Hi, I have a few questions about gain and noise in the transimpedance amplifier. I've been mainly reading Hobbs (2nd edition) and H&H AoE (3rd Ed). References below are to those editions. Most importantly, thank you to the authors for these wonderful texts! === Rolloff of transimpedance === Hobbs Eqn 18.10 gives an expression for the transimpedance Zm, and states that the very steep rolloff is equivalent to 3 poles. I'm trying to understand why the roll-off is 3-pole and not 2-pole (I may be missing something obvious here). In particular, the denominator of Zm is a quadratic in s, so shouldn't there only be two roots (poles)? Also, Hobbs' plot of Zm (Fig. 18.5) shows a 12dB/octave rolloff, not 18dB/octave. e.g. from that plot, Zm(1MHz) = 5,200 ohm and Zm(2MHz)=1,300 ohm, and 20*log(5,200/1,300) = 12dB. === "e_N-Cin noise" ===- I understand that a varying voltage on Cin must have an associated current (e.g. AoE 8.11.3, and in a May 4, 2015 post to this group by Winfield Hill [1]), and I gather that this problem is mitigated at f >> 1/(2pi Rf Cf) because Cf shunts Rf. But I don't understand why this noise current needs to be calculated/included "by hand". In other words, shouldn't e_N-Cin noise naturally fall out of an analysis of Avcl (the non-inverting closed loop gain)? Specifically, if Avcl is used to find the output-referred noise voltage, which is then converted to an input-referred noise current, shouldn't that noise current include the "e_N-Cin" contribution? Hobbs Eqn 18.11 and 18.12, for example, seems to suggest so. Hobbs 18.11 gives an expression for Avcl, and the following sentence says "For frequencies well within the loop bandwidth, the resulting equivalent noise current is approximately i_n = (2pi f Cd) e_N." But, if so, then I'm confused because both Avcl and Zm are flat at low frequency (see Hobbs Fig 18.5), so won't the resulting input-referred noise current also be flat (not rising with frequency, as required by the e_N-Cin noise)? Maybe my confusion lies in how to convert from output noise voltage to input noise current. Which brings us to the next and final question that will further reveal my ignorance ... === Converting from e_o to i_Nin === To convert from the output-referred noise voltage density in a TIA (e.g. e_N times Avcl) to the input-referred noise current density, do you simply divide the output noise voltage by the transimpedance: i_Nin = e_o/Zm? Many thanks for your help and patience, James [1] W. Hill on e_N-Cin noise: "The explanation is easy enough: the voltage noise on the summing junction causes it to move up and down, and the capacitance on that node needs current to do this, which is supplied by the op-amp's output, and appears as signal." === Notation === e_N = op-amp voltage noise density Cin = total shunt capacitance at input to opamp (called Cd in Hobbs) Rf, Cf = feedback R and C Zm = TIA transimpedance Avcl = non-inverting closed-loop voltage gain (Hobbs Eqn 18.11) i_Nin = input-referred current noise density e_o = output-referred voltage noise density s = j omega
Noise and gain in transimpedance amplifiers
Started by ●December 10, 2015
Reply by ●December 10, 20152015-12-10
On 12/10/2015 12:05 AM, jbattat@gmail.com wrote:> Hi, > > I have a few questions about gain and noise in the transimpedance > amplifier.. I've been mainly reading Hobbs (2nd edition) and H&H AoE > (3rd Ed). References below are to those editions. Most importantly, > thank you to the authors for these wonderful texts! > > === Rolloff of transimpedance === Hobbs Eqn 18.10 gives an expression > for the transimpedance Zm, and states that the very steep rolloff is > equivalent to 3 poles. I'm trying to understand why the roll-off is > 3-pole and not 2-pole (I may be missing something obvious here). In > particular, the denominator of Zm is a quadratic in s, so shouldn't > there only be two roots (poles)? Also, Hobbs' plot of Zm (Fig. 18.5) > shows a 12dB/octave rolloff, not 18dB/octave. e.g. from that plot, > Zm(1MHz) = 5,200 ohm and Zm(2MHz)=1,300 ohm, and 20*log(5,200/1,300) > = 12dB.It's asymptotically 3 poles because of the rolloff of A_VOL. I could have put that better, it's true.> > === "e_N-Cin noise" ===- I understand that a varying voltage on Cin > must have an associated current (e.g. AoE 8.11.3, and in a May 4, > 2015 post to this group by Winfield Hill [1]), and I gather that this > problem is mitigated at f >> 1/(2pi Rf Cf) because Cf shunts Rf.No, that doesn't help the input current noise, it just cuts down the gain seen by the higher-frequency noise currents.> > But I don't understand why this noise current needs to be > calculated/included "by hand". In other words, shouldn't e_N-Cin > noise naturally fall out of an analysis of Avcl (the non-inverting > closed loop gain)?Well, because a gain analysis isn't a noise analysis, for one thing.> Specifically, if Avcl is used to find the > output-referred noise voltage, which is then converted to an > input-referred noise current, shouldn't that noise current include > the "e_N-Cin" contribution?Not if you don't put it in. SPICE will get it right, except of course that all the op amp models you can get are horrible.> > Hobbs Eqn 18.11 and 18.12, for example, seems to suggest so. Hobbs > 18.11 gives an expression for Avcl, and the following sentence says > "For frequencies well within the loop bandwidth, the resulting > equivalent noise current is approximately i_n = (2pi f Cd) e_N."It isn't exactly Cd*e_N, because at higher frequencies the op amp doesn't have enough gain to hold the summing junction exactly still (its idea of still, i.e. impressing its input noise voltage across C_d). But it's close enough for noise purposes.> > But, if so, then I'm confused because both Avcl and Zm are flat at > low frequency (see Hobbs Fig 18.5), so won't the resulting > input-referred noise current also be flat (not rising with frequency, > as required by the e_N-Cin noise)?Gain analysis and noise analysis are different. They're often confused, because we're very accustomed to noise sources that are white, e.g. Johnson noise and shot noise. 1/f noise goes up at low frequency, and e_N*C_d noise goes up at high frequency. They don't have the same shape as the gain curves at all. Above the 1/f region, white noise dominates until the differentiated voltage noise becomes large.> > Maybe my confusion lies in how to convert from output noise voltage > to input noise current. Which brings us to the next and final > question that will further reveal my ignorance ... > > === Converting from e_o to i_Nin === To convert from the > output-referred noise voltage density in a TIA (e.g. e_N times Avcl) > to the input-referred noise current density, do you simply divide the > output noise voltage by the transimpedance: i_Nin = e_o/Zm?Yup. Input-referred means "what input signal would generate this output, given a noiseless amplifier with the same gain?" Good on you for drilling through all that. It's amazing to me how many people refuse to do their own math, even when their livelihoods are riding on the success of their projects. In optical measurements, it's easy for one's intuition to be off by 5 orders of magnitude. And calculating this stuff is fun once you get used to it. I did a TIA for a biochip application a few years ago (for the research arm of a Very Very Large Korean Electronics Company). It got within (iirc) 6 dB of the shot noise limit for a 1-nA signal out to 70 MHz. It used a cascoded pHEMT front end and a super small wire bonded input connection so that the total input capacitance was about 0.7 pF. It used capacitive feedback for AC and an active current source for DC, and fixed up the resulting frequency response funnies in the third second stage. I definitely couldn't have done that without doing the math first. (Of course 1 nA is only 16 dB above its own shot noise in 70 MHz, and the signal couldn't be made repetitive, so it was a pretty marginal measurement to begin with.) Cheers Phil Hobbs> > Many thanks for your help and patience, James > > [1] W. Hill on e_N-Cin noise: "The explanation is easy enough: the > voltage noise on the summing junction causes it to move up and down, > and the capacitance on that node needs current to do this, which is > supplied by the op-amp's output, and appears as signal." > > === Notation === e_N = op-amp voltage noise density Cin = total > shunt capacitance at input to opamp (called Cd in Hobbs) Rf, Cf = > feedback R and C Zm = TIA transimpedance Avcl = non-inverting > closed-loop voltage gain (Hobbs Eqn 18.11) i_Nin = input-referred > current noise density e_o = output-referred voltage noise density > s = j omega >-- 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 by ●December 10, 20152015-12-10
On Thursday, December 10, 2015 at 12:05:59 AM UTC-5, jba...@gmail.com wrote:> Hi, > > I have a few questions about gain and noise in the transimpedance amplifier. I've been mainly reading Hobbs (2nd edition) and H&H AoE (3rd Ed). References below are to those editions. Most importantly, thank you to the authors for these wonderful texts! > > === Rolloff of transimpedance === > Hobbs Eqn 18.10 gives an expression for the transimpedance Zm, and states that the very steep rolloff is equivalent to 3 poles. I'm trying to understand why the roll-off is 3-pole and not 2-pole (I may be missing something obvious here). In particular, the denominator of Zm is a quadratic in s, so shouldn't there only be two roots (poles)? Also, Hobbs' plot of Zm (Fig. 18.5) shows a 12dB/octave rolloff, not 18dB/octave. e.g. from that plot, Zm(1MHz) = 5,200 ohm and Zm(2MHz)=1,300 ohm, and 20*log(5,200/1,300) = 12dB. > > === "e_N-Cin noise" ===- > I understand that a varying voltage on Cin must have an associated current (e.g. AoE 8.11.3, and in a May 4, 2015 post to this group by Winfield Hill [1]), and I gather that this problem is mitigated at f >> 1/(2pi Rf Cf) because Cf shunts Rf. > > But I don't understand why this noise current needs to be calculated/included "by hand". In other words, shouldn't e_N-Cin noise naturally fall out of an analysis of Avcl (the non-inverting closed loop gain)? Specifically, if Avcl is used to find the output-referred noise voltage, which is then converted to an input-referred noise current, shouldn't that noise current include the "e_N-Cin" contribution? > > Hobbs Eqn 18.11 and 18.12, for example, seems to suggest so. Hobbs 18.11 gives an expression for Avcl, and the following sentence says "For frequencies well within the loop bandwidth, the resulting equivalent noise current is approximately i_n = (2pi f Cd) e_N." > > But, if so, then I'm confused because both Avcl and Zm are flat at low frequency (see Hobbs Fig 18.5), so won't the resulting input-referred noise current also be flat (not rising with frequency, as required by the e_N-Cin noise)? > > Maybe my confusion lies in how to convert from output noise voltage to input noise current. Which brings us to the next and final question that will further reveal my ignorance ... > > === Converting from e_o to i_Nin === > To convert from the output-referred noise voltage density in a TIA (e.g. e_N times Avcl) to the input-referred noise current density, do you simply divide the output noise voltage by the transimpedance: i_Nin = e_o/Zm? > > Many thanks for your help and patience, > James > > [1] W. Hill on e_N-Cin noise: "The explanation is easy enough: the voltage noise on the summing junction causes it to move up and down, and the capacitance on that node needs current to do this, which is supplied by the op-amp's output, and appears as signal." > > === Notation === > e_N = op-amp voltage noise density > Cin = total shunt capacitance at input to opamp (called Cd in Hobbs) > Rf, Cf = feedback R and C > Zm = TIA transimpedance > Avcl = non-inverting closed-loop voltage gain (Hobbs Eqn 18.11) > i_Nin = input-referred current noise density > e_o = output-referred voltage noise density > s = j omegaGood I see Phil responded. S. Franco in "Design with Operational Amp...." does a nice job of doing the noise analysis in the TIA photodiode circuit. (Lots of other good stuff too.) As Phil says it's good to work through it yourself. With Franco you've got someone to hold your hand. George H.
Reply by ●December 10, 20152015-12-10
Thanks for the reply Phil. I'm really enjoying working through these TIA calculations. Stretches the mind. Always good to get some help from experts, too -- some follow-up questions below:> > particular, the denominator of Zm is a quadratic in s, so shouldn't > > there only be two roots (poles)? Also, Hobbs' plot of Zm (Fig. 18.5) > > shows a 12dB/octave rolloff, not 18dB/octave. e.g. from that plot, > > Zm(1MHz) = 5,200 ohm and Zm(2MHz)=1,300 ohm, and 20*log(5,200/1,300) > > = 12dB. > > It's asymptotically 3 poles because of the rolloff of A_VOL. I could > have put that better, it's true.I did include A_VOL to get the quadratic. A_VOL gives a factor of 1/s and so the quadratic comes from the product of A_VOL and Z_f (which goes like 1/(1+s)). I don't see where the cubic comes from. Ah -- I think I now know. Below fc, Aol ~ 1/s, but above fc, Aol has a 2-pole roll-off, so the Aol ~ 1/s assumption eventually breaks down. Is that what you mean by "asymptotically 3 poles"? If so, then that makes sense to me, and is consistent with Fig. 18.5. You'd expect to see the transition from 2-pole to 3-pole rolloff above 2MHz, which is outside of the frequency range for Zm shown in Fig. 18.5 (assuming an LF356 with fc=4MHz, as indicated in Fig. 18.4).> > === "e_N-Cin noise" === > > <...snip...> > > But I don't understand why this noise current needs to be > > calculated/included "by hand". In other words, shouldn't e_N-Cin > > noise naturally fall out of an analysis of Avcl (the non-inverting > > closed loop gain)? > > Well, because a gain analysis isn't a noise analysis, for one thing.But the "noise gain" *is* Avcl. (i.e. Avcl is dVout/dVin, measured at the noninverting input). e.g. from the discussion leading up to Hobbs Eqn 18.11 "clearly e_NAmp will be multiplied by the noninverting gain of the amplifier, Avcl -- which is therefore the noise gain of the stage." Or do you mean something different by "noise analysis?"> > Specifically, if Avcl is used to find the > > output-referred noise voltage, which is then converted to an > > input-referred noise current, shouldn't that noise current include > > the "e_N-Cin" contribution? > > Not if you don't put it in.OK, this is getting to the heart of the question. I'm arguing that I have put it in. I compute the noise gain (Avcl) for e_N by applying Kirchoff's Current Law at the summing junction (inverting input). Current through Cin is included in this model. Even the current driven through Cin by the op-amp voltage noise. Why would I need to add, separately, another contribution "e_N-Cin" current? Avcl *does* account for all of the current through Cin that is caused by e_N. See derivations in URL below, along with claim that the noise current is only proportional to frequency (i.e. the so-called "e_N-Cin noise") in a limited region of frequency space, between fz=1/[2*pi*(Cin+Cf)] and fp=1/(2*pi*Cf): http://academics.wellesley.edu/Physics/jbattat/electronics/tia_noiseGain1.jpg http://academics.wellesley.edu/Physics/jbattat/electronics/tia_noiseGain2.jpg Yes, this derivation (1) ignores the rolloff at very high frequency caused by op-amp open-loop gain rolloff, and (2) assumes white op-amp noise voltage. But these assumptions can easily be included -- and they don't change my basic conclusion.> > Hobbs Eqn 18.11 and 18.12, for example, seems to suggest so. Hobbs > > 18.11 gives an expression for Avcl, and the following sentence says > > "For frequencies well within the loop bandwidth, the resulting > > equivalent noise current is approximately i_n = (2pi f Cd) e_N." > > It isn't exactly Cd*e_N, because at higher frequencies the op amp > doesn't have enough gain to hold the summing junction exactly still (its > idea of still, i.e. impressing its input noise voltage across C_d). But > it's close enough for noise purposes.What I meant was that the book implies that 18.12 comes out of 18.11 (i.e. that indeed the Avcl expression contains the phenomenon of "e_N-Cin noise"). Which I agree with.> > But, if so, then I'm confused because both Avcl and Zm are flat at > > low frequency (see Hobbs Fig 18.5), so won't the resulting > > input-referred noise current also be flat (not rising with frequency, > > as required by the e_N-Cin noise)? > > Gain analysis and noise analysis are different. They're often confused, > because we're very accustomed to noise sources that are white, e.g. > Johnson noise and shot noise. 1/f noise goes up at low frequency, and > e_N*C_d noise goes up at high frequency. They don't have the same shape > as the gain curves at all. Above the 1/f region, white noise dominates > until the differentiated voltage noise becomes large.I'm on board with the frequency dependences of Johnson, 1/f, shot, etc. But the expression for e_N-Cin noise states that this noise current grows linearly with frequency even for white op-amp voltage noise (e_N). I'm trying to find the physical origin of this "effect" and claim that it's fundamental origin is the linear rise of the noise voltage with frequency (over a limited range of frequencies). James
Reply by ●December 10, 20152015-12-10
> Good I see Phil responded. > S. Franco in "Design with Operational Amp...." does a nice job of doing the noise > analysis in the TIA photodiode circuit. (Lots of other good stuff too.)Thanks for pointing out this reference. I hadn't seen it and look forward to reading it. I've found it very helpful to read about the same idea from multiple sources (so far Hobbs, AoE and Graeme have been very complementary).
Reply by ●December 10, 20152015-12-10
On 12/10/2015 04:04 PM, jbattat@gmail.com wrote:> Thanks for the reply Phil. I'm really enjoying working through these > TIA calculations. Stretches the mind. Always good to get some help > from experts, too -- some follow-up questions below: > >>> particular, the denominator of Zm is a quadratic in s, so >>> shouldn't there only be two roots (poles)? Also, Hobbs' plot of >>> Zm (Fig. 18.5) shows a 12dB/octave rolloff, not 18dB/octave. >>> e.g. from that plot, Zm(1MHz) = 5,200 ohm and Zm(2MHz)=1,300 ohm, >>> and 20*log(5,200/1,300) = 12dB. >> >> It's asymptotically 3 poles because of the rolloff of A_VOL. I >> could have put that better, it's true. > > I did include A_VOL to get the quadratic. A_VOL gives a factor of > 1/s and so the quadratic comes from the product of A_VOL and Z_f > (which goes like 1/(1+s)). I don't see where the cubic comes from. > > Ah -- I think I now know. Below fc, Aol ~ 1/s, but above fc, Aol has > a 2-pole roll-off, so the Aol ~ 1/s assumption eventually breaks > down. Is that what you mean by "asymptotically 3 poles"?Once the op amp isn't holding its input still any longer, the diode capacitance makes Vin go as 1/f, which gives you an extra pole. I clarified the discussion in the draft third edition, thanks!> > If so, then that makes sense to me, and is consistent with Fig. 18.5. > You'd expect to see the transition from 2-pole to 3-pole rolloff > above 2MHz, which is outside of the frequency range for Zm shown in > Fig. 18.5 (assuming an LF356 with fc=4MHz, as indicated in Fig. > 18.4). > >>> === "e_N-Cin noise" === <...snip...> But I don't understand why >>> this noise current needs to be calculated/included "by hand". In >>> other words, shouldn't e_N-Cin noise naturally fall out of an >>> analysis of Avcl (the non-inverting closed loop gain)? >> >> Well, because a gain analysis isn't a noise analysis, for one >> thing. > > But the "noise gain" *is* Avcl. (i.e. Avcl is dVout/dVin, measured > at the noninverting input). e.g. from the discussion leading up to > Hobbs Eqn 18.11 "clearly e_NAmp will be multiplied by the > noninverting gain of the amplifier, Avcl -- which is therefore the > noise gain of the stage." > > Or do you mean something different by "noise analysis?"OK, there are two gains in the problem--the noise gain and the transimpedance gain. I agree that the noise gain is A_VCL (noninverting).> >>> Specifically, if Avcl is used to find the output-referred noise >>> voltage, which is then converted to an input-referred noise >>> current, shouldn't that noise current include the "e_N-Cin" >>> contribution? >> >> Not if you don't put it in. > > OK, this is getting to the heart of the question. I'm arguing that I > have put it in. I compute the noise gain (Avcl) for e_N by applying > Kirchoff's Current Law at the summing junction (inverting input). > Current through Cin is included in this model. Even the current > driven through Cin by the op-amp voltage noise. Why would I need to > add, separately, another contribution "e_N-Cin" current? Avcl *does* > account for all of the current through Cin that is caused by e_N.Sure. I was thinking of the transimpedance as the gain under discussion.> > See derivations in URL below, along with claim that the noise current > is only proportional to frequency (i.e. the so-called "e_N-Cin > noise") in a limited region of frequency space, between > fz=1/[2*pi*(Cin+Cf)] and fp=1/(2*pi*Cf): > http://academics.wellesley.edu/Physics/jbattat/electronics/tia_noiseGain1.jpg > >http://academics.wellesley.edu/Physics/jbattat/electronics/tia_noiseGain2.jpg> > Yes, this derivation (1) ignores the rolloff at very high frequency > caused by op-amp open-loop gain rolloff, and (2) assumes white op-amp > noise voltage. But these assumptions can easily be included -- and > they don't change my basic conclusion. > >>> Hobbs Eqn 18.11 and 18.12, for example, seems to suggest so. >>> Hobbs 18.11 gives an expression for Avcl, and the following >>> sentence says "For frequencies well within the loop bandwidth, >>> the resulting equivalent noise current is approximately i_n = >>> (2pi f Cd) e_N." >> >> It isn't exactly Cd*e_N, because at higher frequencies the op amp >> doesn't have enough gain to hold the summing junction exactly still >> (its idea of still, i.e. impressing its input noise voltage across >> C_d). But it's close enough for noise purposes. > > What I meant was that the book implies that 18.12 comes out of 18.11 > (i.e. that indeed the Avcl expression contains the phenomenon of > "e_N-Cin noise"). Which I agree with. > >>> But, if so, then I'm confused because both Avcl and Zm are flat >>> at low frequency (see Hobbs Fig 18.5), so won't the resulting >>> input-referred noise current also be flat (not rising with >>> frequency, as required by the e_N-Cin noise)? >> >> Gain analysis and noise analysis are different. They're often >> confused, because we're very accustomed to noise sources that are >> white, e.g. Johnson noise and shot noise. 1/f noise goes up at low >> frequency, and e_N*C_d noise goes up at high frequency. They don't >> have the same shape as the gain curves at all. Above the 1/f >> region, white noise dominates until the differentiated voltage >> noise becomes large. > > I'm on board with the frequency dependences of Johnson, 1/f, shot, > etc. But the expression for e_N-Cin noise states that this noise > current grows linearly with frequency even for white op-amp voltage > noise (e_N).It does. What I'm trying to get across there is that the noise current going through C_d is a real current that you could in principle measure with a meter. Multiply that by the transconductance, and you get the C_d*e_N contribution to the output noise. It's a way of supplying physical motivation for what the A_VCL calculation predicts. I'm trying to find the physical origin of this "effect"> and claim that it's fundamental origin is the linear rise of the > noise voltage with frequency (over a limited range of frequencies). > > James >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 by ●December 10, 20152015-12-10
Thanks Phil, very helpful responses. My apologies that my first question about the transimpedance gain (Zm) created confusion about what gain I was using for the noise analysis. On the "e_N-Cin noise" question, it seems like we're in agreement then that the rise in input-referred noise current density with frequency is indeed caused by the ramp up of Avcl between fz and fp, and since that happens over a finite range of frequencies the "e_N-Cin noise" (meaning a current proportional to frequency) only exists over that limited range of frequencies. i.e. the "e_N-Cin noise" is not a separate phenomenon -- it's a consequence of the shape of Avcl. Below fz, the input noise current is white (until you go down in frequency far enough to reach the 1/f rise in e_NAmp -- e.g. ~100Hz for the LF411), and above fp the input current noise is also white (until you get hammered by the rolloff of Zm and/or Avol). James> > Why would I need to > > add, separately, another contribution "e_N-Cin" current? Avcl *does* > > account for all of the current through Cin that is caused by e_N. > > Sure. I was thinking of the transimpedance as the gain under discussion.> > I'm on board with the frequency dependences of Johnson, 1/f, shot, > > etc. But the expression for e_N-Cin noise states that this noise > > current grows linearly with frequency even for white op-amp voltage > > noise (e_N). > > It does. What I'm trying to get across there is that the noise current > going through C_d is a real current that you could in principle measure > with a meter. Multiply that by the transconductance, and you get the > C_d*e_N contribution to the output noise. > > It's a way of supplying physical motivation for what the A_VCL > calculation predicts.
Reply by ●December 10, 20152015-12-10
>On the "e_N-Cin noise" question, it seems like we're in agreement then that the rise in input- >referred noise current density with frequency is indeed caused by the ramp up of Avcl between >fz and fp, and since that happens over a finite range of frequencies the "e_N-Cin noise" >(meaning a current proportional to frequency) only exists over that limited range of >frequencies. �i.e. the "e_N-Cin noise" is not a separate phenomenon -- it's a consequence of >the shape of Avcl.It's an input vs output question. e_N causes a real current to flow in the external circuit, more or less independent of the details of the feedback network. That means that its effect on the SNR (which is what we mostly care about) hardly depends on the feedback details. I'm in favour of keeping it that way and multiplying it by the transconductance, because that keeps closer to the physics and makes it much easier to compare different designs in my head. So no, I wouldn't agree that A_VCL causes the effect. I'd say that C_d causes both A_VCL and the e_N*C_d noise, but that the latter is more fundamental. Cheers Phil Hobbs
Reply by ●December 10, 20152015-12-10
> It's an input vs output question. e_N causes a real current to flow in the external circuit, more or less independent of the details of the feedback network. That means that its effect on the SNR (which is what we mostly care about) hardly depends on the feedback details. I'm in favour of keeping it that way and multiplying it by the transconductance, because that keeps closer to the physics and makes it much easier to compare different designs in my head. > > So no, I wouldn't agree that A_VCL causes the effect. I'd say that C_d causes both A_VCL and the e_N*C_d noise, but that the latter is more fundamental.Now you've lost me. Maybe because of semantics, maybe something more fundamental. I don't see how e_N*C_d noise is a separate phenomenon from A_VCL (which already accounts for current flow through C_d). Let's think about the low-frequency limit here. And let's simplify the feedback network to be a resistor (the impedance of Cf at low frequency is very high anyway). Let's also assume that eN is white for now. In that scenario, if the current through C_d is indeed i ~ 2pi*Cd*eN*f, then this current must also flow through Rf, to produce an output voltage noise density that grows linearly with frequency. But that conflicts with the result that I think you agree with: e_out = Avcl*e_NAmp which implies that e_out is white, because e_NAmp is white, and Avcl is flat at low frequency. Do you agree that these two analyses disagree (current through C_d ~ f, vs. e_out = Avcl*e_N)? This is one of those times when a brief conversation at a whiteboard is priceless! Thanks for your patience -- I'm learning a lot here.
Reply by ●December 10, 20152015-12-10
On 12/10/2015 06:37 PM, jbattat@gmail.com wrote:>> It's an input vs output question. e_N causes a real current to flow >> in the external circuit, more or less independent of the details of >> the feedback network. That means that its effect on the SNR (which >> is what we mostly care about) hardly depends on the feedback >> details. I'm in favour of keeping it that way and multiplying it by >> the transconductance, because that keeps closer to the physics and >> makes it much easier to compare different designs in my head. >> >> So no, I wouldn't agree that A_VCL causes the effect. I'd say that >> C_d causes both A_VCL and the e_N*C_d noise, but that the latter is >> more fundamental. > > Now you've lost me. Maybe because of semantics, maybe something more > fundamental. I don't see how e_N*C_d noise is a separate phenomenon > from A_VCL (which already accounts for current flow through C_d). > > Let's think about the low-frequency limit here. And let's simplify > the feedback network to be a resistor (the impedance of Cf at low > frequency is very high anyway). Let's also assume that eN is white > for now. > > In that scenario, if the current through C_d is indeed i ~ > 2pi*Cd*eN*f, then this current must also flow through Rf, to produce > an output voltage noise density that grows linearly with frequency. > But that conflicts with the result that I think you agree with: e_out > = Avcl*e_NAmp which implies that e_out is white, because e_NAmp is > white, and Avcl is flat at low frequency. Do you agree that these > two analyses disagree (current through C_d ~ f, vs. e_out = > Avcl*e_N)? > > This is one of those times when a brief conversation at a whiteboard > is priceless! Thanks for your patience -- I'm learning a lot here. >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. 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
