Hi, <background> This really isn't a circuit question but people here likely understand what I'm finding so confusing. I've built several AC coupled photodetectors to play around with. Basic transimpedance amp using a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 with 30pf comp cap. I've also added a small bypass cap on the feedback resistor to flatten out the response at 500kHz. The output I feed through a bypass cap into my SDR-IQ software defined radio. Lot's of fun playing with these things and they seem to work real well. Good for me. </background> <question> If I shine an incandescent bulb on the detector and crank up the light so I get roughly 100 uA photo current why is the "wave noise" so small compared to the DC output of the detector? From theory the power density spectrum, S(w), of the optical field is the Fourier transform of the field time correlation function <E(t)E(t-t')>. The field correlation function as far as my detector goes is a delta function in time whose integral should be the light power flux at DC. What happens in reality is S(w) has a huge spike at DC (the constant light flux) and then a really small amount of noise off DC (probably shot noise). The math, which I think I understand but can't connect with the physics is that S(0) should be equal to S(100kHz). Clearly this isn't the case. Why? </question> Thanks Paul C.

# Optical noise

Started by ●November 29, 2012

Reply by ●November 29, 20122012-11-29

Simple way to look at it: a random baseball pitch is a single event, of a relatively large object. It's very "impulsive", in that a baseball hits with a large impulse, infrequently, which you would expect to obey a Poisson distribution. Suppose you upped the rate, while reducing the size (so that the momentum or mass flow or something like that is more or less the same), which increases the number of particles. Consider a sand blaster: a whole lot of tiny sand particles come out, but when they individually whack into a surface, they don't make the surface go SLAP, SLAP every time, but rather it's the din of a million microscopic twacks, resulting in a dull hiss. The momentum delivered to the surface is rather constant (of course, a regular sandblaster has a lot of compressed air behind it, in addition to the sand, but hold that thought). If you remove the sand from the compressed air, so it's just air blowing, the momentum is due entirely to the (extremely tiny, extremely numerous) molecules hitting (not even hitting at this point, rather, piling up against it and sliding away), and the amount of noise very small. In general, the noise in a random variable goes as 1/sqrt(N) for N particles in the system (whatever that happens to mean). Examples: A sandblaster delivering 10^6 sand grains/second is louder than a sandblaster delivering only 10^4, but not by 100 times, only sqrt(10^6/10^4) = 10 times. A number of equally powerful white noise sources, added together, has a total amplitude of sqrt(N) times (i.e., the noise per source is 1/sqrt(N)). More generally, the noise of N sources with amplitude a_i is sqrt(SUM(a_i^2)) (sum from i=1 to N), the R^N-vector sum of all independent components. You may already known all this... Now, applying this to your case, putting more noisy current sources in parallel (equivalent to noisy voltage sources in series; either way, they add) makes the current more even. Noise is increased by a factor of sqrt(N), but the correlated signal increases by N, so the SNR goes as 1/sqrt(N). In terms of correlation, the DC term is correlated, while all other frequencies are uncorrelated. So the DC term adds, while the others drop out. Tim -- Deep Friar: a very philosophical monk. Website: http://seventransistorlabs.com "Paul Colby" <paulccolby@earthlink.net> wrote in message news:2012112916493113180-paulccolby@earthlinknet...> Hi, > <background> > This really isn't a circuit question but people here likely understand > what I'm finding so confusing. > I've built several AC coupled photodetectors to play around with. Basic > transimpedance amp using > a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 > with 30pf comp cap. > I've also added a small bypass cap on the feedback resistor to flatten > out the response at 500kHz. > The output I feed through a bypass cap into my SDR-IQ software defined > radio. Lot's of fun playing > with these things and they seem to work real well. Good for me. > </background> > > <question> > If I shine an incandescent bulb on the detector and crank up the light > so I get roughly 100 uA photo > current why is the "wave noise" so small compared to the DC output of > the detector? From theory > the power density spectrum, S(w), of the optical field is the Fourier > transform of the field time > correlation function <E(t)E(t-t')>. The field correlation function as > far as my detector goes is a delta > function in time whose integral should be the light power flux at DC. > What happens in reality is S(w) > has a huge spike at DC (the constant light flux) and then a really small > amount of noise off DC (probably > shot noise). The math, which I think I understand but can't connect with > the physics is that S(0) > should be equal to S(100kHz). Clearly this isn't the case. Why? > </question> > > Thanks > Paul C. >

Reply by ●November 30, 20122012-11-30

On 2012-11-29 19:38:13 -0800, Tim Williams said:> Simple way to look at it: a random baseball pitch is a single event, of > a relatively large object. It's very "impulsive", in that a baseball > hits with a large impulse, infrequently, which you would expect to obey > a Poisson distribution. > > Suppose you upped the rate, while reducing the size (so that the > momentum or mass flow or something like that is more or less the same), > which increases the number of particles. Consider a sand blaster: a > whole lot of tiny sand particles come out, but when they individually > whack into a surface, they don't make the surface go SLAP, SLAP every > time, but rather it's the din of a million microscopic twacks, > resulting in a dull hiss. The momentum delivered to the surface is > rather constant (of course, a regular sandblaster has a lot of > compressed air behind it, in addition to the sand, but hold that > thought). > > If you remove the sand from the compressed air, so it's just air > blowing, the momentum is due entirely to the (extremely tiny, extremely > numerous) molecules hitting (not even hitting at this point, rather, > piling up against it and sliding away), and the amount of noise very > small. > > In general, the noise in a random variable goes as 1/sqrt(N) for N > particles in the system (whatever that happens to mean). > > Examples: > A sandblaster delivering 10^6 sand grains/second is louder than a > sandblaster delivering only 10^4, but not by 100 times, only > sqrt(10^6/10^4) = 10 times. > A number of equally powerful white noise sources, added together, has a > total amplitude of sqrt(N) times (i.e., the noise per source is > 1/sqrt(N)). More generally, the noise of N sources with amplitude a_i > is sqrt(SUM(a_i^2)) (sum from i=1 to N), the R^N-vector sum of all > independent components. > > You may already known all this... > > Now, applying this to your case, putting more noisy current sources in > parallel (equivalent to noisy voltage sources in series; either way, > they add) makes the current more even. Noise is increased by a factor > of sqrt(N), but the correlated signal increases by N, so the SNR goes > as 1/sqrt(N). > > In terms of correlation, the DC term is correlated, while all other > frequencies are uncorrelated. So the DC term adds, while the others > drop out. > > TimIf I understand things correctly what you describe is just photon shot noise or counting statistics. "Wave noise" refers to a classical intensity modulation due to beat wave frequencies. For example 10 MHz + 10.01MHz will beat at 0.01 MHz. Same thing happens in light and it should cause the intensity to have a fluctuation spectrum. A (likely too) simple argument implies that the intensity power spectrum at DC should be the same magnitude as the power spectrum at 100kHz. Now, your comment on N parallel and random current sources is noted. I thought the current output of a photodiode is a square law detection of the impinging E-field. Is this false?

Reply by ●November 30, 20122012-11-30

"Paul Colby" <paulccolby@paulccolby.com> wrote in message news:2012112920200471336-paulccolby@paulccolbycom...> If I understand things correctly what you describe is just photon shot > noise or counting statistics. "Wave noise" refers to a classical > intensity > modulation due to beat wave frequencies. For example 10 MHz + 10.01MHz > will beat at 0.01 MHz. Same thing happens in light and it should > cause the intensity to have a fluctuation spectrum. A (likely too) > simple argument implies that the intensity power spectrum at DC should > be > the same magnitude as the power spectrum at 100kHz. Now, your comment on > N parallel and random current sources is noted. I thought > the current output of a photodiode is a square law detection of the > impinging E-field. Is this false?Ahh, so the waves... Well, one would expect that, if the wavefronts are uncorrelated ("white" light, after all!), then the wave noise would be similarly uncorrelated, and fluctuations would occur with a rate comparable to the bandwidth of the optical signal itself (i.e., about 300THz wide = femtosecond fringing). Squaring is a time-domain multiplication, which means a frequency-domain autoconvolution -- essentially, the signal is shifted from "radio frequency" down to baseband, and the bandwidth doubled (which isn't really, because it goes from +BW to -BW, so it really has a bandwidth of BW again). If we start with a rectangular spectrum of "white" light from, say, red (~450THz) to blue (~750THz), it has a flat bandwidth of 300THz. Autoconvolution smears out the original rectangular spectrum into a *triangular* baseband signal, from -300THz to +300THz (infrared). In other words, "pink light". This baseband signal is further filtered by junction capacitance and circuitry bandwidth, cutting off all but a paltry -- maybe 1MHz or so, whatever you're actually getting. So, ignoring the shot noise effect, in the large-signal limit I suppose, you should get something like pink noise? Which, of course, will have a strong DC component, but the surrounding noise (assuming it's not dominated by shot noise, which is flat) should be 1/f style. Hopefully Phil Hobbs will drop in, correcting my likely many errors. :) Tim -- Deep Friar: a very philosophical monk. Website: http://seventransistorlabs.com

Reply by ●November 30, 20122012-11-30

On 2012-11-29 21:09:02 -0800, Tim Williams said:> "Paul Colby" <paulccolby@paulccolby.com> wrote in message > news:2012112920200471336-paulccolby@paulccolbycom... >> If I understand things correctly what you describe is just photon shot >> noise or counting statistics. "Wave noise" refers to a classical >> intensity >> modulation due to beat wave frequencies. For example 10 MHz + 10.01MHz >> will beat at 0.01 MHz. Same thing happens in light and it should >> cause the intensity to have a fluctuation spectrum. A (likely too) >> simple argument implies that the intensity power spectrum at DC should >> be >> the same magnitude as the power spectrum at 100kHz. Now, your comment >> on N parallel and random current sources is noted. I thought >> the current output of a photodiode is a square law detection of the >> impinging E-field. Is this false?> > Ahh, so the waves... > > Well, one would expect that, if the wavefronts are uncorrelated > ("white" light, after all!), then the wave noise would be similarly > uncorrelated, and fluctuations would occur with a rate comparable to > the bandwidth of the optical signal itself (i.e., about 300THz wide = > femtosecond fringing). > > Squaring is a time-domain multiplication, which means a > frequency-domain autoconvolution -- essentially, the signal is shifted > from "radio frequency" down to baseband, and the bandwidth doubled > (which isn't really, because it goes from +BW to -BW, so it really has > a bandwidth of BW again). If we start with a rectangular spectrum of > "white" light from, say, red (~450THz) to blue (~750THz), it has a flat > bandwidth of 300THz. Autoconvolution smears out the original > rectangular spectrum into a *triangular* baseband signal, from -300THz > to +300THz (infrared). In other words, "pink light". This baseband > signal is further filtered by junction capacitance and circuitry > bandwidth, cutting off all but a paltry -- maybe 1MHz or so, whatever > you're actually getting.Thanks Tim This is just the argument I'm being confused by. Your pink noise = white when you chop 300 THz down to 1MHz and the power spectrum at 0Hz should for all intents and purposes be the same as at 100kHz. This clearly isn't the case for way white light is detected.> > So, ignoring the shot noise effect, in the large-signal limit I > suppose, you should get something like pink noise? Which, of course, > will have a strong DC component, but the surrounding noise (assuming > it's not dominated by shot noise, which is flat) should be 1/f style.Okay, how do we sneak the large DC component in? A triangle shaped noise spectrum with a frequency base 300 THz wide is what follows from the above argument. There is at least six orders of magnitude between the power spectrum at 0Hz and that at 100kHz which isn't a triangle. Hence my confusion.> > Hopefully Phil Hobbs will drop in, correcting my likely many errors. :)Don't think you've made any. I'm likely missing something basic.> > Tim

Reply by ●November 30, 20122012-11-30

On 11/29/2012 07:49 PM, Paul Colby wrote:> Hi, > <background> > This really isn't a circuit question but people here likely understand > what I'm finding so confusing. > I've built several AC coupled photodetectors to play around with. Basic > transimpedance amp using > a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 > with 30pf comp cap. > I've also added a small bypass cap on the feedback resistor to flatten > out the response at 500kHz. > The output I feed through a bypass cap into my SDR-IQ software defined > radio. Lot's of fun playing > with these things and they seem to work real well. Good for me. > </background> > > <question> > If I shine an incandescent bulb on the detector and crank up the light > so I get roughly 100 uA photo > current why is the "wave noise" so small compared to the DC output of > the detector? From theory > the power density spectrum, S(w), of the optical field is the Fourier > transform of the field time > correlation function <E(t)E(t-t')>. The field correlation function as > far as my detector goes is a delta > function in time whose integral should be the light power flux at DC. > What happens in reality is S(w) > has a huge spike at DC (the constant light flux) and then a really small > amount of noise off DC (probably > shot noise). The math, which I think I understand but can't connect with > the physics is that S(0) > should be equal to S(100kHz). Clearly this isn't the case. Why? > </question> > > Thanks > Paul C. >The fluctuations are white, but the DC is separate. 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 ●November 30, 20122012-11-30

On Nov 29, 7:49=A0pm, Paul Colby <paulcco...@earthlink.net> wrote:> Hi, > <background> > This really isn't a circuit question but people here likely understand > what I'm finding so confusing. > I've built several AC coupled photodetectors to play around with. Basic > transimpedance amp using > a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 > with 30pf comp cap. > I've also added a small bypass cap on the feedback resistor to flatten > out the response at 500kHz. > The output I feed through a bypass cap into my SDR-IQ software defined > radio. Lot's of fun playing > with these things and they seem to work real well. Good for me. > </background> > > <question> > If I shine an incandescent bulb on the detector and crank up the light > so I get roughly 100 uA photo > current why is the "wave noise" so small compared to the DC output of > the detector? From theory > the power density spectrum, S(w), of the optical field is the Fourier > transform of the field time > correlation function <E(t)E(t-t')>. The field correlation function as > far as my detector goes is a delta > function in time whose integral should be the light power flux at DC. > What happens in reality is S(w) > has a huge spike at DC (the constant light flux) and then a really > small amount of noise off DC (probably > shot noise). The math, which I think I understand but can't connect > with the physics is that S(0) > should be equal to S(100kHz). Clearly this isn't the case. Why? > </question> > > Thanks > Paul C.Hi Paul, If you are talking about wave noise ala Hanburry-Brown. Then this is very hard to see. (It would take me a bit to write down the correct equations... And then I'd still likely make a mistake :^) But the parameters include, BW of source, source size, wavlength of light, and source to detector distance. Anyway when all said and done I think this leads to (at best) a doubling of the shot noise. Your incandescent light is too broad in spectrum, and too large in area. I once spent a day trying to see the excess noise with our Rubidium lamp shinning through a very small hole from across the lab.... Didn't work. (For reasons I now uderstand.) I've got Hanburry-Brown's book at home and I can pull the equations from that if you would like. (Maybe you can find them on the web?) George H. (and indeed this is one of Phil H's favorite subjects.. so he may fill in the details.)

Reply by ●November 30, 20122012-11-30

On 2012-11-30 07:58:56 -0800, George Herold said:> On Nov 29, 7:49�pm, Paul Colby <paulcco...@earthlink.net> wrote: >> Hi, >> <background> >> This really isn't a circuit question but people here likely understand >> what I'm finding so confusing. >> I've built several AC coupled photodetectors to play around with. Basic >> transimpedance amp using >> a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 >> with 30pf comp cap. >> I've also added a small bypass cap on the feedback resistor to flatten >> out the response at 500kHz. >> The output I feed through a bypass cap into my SDR-IQ software defined >> radio. Lot's of fun playing >> with these things and they seem to work real well. Good for me. >> </background> >> >> <question> >> If I shine an incandescent bulb on the detector and crank up the light >> so I get roughly 100 uA photo >> current why is the "wave noise" so small compared to the DC output of >> the detector? From theory >> the power density spectrum, S(w), of the optical field is the Fourier >> transform of the field time >> correlation function <E(t)E(t-t')>. The field correlation function as >> far as my detector goes is a delta >> function in time whose integral should be the light power flux at DC. >> What happens in reality is S(w) >> has a huge spike at DC (the constant light flux) and then a really >> small amount of noise off DC (probably >> shot noise). The math, which I think I understand but can't connect >> with the physics is that S(0) >> should be equal to S(100kHz). Clearly this isn't the case. Why? >> </question> >> >> Thanks >> Paul C. > > Hi Paul, If you are talking about wave noise ala Hanburry-Brown. > Then this is very hard to see. (It would take me a bit to write down > the correct equations... And then I'd still likely make a mistake :^) > But the parameters include, BW of source, source size, wavlength of > light, and source to detector distance. Anyway when all said and done > I think this leads to (at best) a doubling of the shot noise. > Your incandescent light is too broad in spectrum, and too large in > area. > I once spent a day trying to see the excess noise with our Rubidium > lamp shinning through a very small hole from across the lab.... Didn't > work. (For reasons I now uderstand.) > > I've got Hanburry-Brown's book at home and I can pull the equations > from that if you would like. (Maybe you can find them on the web?) > > George H. > > (and indeed this is one of Phil H's favorite subjects.. so he may fill > in the details.)Yes, the light slowly dawns. I think someone here posted a pdf of Hanbury Brown's book which is a real good read. I've been using octave to jinn up some band limited random phase signals. The resulting intensity power spectrum indeed has a massive DC spike for reasons that are now apparent. I think it's somewhere around chapter 4 that HB goes through wave versus shot noise in a plane wave. Based on what I've read there shot and wave noise should be about equal for a DC photocurrent of 300 uA. But alas this is for a plane wave and not my light bulb (or the sun for that matter) because light from different directions add by intensity and not by field. So, would one stand a chance at seeing white light wave noise if one first couples the light into a fiber before going to the diode? The thought here is that I would have a spatially coherent spot on my diode so the HB plane wave argument would be closer to truth?? Thanks Paul C.

Reply by ●November 30, 20122012-11-30

On 2012-11-30 07:47:35 -0800, Phil Hobbs said:>> <question> >> If I shine an incandescent bulb on the detector and crank up the light >> so I get roughly 100 uA photo >> current why is the "wave noise" so small compared to the DC output of >> the detector? From theory >> the power density spectrum, S(w), of the optical field is the Fourier >> transform of the field time >> correlation function <E(t)E(t-t')>. The field correlation function as >> far as my detector goes is a delta >> function in time whose integral should be the light power flux at DC. >> What happens in reality is S(w) >> has a huge spike at DC (the constant light flux) and then a really small >> amount of noise off DC (probably >> shot noise). The math, which I think I understand but can't connect with >> the physics is that S(0) >> should be equal to S(100kHz). Clearly this isn't the case. Why? >> </question> >> >> Thanks >> Paul C. >> > > The fluctuations are white, but the DC is separate. > > Cheers > > Phil HobbsActually, is it? The arithmetic should work at DC if it's the right arithmetic. I've done some numerical experimenting with random phase band limited signals and indeed there is a large DC spike in the power fluctuation spectrum for reasons that are apparent. My argument above must be confused somehow... Thanks Paul C.

Reply by ●November 30, 20122012-11-30

On Nov 30, 11:40=A0am, Paul Colby <paulcco...@paulccolby.com> wrote:> On 2012-11-30 07:58:56 -0800, George Herold said: > > > > > > > On Nov 29, 7:49 pm, Paul Colby <paulcco...@earthlink.net> wrote: > >> Hi, > >> <background> > >> This really isn't a circuit question but people here likely understand > >> what I'm finding so confusing. > >> I've built several AC coupled photodetectors to play around with. Basi=c> >> transimpedance amp using > >> a 0.8 m^2 diode with 150K feedback resistor. The opamp is an old LM301 > >> with 30pf comp cap. > >> I've also added a small bypass cap on the feedback resistor to flatten > >> out the response at 500kHz. > >> The output I feed through a bypass cap into my SDR-IQ software defined > >> radio. Lot's of fun playing > >> with these things and they seem to work real well. Good for me. > >> </background> > > >> <question> > >> If I shine an incandescent bulb on the detector and crank up the light > >> so I get roughly 100 uA photo > >> current why is the "wave noise" so small compared to the DC output of > >> the detector? From theory > >> the power density spectrum, S(w), of the optical field is the Fourier > >> transform of the field time > >> correlation function <E(t)E(t-t')>. The field correlation function as > >> far as my detector goes is a delta > >> function in time whose integral should be the light power flux at DC. > >> What happens in reality is S(w) > >> has a huge spike at DC (the constant light flux) and then a really > >> small amount of noise off DC (probably > >> shot noise). The math, which I think I understand but can't connect > >> with the physics is that S(0) > >> should be equal to S(100kHz). Clearly this isn't the case. Why? > >> </question> > > >> Thanks > >> Paul C. > > > Hi Paul, =A0If you are talking about wave noise ala Hanburry-Brown. > > Then this is very hard to see. =A0(It would take me a bit to write down > > the correct equations... And then I'd still likely make a mistake :^) > > But the parameters include, BW of source, source size, wavlength of > > light, and source to detector distance. =A0Anyway when all said and don=e> > I think this leads to (at best) a doubling of the shot noise. > > Your incandescent light is too broad in spectrum, and too large in > > area. > > I once spent a day trying to see the excess noise with our Rubidium > > lamp shinning through a very small hole from across the lab.... Didn't > > work. =A0(For reasons I now uderstand.) > > > I've got Hanburry-Brown's book at home and I can pull the equations > > from that if you would like. =A0(Maybe you can find them on the web?) > > > George H. > > > (and indeed this is one of Phil H's favorite subjects.. so he may fill > > in the details.) > > Yes, the light slowly dawns. I think someone here posted a pdf > of Hanbury Brown's book which is a real good read. I've been > using octave to jinn up some band limited random phase signals. > The resulting intensity power spectrum indeed has a massive > DC spike for reasons that are now apparent. I think it's somewhere > around chapter 4 that HB goes through wave versus shot noise > in a plane wave. Based on what I've read there shot and wave > noise should be about equal for a DC photocurrent of 300 uA. > But alas this is for a plane wave and not my light bulb (or the sun > for that matter) because light from different directions add by > intensity and not by field.Chapter 4 sound right. I don't think there's any absolute current that is needed. But there is a relation between the photon flux and the bandwidth. And the signal goes as the probability that two photons arrive at the same 'time'. For this case I take the photon 'length' to be given by ~ 1 over the bandwidth.. with the appropriate scaling factors. There's also a spatial 'overlap' condition that has to be satisfied. So that goes as the QM probability that an event (the detection of two photons) happens in two different ways. I've pictured one, one photon leaves S1 and arrives at det1 and 2 to 2. The other is that S1 goes to det2 and S2 to det1. (harder to draw.) ^ S1..........................det1 ^ | | |D d | | V S2..........................det2 V <-----------L--------------> When you work all that out. (several pages) You get a condition that D*d/L*lambda < 1 (or 1/2 maybe?) (lambda is the wavelength) Now you can think of the S's as different sources and the det's as different detectors or just the size of one source (detector.)> > So, would one stand a chance at seeing white light wave noise > if one first couples the light into a fiber before going to the diode? > The thought here is that I would have a spatially coherent spot > on my diode so the HB plane wave argument would be closer to > truth??Well I'm sorry to say but I don't think so. (But I'm far from an expert.) I think the end of your fiber becomes the new source.... or converesly the input to the fiber is your detector. In my case I thought is was permissible to put a lens between the source and detector. I was able to get lots of light, but no excess noise. I'd still like to do this in the lab some day. Phil H. claims it's not that hard, but I think we each have different 'hardness scales'. His talc is my diamond :^) George H.> > Thanks > Paul C.- Hide quoted text - > > - Show quoted text -