grows linear ly with frequency. e_N*A_VCL predicts that e_out is flat to second order i n frequency.

Also, what "white noise floor"? There are not two separate noise sources h ere. Just e_N.

No offense taken, but I'd actually argue the opposite. To me, the "physics of the problem" means finding the root cause (like the shape of Avcl) that explains a bunch of individual "effects" (like e_N*C_d noise). At it's co re, there's just an input noise voltage and a voltage gain. It may be more handy and quick to keep the term "e_N*C_d noise" in your pocket, but it is not the "physics of the problem". e_N*C_d noise doesn't extend over an ar bitrary range of frequencies. It's a label to place on the shape of Avcl b etween fz and fp.

I couldn't agree more. But e_N*C_d noise is simply not fundamental to the problem.

It's like saying that there are "hyperbolic" and "elliptical" effects in gr avitational orbits, when really there's just *gravity*, which in a two body system can produce either hyperbolic or elliptical orbits, depending on th e total energy of the particle. In this op-amp problem, I'd say that there 's just e_N and a voltage gain Avcl, and depending on your frequency, the e quivalent input current is either flat or rising with frequency.

Reply to
jbattat

cs of the problem" means finding the root cause (like the shape of Avcl) th at explains a bunch of individual "effects" (like e_N*C_d noise). At it's core, there's just an input noise voltage and a voltage gain. It may be mo re handy and quick to keep the term "e_N*C_d noise" in your pocket, but it is not the "physics of the problem". e_N*C_d noise doesn't extend over an arbitrary range of frequencies. It's a label to place on the shape of Avcl between fz and fp.

How about this -- AoE explains that the "e_N*C_d noise" rolls off above fp. This is exactly the frequency that Avcl rolls off. All I'm claiming is t hat as you go down in frequency, the e_N*C_d noise also rolls off below fz. Does my phrasing it this way help?

Reply to
jbattat

On Thursday, December 10, 2015 at 10:24:58 PM UTC-5, snipped-for-privacy@gmail.com wrote :

sics of the problem" means finding the root cause (like the shape of Avcl) that explains a bunch of individual "effects" (like e_N*C_d noise). At it' s core, there's just an input noise voltage and a voltage gain. It may be more handy and quick to keep the term "e_N*C_d noise" in your pocket, but i t is not the "physics of the problem". e_N*C_d noise doesn't extend over a n arbitrary range of frequencies. It's a label to place on the shape of Av cl between fz and fp.

p. This is exactly the frequency that Avcl rolls off. All I'm claiming is that as you go down in frequency, the e_N*C_d noise also rolls off below f z. Does my phrasing it this way help?

Oh, and the reason that the e_N*C_d noise rolls off below fz is because Avc l is flat below fz.

Reply to
jbattat

No, it isn't inconsistent. The e_N*C_d contribution is one of several. It isn't dominant at low frequency, but it often is at high frequency.

There are a whole bunch of sources. Johnson, shot, e_N, i_N, what you had for breakfast....

Have it your way. Clearly I'm not going to be able to help much further.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

I've only done bootstraps w/ opamps. pHEMTs are even

OK I guess that's my case. I've got enough photons at some BW such that the shot noise is above resistor noise (>50 mV of sig.) I mean there is always a resistor somewhere. (For us mere mortals. :^)

Of course when you're making a measurement you typically don't want to throw away BW.

Sure, But many of us design in the simpler world, your bread and butter is my bell and whistle.

George H.

Reply to
George Herold

But that's just it: in the model I'm talking about, there's only e_N. I'm isolating a single noise term for study. There's no Johnson, no shot, no scrambled eggs. So unless e_N can give rise to two separate noise currents , then there's an inconsistency.

Quite the contrary, you've been very helpful! I appreciate your taking the time on this thread to help me out. I can see that I've worn out my welco me -- let me think more about this silently and perhaps bug other folks els ewhere.

Many thanks and all the best, James

Reply to
jbattat

ns except e_NAmp, and to show that there's an inconsistency in the e_N-C_d current argument.

o signal in this example).

, then the output voltage grows with frequency (since Zm is flat at low fre quency). But that cannot be right since e_output = Avcl * e_NAmp and the two terms on the right hand side are white, so e_output must also be white .

o don't exist in this example.

Huh... no you need to go through the math, there are two terms that give rise to the noise peak. Cin and the pole in the opamp gain roll off (It's not too hard to measure (the noise peak) if you build a TIA.)

Put white noise into a high Q LC and you get a big noise peak on the output .

George H.

Reply to
George Herold

OK, here's one specific question that would really help me understand the e_N*C_d current better.

The e_N-C_d noise current grows linearly with frequency.

At low frequencies the noise gain (Avcl) is flat to second order in frequency, which suggests no current flow (to first order in frequency) in the feedback network.

So what sources the e_N*C_d current? The op-amp inverting terminal?

Reply to
jbattat

OK try this, Forget about the noise current and noise voltage. Imagine you've got a non-inverting amp. R feedback, but the impedance from the inverting input to ground is a capacitor. What's the voltage gain for signals at the non-inverting input?

Then the noise voltage is just like an applied voltage.

The noise current thing (e_N*C_d) is nice once you've done a few TIA's cause it's an easy way to estimate that piece of the noise.

George H.

Reply to
George Herold

There are some useful references available online originally from the Burr-Brown Applications Handbook (my 1994 copy). They are now also available from the TI website (TI.com). There are three that may be of use: Noise analysis of FET transimpedance amplifiers: AB-076, or TI sboa060 Photodiode monitoring with opamps:AB-075, or TI sboa035 Compensate transimpedance amplifiers intuitively: AB-050, or TI sboa055a

Hope these may be of some assistance. Scott.

Reply to
Prof78

OK, I've sorted out my error in thinking.

I had in my mind that: e_out = i*Rf But of course that's wrong. The output voltage has a constant term as well : e_out = e_N + i*Rf

Indeed, there's no inconsistency between an "e_N*C_in" noise current that g rows in proportion to frequency, with a noise gain (Avcl) that is flat at l ow frequency (to second order in frequency). At low frequencies, i*Rf term is small perturbation compared to e_N, so the noise gain is flat, even tho ugh i (and therefore i*Rf) grows proportional to frequency.

Said another way: As frequency increases from DC, the output voltage rises from e_N to (e_N + epsilon) where epsilon

Reply to
jbattat

Another way of looking at it is that the rising i_N contribution gets flattened out by the Rf*Cf rolloff.

The reason the noise gain is flat to first order (not to second order) at low frequency is that the linearly growing C_d term is in quadrature with the constant term.

|1 + j omega C_d | = sqrt(1 + (omega C_D)**2) ~ 1 + 0.5*(omega C_D)**2

  • h. o. t.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Agreed.

I may have used a term incorrectly. When I say "flat to second order" I meant that the noise gain is flat (at unity) unless you consider terms that are second order (and higher) in frequency (as in your expansion above).

By the way, in your first post, you alluded to a 3rd edition of your text. Is the timeline for that public/known yet?

Reply to
jbattat

Nope. Probably 2017 sometime--I'm down to a couple of hundred FIXMEs, but t here's some condensing to be done so that it doesn't wind up being a phone book.

I've been scribbling things in there pretty continually since 1994.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.