> 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
Noise and gain in transimpedance amplifiers
Started by ●December 10, 2015
Reply by ●December 10, 20152015-12-10
Reply by ●December 10, 20152015-12-10
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. 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
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. 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
> > SNR = i_signal * Zm / (e_N*Avcl)...snip...> You left out the signal.Why do you say I left out the signal? I wrote down the SNR as a ratio of voltages at the output. The signal is i_signal * Zm (i_signal is the photodiode current). The noise (again, voltage at the output) is e_N*Avcl. Also, can you comment on the low-frequency hypothetical?
Reply by ●December 10, 20152015-12-10
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.> > 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
On 12/10/2015 09:29 PM, jbattat@gmail.com wrote:>>> SNR = i_signal * Zm / (e_N*Avcl) > ...snip... >> You left out the signal. > > Why do you say I left out the signal? I wrote down the SNR as a ratio > of voltages at the output. The signal is i_signal * Zm (i_signal is > the photodiode current). The noise (again, voltage at the output) is > e_N*Avcl.I clarified that in my follow-up. Once you mix the signal and noise currents together, there's no getting them apart again. The physics is key.> > > 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. If you're more comfortable with the closed-form expression for A_VCL, that's certainly OK with me. I find that keeping close to the physics makes it a whole lot easier to do tradeoffs and find new topologies. There's lots of life after op amp TIAs. ;) 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
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. 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.) 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. ;) 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
> > 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.
Reply by ●December 10, 20152015-12-10
On 12/10/2015 09:50 PM, jbattat@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. >It isn't inconsistent. A_VCL is asymptotically flat at low frequency (neglecting 1/f noise), but that doesn't mean it's exactly flat. The e_N*C_d noise doesn't magically disappear when it drops below the white noise floor. With all due respect, it seems to me that you don't care enough about the physics of the problem. What's so hard to understand about the origin of the e_N*C_d contribution? Is it too untidy? Some folks like to tie everything up in one neat mathematical expression (radar books are full of those, for instance). I'm not in that camp. What I mostly care about is knowing what will happen if I change things, because I do that a lot. Hanging on to things that have to be true, e.g. the physical origins of the noise, is a big help. 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
On Thursday, December 10, 2015 at 9:35:55 PM UTC-5, Phil Hobbs wrote:> On 12/10/2015 09:29 PM, jbattat@gmail.com wrote: > >>> SNR = i_signal * Zm / (e_N*Avcl) > > ...snip... > >> You left out the signal. > > > > Why do you say I left out the signal? I wrote down the SNR as a ratio > > of voltages at the output. The signal is i_signal * Zm (i_signal is > > the photodiode current). The noise (again, voltage at the output) is > > e_N*Avcl. > > I clarified that in my follow-up. Once you mix the signal and noise > currents together, there's no getting them apart again. The physics is key.Well, you can do those correlation things.. multiplying two "equivalent" signal chains can get rid of the uncorrelated amp noise... slowly. It's a case when the noise is the signal. :^) (I've never done it.... seems like a lot of work.) George H.> > > > > > > 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. > > If you're more comfortable with the closed-form expression for A_VCL, > that's certainly OK with me. I find that keeping close to the physics > makes it a whole lot easier to do tradeoffs and find new topologies. > There's lots of life after op amp TIAs. ;) > > 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
