On 12/14/2015 02:17 PM, jbattat@gmail.com wrote:
> 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 grows in proportion to frequency, with a noise gain (Avcl) that
> is flat at low 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 though 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 << e_N.   The epsilon
> term is indeed proportional to frequency (it's given by i_N*Rf where
> i_N is the current through C_in).  But epsilon << e_N for frequencies
> below fz=1/(2*pi*Rf*Cd), and so the noise gain stays approximately
> flat with frequency.  In this approximation, the noise gain is given
> by (e_out / e_N) ~ 1+f/fz.
> 
> For those interested, I've made some plots of the actual noise gain
> (Avcl = non-inverting closed-loop gain), along with this linear
> approximation: Avcl ~ 1+f/fz, to explore how well the linear
> approximation holds:
> 
> http://academics.wellesley.edu/Physics/jbattat/electronics/tia.html
> 
> The bottom line is that the linear approximation holds very well for
> frequencies below 1/(2*pi*Rf*Cf).  This agrees with the discussion in
> AoE (Section 8.11.4, p. 539) [1], with one possible wording
> clarification (see footnote below).
> 
> So I'm comfortable now with thinking of the op-amp voltage noise as
> creating a current through C_in that is proportional to frequency.  I
> see that this current is linear in frequency even well above
> 1/(2*pi*Rf*Cf).  And I see that the noise gain Avcl is well-modeled
> by the linear approximation (e_N+i*Rf) for frequencies below
> 1/(2*pi*Rf*Cf).
> 
> Many thanks for all of your input, especially the many replies from
> the ever-patient Phil, James
> 
> 
> [1] AoE discussing the e_N*Cin current: "It would continue to rise
> forever, except for the effect of the parallel capacitance Cf, which
> causes the noise current to flatten off at a frequency
> fc=1/(2*pi*Rf*Cf)"
> 
> 2 notes about this quote from AoE:
> 
> (1) their fc is not the same my fc in the linked webpage above.  I
> use fc to represent the unity-gain frequency of the op-amp.  They use
> fc for the TIA usable bandwidth.
> 
> (2) The wording in AoE implies that the e_N*Cin current levels off,
> but I find that the noise current continues to increase with
> frequency well above 1/(2*pi*Rf*Cf).  It's true that the noise *gain*
> levels off above this frequency because nearly all of the current
> goes through Cf, not Rf (see also Phil's first response in this
> thread, which I now understand more carefully than on my first
> reading!).  In other words, something does level off, but it's the
> noise gain, not the current through Cin.  Presumably the current
> through Cin should remain linear in frequency until the op-amp is no
> longer able to make the inverting input follow the non-inverting
> input.
> 

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

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 grows in proportion to frequency, with a noise gain (Avcl) that is flat at low 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 though 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 << e_N.   The epsilon term is indeed proportional to frequency (it's given by i_N*Rf where i_N is the current through C_in).  But epsilon << e_N for frequencies below fz=1/(2*pi*Rf*Cd), and so the noise gain stays approximately flat with frequency.  In this approximation, the noise gain is given by (e_out / e_N) ~ 1+f/fz.

For those interested, I've made some plots of the actual noise gain (Avcl = non-inverting closed-loop gain), along with this linear approximation: Avcl ~ 1+f/fz, to explore how well the linear approximation holds:

  http://academics.wellesley.edu/Physics/jbattat/electronics/tia.html

The bottom line is that the linear approximation holds very well for frequencies below 1/(2*pi*Rf*Cf).  This agrees with the discussion in AoE (Section 8.11.4, p. 539) [1], with one possible wording clarification (see footnote below).

So I'm comfortable now with thinking of the op-amp voltage noise as creating a current through C_in that is proportional to frequency.  I see that this current is linear in frequency even well above 1/(2*pi*Rf*Cf).  And I see that the noise gain Avcl is well-modeled by the linear approximation (e_N+i*Rf) for frequencies below 1/(2*pi*Rf*Cf).

Many thanks for all of your input, especially the many replies from the ever-patient Phil,
James


[1] AoE discussing the e_N*Cin current: "It would continue to rise forever, except for the effect of the parallel capacitance Cf, which causes the noise current to flatten off at a frequency fc=1/(2*pi*Rf*Cf)"

2 notes about this quote from AoE:  

(1) their fc is not the same my fc in the linked webpage above.  I use fc to represent the unity-gain frequency of the op-amp.  They use fc for the TIA usable bandwidth. 

(2) The wording in AoE implies that the e_N*Cin current levels off, but I find that the noise current continues to increase with frequency well above 1/(2*pi*Rf*Cf).  It's true that the noise *gain* levels off above this frequency because nearly all of the current goes through Cf, not Rf (see also Phil's first response in this thread, which I now understand more carefully than on my first reading!).  In other words, something does level off, but it's the noise gain, not the current through Cin.  Presumably the current through Cin should remain linear in frequency until the op-amp is no longer able to make the inverting input follow the non-inverting input.

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.

On Friday, December 11, 2015 at 10:58:35 AM UTC-5, jba...@gmail.com wrote:
> 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?

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.

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?

On Thursday, December 10, 2015 at 9:51:15 PM UTC-5, jba...@gmail.com wrote:
> > > Also, can you comment on the low-frequency hypothetical?
> > 
> > At low frequency, neglecting 1/f noise in the op amp, the e_N*C_d noise 
> > is swamped by white noise from the amplifier, feedback resistor, and 
> > (hopefully) shot noise.  It's still there, though.
> 
> But the point of this hypothetical is to strip away all noise contributions except e_NAmp, and to show that there's an inconsistency in the e_N-C_d current argument.
> 
> So let's ignore Johnson noise (Rf is noiseless) and shot noise (there's no signal in this example).
> 
> If there really is a noise current from e_N-C_d proportional to frequency, then the output voltage grows with frequency (since Zm is flat at low frequency).  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.
> 
> We can't bury this inconsistency in Johnson or shot noise.  The latter two 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.

> 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.
> >
> > Also, what "white noise floor"?  There are not two separate noise
> > sources here.  Just e_N.
> 
> There are a whole bunch of sources.  Johnson, shot, e_N, i_N, what you 
> had for breakfast....

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.

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

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 welcome -- let me think more about this silently and perhaps bug other folks elsewhere.

Many thanks and all the best,
James

On Thursday, December 10, 2015 at 9:49:15 PM UTC-5, Phil Hobbs wrote:
> On 12/10/2015 09:35 PM, George Herold wrote:
> > On Thursday, December 10, 2015 at 9:07:09 PM UTC-5, Phil Hobbs wrote:
> >> On 12/10/2015 09:04 PM, Phil Hobbs wrote:
> >>> On 12/10/2015 08:33 PM, jbattat@gmail.com wrote:
> >>>>> Imagine putting a big variable cap in parallel with R_F.  By cranking it
> >>>>> this way and that, you can make A_VCL do whatever you want, but the SNR
> >>>>> basically stays still, because e_N and C_d stay still.
> >>>>
> >>>> Yes, I agree that Avcl changes when you change Cf (well, above
> >>>> f=1/(2pi*Rf*Cf).  Below that frequency, Avcl is unity independent of
> >>>> Cf).  But the SNR is given by:
> >>>>     SNR = i_signal * Zm / (e_N*Avcl)
> >>>> and if you fiddle Cf, you're changing both Avcl and Zm.  It's not
> >>>> obvious to me that their ratio is unchanged.
> >>>>
> >>>> It would be helpful to hear your thoughts on the low-frequency
> >>>> hypothetical from my previous post.  i.e. if the noise current really
> >>>> grows with frequency from DC on up to fp, then at low frequency the
> >>>> output noise voltage would grow with frequency.  But e_N * Avcl is
> >>>> white at low frequency.
> >>>>
> >>>> James
> >>>>
> >>>
> >>> You left out the signal.  The signal current comes in via the
> >>> photodiode, just like the e_N*C_d current.   Once they're mixed
> >>> together, all the amp can do is change the frequency response.  The
> >>> maximum SNR is fixed.
> >>>
> >>
> >> I should add that the SNR discussion earlier in the chapter, where we
> >> start with a simple load resistor and work from there, is highly
> >> relevant.  Instruments live and die by their SNR and stability.
> >> Frequency response, you can fix afterwards.
> >
> > Re: SNR, I'm not sure this is true or not. But my gut says
> > that somewhere in the source size, (PD area), light level,
> > (detector R.), signal frequency, available opamp, space.
> > That a PD reversed biased (from a clean source) into an R,
> > with a opamp looking at the R voltage is as good as anything.
> > (OK I'm thinking simple and not adding any transistor jiu-jitsu.)
> >
> > George H.
> 
> Nope.  Op amps are a good 20 dB off the pace in very many instances, and 
> even further if you really tweak things to the eyeballs. For instance, a 
> bootstrap made from a BF862 JFET and a couple of BJTs can reduce the 
> effective capacitance of a photodiode by a factor of 10**3, and replace 
> the op amp's noise with the ~0.7 nV of the BF862, with a bias current of 
> 2 pA.  They don't make op amps anywhere near that good. 
I've only done bootstraps w/ opamps.  
pHEMTs are even 
> better at frequencies above about 2 MHz, though they take a bit more TLC 
> than BF862s.
> 
> Photons are often very expensive, which makes extra design effort on the 
> front end very worthwhile.  (It's also fun, once you've done it once or 
> twice.)
> 
> That said, of course there are plenty of easy cases, where the light is 
> bright and the bandwidth smallish, and the by-the-book approach works 
> fine.  (I give 5 rules for opamp-based TIA design in Section 18.4.3 of 
> the second edition, which a number of people have told me were very 
> helpful.)

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.     
> 
> I certainly don't advocate adding bells and whistles you don't need, but 
> then people don't phone me up for the simple ones. ;)
Sure,  But many of us design in the simpler world, 
your bread and butter is my bell and whistle. 

George H.  
  
> 
> 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