TIA Photodiode Bootstrap at 10MHz
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From: Phil Hobbs on 4 Nov 2009 23:12 Artist wrote: > Phil Hobbs wrote: >> Artist wrote: >>> Phil Hobbs wrote: >>>> Artist wrote: >>>>> I need to bootstrap a photodiode in a TIA circuit similar to the way >>>>> it is done as shown on page 18 of: >>>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >>>>> This example is much too limited in bandwidth. I need a 10MHz >>>>> bandwidth. >>>>> >>>>> The bootstrapping is needed because of the low impedance of the >>>>> photodiode. This is 150pF in parallel with 1 Kohm. The problem is one >>>>> of designing a 10MHz unity gain amplifier with high impedance input, >>>>> low noise, negligible phase change, and unity gain. >>>>> >>>>> Does anyone have any ideas? I am not sure it can be done. >>>>> >>>> >>>> One method is to connect the PD directly to the input of a nice quiet >>>> 50-ohm amplifier. If you have at least 200 uA of photocurrent, this >>>> will >>>> work very well--you can get to the shot noise limit that way. >>>> >>>> At lower photocurrents, life gets a bit harder. Your particular problem >>>> gets quite difficult below about 20 uA--at that point you have to start >>>> trading away SNR or reducing that capacitance. The best Si PIN diodes >>>> have a capacitance of 40-100 pF/cm**2 when reverse biased, so if >>>> your PD >>>> isn't at least a half inch square, you can reduce the capacitance by >>>> choosing a different PD and/or reverse biasing. >>>> >>>> So how big a photocurrent are you expecting, and what's your SNR >>>> target? >>>> >>>> Cheers >>>> >>>> Phil Hobbs >>>> >>> The peak current is expected to be 1 uA. >> >> If there's a way to make that 10 uA, your life will be much easier. >>> >>> The latest value for the capacitance I have is now 30pF. >>> >>> I do not have a choice on photodiodes. The detector I have been >>> assigned to make work for this project is not actually a photodiode in >>> the conventional sense. It is a custom made photoelectromotive force >>> detector for use in a laser ultrasonics application. This device >>> cannot be reverse biased like a PIN diode. >> >>> >>> A major concern about the low series resistance is that it will create >>> a high gain noninverting amplifier with the feedback resistor for the >>> equivalent input noise on the inverting input. This gain will also >>> reduce the bandwidth of the opamp circuit. >>> >>> The zero the capacitance will make is another reason I am looking to >>> bootstrap this. >>> >> >> Bootstraps have the same noise multiplication problem as TIAs, for the >> same reason: they put their own noise voltage across the PD capacitance. >> With equivalent devices, you can get a 3 dB improvement by using both, >> but bootstrapping is not a slam dunk. One good thing about it is that >> you can AC-couple the bootstrap, which means it can be single-ended >> rather than differential. >> >> You can get the same 3 dB improvement by putting a TIA on each end of >> the PD. >> >> If it's a photoacoustic measurement, you may not need DC-10 MHz. What's >> the actual measurement bandwidth? >> >> Cheers >> >> Phil Hobbs >> > The low end of the bandwidth is 100kHz. > > A bootstrap does not necessarily add noise to the circuit. If the > bootstrap amplifier has less noise than the op amp its noise is swapped > for that of the op amp. This measured reduction in noise is documented > in figures 4b, 5b, and 6b on pages 17 and 18 of: > http://cds.linear.com/docs/Datasheet/6244fa.pdf > > Suppose I returned the photoelectromotive detector to ground. Since the > expected detector current is 1uA, to get some reasonable output from > this first TIA stage I would need a feedback resistor on the order of > 1Mohm. Since the series resistance of the detector is 1kOhm there is a > noise gain of 1001 before the zero from the 30pF capacitance kicks in at > 5.3MHz. Taking into account the DC noise gain, the capacitance on the > virtual ground (dominated by detector capacitance) and the needed > bandwidth of the TIA (10MHz) I would need an opamp with an 18.8THz GBW. > If it weren't for the noise gain I would have needed only 18.8MHz. The > only way this is going to be done is by means of bootstrapping to reduce > the effective impedance of the detector. > In general you can make the TIA out of the same type of device as the bootstrap, which is what I'm assuming. If you have a noise problem, then reducing the TIA noise is job 1, followed by bootstrapping. Using the same device type in the TIA and the bootstrap gives a noise level about 3 dB better than either one alone. A quiet bootstrap plus a noisy TIA is a lot better than just the noisy TIA barefoot, but not as good as a quiet bootstrap and a quiet TIA. The resistance of the detector is in series with the capacitance, not in parallel, so it looks like a lead-lag network. The capacitance makes a feedback zero, and the resistance puts in a zero. The noise gain may be 1000 at high frequency, but not at low frequency. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
From: Artist on 5 Nov 2009 12:40 Phil Hobbs wrote: > Artist wrote: >> Phil Hobbs wrote: >>> Artist wrote: >>>> Phil Hobbs wrote: >>>>> Artist wrote: >>>>>> I need to bootstrap a photodiode in a TIA circuit similar to the way >>>>>> it is done as shown on page 18 of: >>>>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >>>>>> This example is much too limited in bandwidth. I need a 10MHz >>>>>> bandwidth. >>>>>> >>>>>> The bootstrapping is needed because of the low impedance of the >>>>>> photodiode. This is 150pF in parallel with 1 Kohm. The problem is one >>>>>> of designing a 10MHz unity gain amplifier with high impedance input, >>>>>> low noise, negligible phase change, and unity gain. >>>>>> >>>>>> Does anyone have any ideas? I am not sure it can be done. >>>>>> >>>>> >>>>> One method is to connect the PD directly to the input of a nice quiet >>>>> 50-ohm amplifier. If you have at least 200 uA of photocurrent, this >>>>> will >>>>> work very well--you can get to the shot noise limit that way. >>>>> >>>>> At lower photocurrents, life gets a bit harder. Your particular >>>>> problem >>>>> gets quite difficult below about 20 uA--at that point you have to >>>>> start >>>>> trading away SNR or reducing that capacitance. The best Si PIN diodes >>>>> have a capacitance of 40-100 pF/cm**2 when reverse biased, so if >>>>> your PD >>>>> isn't at least a half inch square, you can reduce the capacitance by >>>>> choosing a different PD and/or reverse biasing. >>>>> >>>>> So how big a photocurrent are you expecting, and what's your SNR >>>>> target? >>>>> >>>>> Cheers >>>>> >>>>> Phil Hobbs >>>>> >>>> The peak current is expected to be 1 uA. >>> >>> If there's a way to make that 10 uA, your life will be much easier. >>>> >>>> The latest value for the capacitance I have is now 30pF. >>>> >>>> I do not have a choice on photodiodes. The detector I have been >>>> assigned to make work for this project is not actually a photodiode in >>>> the conventional sense. It is a custom made photoelectromotive force >>>> detector for use in a laser ultrasonics application. This device >>>> cannot be reverse biased like a PIN diode. >>> >>>> >>>> A major concern about the low series resistance is that it will create >>>> a high gain noninverting amplifier with the feedback resistor for the >>>> equivalent input noise on the inverting input. This gain will also >>>> reduce the bandwidth of the opamp circuit. >>>> >>>> The zero the capacitance will make is another reason I am looking to >>>> bootstrap this. >>>> >>> >>> Bootstraps have the same noise multiplication problem as TIAs, for the >>> same reason: they put their own noise voltage across the PD capacitance. >>> With equivalent devices, you can get a 3 dB improvement by using both, >>> but bootstrapping is not a slam dunk. One good thing about it is that >>> you can AC-couple the bootstrap, which means it can be single-ended >>> rather than differential. >>> >>> You can get the same 3 dB improvement by putting a TIA on each end of >>> the PD. >>> >>> If it's a photoacoustic measurement, you may not need DC-10 MHz. What's >>> the actual measurement bandwidth? >>> >>> Cheers >>> >>> Phil Hobbs >>> >> The low end of the bandwidth is 100kHz. >> >> A bootstrap does not necessarily add noise to the circuit. If the >> bootstrap amplifier has less noise than the op amp its noise is >> swapped for that of the op amp. This measured reduction in noise is >> documented in figures 4b, 5b, and 6b on pages 17 and 18 of: >> http://cds.linear.com/docs/Datasheet/6244fa.pdf >> >> Suppose I returned the photoelectromotive detector to ground. Since >> the expected detector current is 1uA, to get some reasonable output >> from this first TIA stage I would need a feedback resistor on the >> order of 1Mohm. Since the series resistance of the detector is 1kOhm >> there is a noise gain of 1001 before the zero from the 30pF >> capacitance kicks in at 5.3MHz. Taking into account the DC noise gain, >> the capacitance on the virtual ground (dominated by detector >> capacitance) and the needed bandwidth of the TIA (10MHz) I would need >> an opamp with an 18.8THz GBW. If it weren't for the noise gain I would >> have needed only 18.8MHz. The only way this is going to be done is by >> means of bootstrapping to reduce the effective impedance of the detector. >> > > In general you can make the TIA out of the same type of device as the > bootstrap, which is what I'm assuming. If you have a noise problem, > then reducing the TIA noise is job 1, followed by bootstrapping. Using > the same device type in the TIA and the bootstrap gives a noise level > about 3 dB better than either one alone. > > A quiet bootstrap plus a noisy TIA is a lot better than just the noisy > TIA barefoot, but not as good as a quiet bootstrap and a quiet TIA. > > The resistance of the detector is in series with the capacitance, not in > parallel, so it looks like a lead-lag network. The capacitance makes a > feedback zero, and the resistance puts in a zero. The noise gain may be > 1000 at high frequency, but not at low frequency. > > Cheers > > Phil Hobbs > I have been told by the project engineer who has given me this assignment the capacitance is a parallel one. It is a custom detector device so there is no data sheet I can link to. I have not even been given a hard copy of one. So I can only go by what he has told me. Do you have knowledge of photoelectromotive force detector physics that causes you to disagree? Right now my main concern is just getting the bandwidth. Once one or more ways to do that are established I can be concerned about minimizing noise. I have not until now considered using another op amp instead of an amp composed of discrete devices to do the bootstrapping. Doing it this way would reduce added capacitance on the virtual ground. Regarding my earlier posting that was 18.8GHz not 18.8THz. -- If you desire to respond directly remove the "sj." from the domain name part of my email address. It is a spam jammer.
From: George Herold on 5 Nov 2009 22:21 On Nov 5, 12:40 pm, Artist <Art...(a)sj.speakeasy.net> wrote: > Phil Hobbs wrote: > > Artist wrote: > >> Phil Hobbs wrote: > >>> Artist wrote: > >>>> Phil Hobbs wrote: > >>>>> Artist wrote: > >>>>>> I need to bootstrap a photodiode in a TIA circuit similar to the way > >>>>>> it is done as shown on page 18 of: > >>>>>>http://cds.linear.com/docs/Datasheet/6244fa.pdf > >>>>>> This example is much too limited in bandwidth. I need a 10MHz > >>>>>> bandwidth. > > >>>>>> The bootstrapping is needed because of the low impedance of the > >>>>>> photodiode. This is 150pF in parallel with 1 Kohm. The problem is one > >>>>>> of designing a 10MHz unity gain amplifier with high impedance input, > >>>>>> low noise, negligible phase change, and unity gain. > > >>>>>> Does anyone have any ideas? I am not sure it can be done. > > >>>>> One method is to connect the PD directly to the input of a nice quiet > >>>>> 50-ohm amplifier. If you have at least 200 uA of photocurrent, this > >>>>> will > >>>>> work very well--you can get to the shot noise limit that way. > > >>>>> At lower photocurrents, life gets a bit harder. Your particular > >>>>> problem > >>>>> gets quite difficult below about 20 uA--at that point you have to > >>>>> start > >>>>> trading away SNR or reducing that capacitance. The best Si PIN diodes > >>>>> have a capacitance of 40-100 pF/cm**2 when reverse biased, so if > >>>>> your PD > >>>>> isn't at least a half inch square, you can reduce the capacitance by > >>>>> choosing a different PD and/or reverse biasing. > > >>>>> So how big a photocurrent are you expecting, and what's your SNR > >>>>> target? > > >>>>> Cheers > > >>>>> Phil Hobbs > > >>>> The peak current is expected to be 1 uA. > > >>> If there's a way to make that 10 uA, your life will be much easier. > > >>>> The latest value for the capacitance I have is now 30pF. > > >>>> I do not have a choice on photodiodes. The detector I have been > >>>> assigned to make work for this project is not actually a photodiode in > >>>> the conventional sense. It is a custom made photoelectromotive force > >>>> detector for use in a laser ultrasonics application. This device > >>>> cannot be reverse biased like a PIN diode. > > >>>> A major concern about the low series resistance is that it will create > >>>> a high gain noninverting amplifier with the feedback resistor for the > >>>> equivalent input noise on the inverting input. This gain will also > >>>> reduce the bandwidth of the opamp circuit. > > >>>> The zero the capacitance will make is another reason I am looking to > >>>> bootstrap this. > > >>> Bootstraps have the same noise multiplication problem as TIAs, for the > >>> same reason: they put their own noise voltage across the PD capacitance. > >>> With equivalent devices, you can get a 3 dB improvement by using both, > >>> but bootstrapping is not a slam dunk. One good thing about it is that > >>> you can AC-couple the bootstrap, which means it can be single-ended > >>> rather than differential. > > >>> You can get the same 3 dB improvement by putting a TIA on each end of > >>> the PD. > > >>> If it's a photoacoustic measurement, you may not need DC-10 MHz. What's > >>> the actual measurement bandwidth? > > >>> Cheers > > >>> Phil Hobbs > > >> The low end of the bandwidth is 100kHz. > > >> A bootstrap does not necessarily add noise to the circuit. If the > >> bootstrap amplifier has less noise than the op amp its noise is > >> swapped for that of the op amp. This measured reduction in noise is > >> documented in figures 4b, 5b, and 6b on pages 17 and 18 of: > >>http://cds.linear.com/docs/Datasheet/6244fa.pdf > > >> Suppose I returned the photoelectromotive detector to ground. Since > >> the expected detector current is 1uA, to get some reasonable output > >> from this first TIA stage I would need a feedback resistor on the > >> order of 1Mohm. Since the series resistance of the detector is 1kOhm > >> there is a noise gain of 1001 before the zero from the 30pF > >> capacitance kicks in at 5.3MHz. Taking into account the DC noise gain, > >> the capacitance on the virtual ground (dominated by detector > >> capacitance) and the needed bandwidth of the TIA (10MHz) I would need > >> an opamp with an 18.8THz GBW. If it weren't for the noise gain I would > >> have needed only 18.8MHz. The only way this is going to be done is by > >> means of bootstrapping to reduce the effective impedance of the detector. > > > In general you can make the TIA out of the same type of device as the > > bootstrap, which is what I'm assuming. If you have a noise problem, > > then reducing the TIA noise is job 1, followed by bootstrapping. Using > > the same device type in the TIA and the bootstrap gives a noise level > > about 3 dB better than either one alone. > > > A quiet bootstrap plus a noisy TIA is a lot better than just the noisy > > TIA barefoot, but not as good as a quiet bootstrap and a quiet TIA. > > > The resistance of the detector is in series with the capacitance, not in > > parallel, so it looks like a lead-lag network. The capacitance makes a > > feedback zero, and the resistance puts in a zero. The noise gain may be > > 1000 at high frequency, but not at low frequency. > > > Cheers > > > Phil Hobbs > > I have been told by the project engineer who has given me this > assignment the capacitance is a parallel one. It is a custom detector > device so there is no data sheet I can link to. I have not even been > given a hard copy of one. So I can only go by what he has told me. Do > you have knowledge of photoelectromotive force detector physics that > causes you to disagree? > > Right now my main concern is just getting the bandwidth. Once one or > more ways to do that are established I can be concerned about minimizing > noise. > > I have not until now considered using another op amp instead of an amp > composed of discrete devices to do the bootstrapping. Doing it this way > would reduce added capacitance on the virtual ground. > > Regarding my earlier posting that was 18.8GHz not 18.8THz. > > -- > If you desire to respond directly remove the "sj." from the domain name > part of my email address. It is a spam jammer.- Hide quoted text - > > - Show quoted text - "I have been told by the project engineer who has given me this > assignment the capacitance is a parallel one." That sounds much worse. It's like some photo-resistor.? It will always be leaking as you bias it. (and leaking as the light shines on it.) ughh. George H.
From: Artist on 5 Nov 2009 23:46 George Herold wrote: > On Nov 5, 12:40 pm, Artist<Art...(a)sj.speakeasy.net> wrote: >> Phil Hobbs wrote: >>> Artist wrote: >>>> Phil Hobbs wrote: >>>>> Artist wrote: >>>>>> Phil Hobbs wrote: >>>>>>> Artist wrote: >>>>>>>> I need to bootstrap a photodiode in a TIA circuit similar to the way >>>>>>>> it is done as shown on page 18 of: >>>>>>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >>>>>>>> This example is much too limited in bandwidth. I need a 10MHz >>>>>>>> bandwidth. >> >>>>>>>> The bootstrapping is needed because of the low impedance of the >>>>>>>> photodiode. This is 150pF in parallel with 1 Kohm. The problem is one >>>>>>>> of designing a 10MHz unity gain amplifier with high impedance input, >>>>>>>> low noise, negligible phase change, and unity gain. >> >>>>>>>> Does anyone have any ideas? I am not sure it can be done. >> >>>>>>> One method is to connect the PD directly to the input of a nice quiet >>>>>>> 50-ohm amplifier. If you have at least 200 uA of photocurrent, this >>>>>>> will >>>>>>> work very well--you can get to the shot noise limit that way. >> >>>>>>> At lower photocurrents, life gets a bit harder. Your particular >>>>>>> problem >>>>>>> gets quite difficult below about 20 uA--at that point you have to >>>>>>> start >>>>>>> trading away SNR or reducing that capacitance. The best Si PIN diodes >>>>>>> have a capacitance of 40-100 pF/cm**2 when reverse biased, so if >>>>>>> your PD >>>>>>> isn't at least a half inch square, you can reduce the capacitance by >>>>>>> choosing a different PD and/or reverse biasing. >> >>>>>>> So how big a photocurrent are you expecting, and what's your SNR >>>>>>> target? >> >>>>>>> Cheers >> >>>>>>> Phil Hobbs >> >>>>>> The peak current is expected to be 1 uA. >> >>>>> If there's a way to make that 10 uA, your life will be much easier. >> >>>>>> The latest value for the capacitance I have is now 30pF. >> >>>>>> I do not have a choice on photodiodes. The detector I have been >>>>>> assigned to make work for this project is not actually a photodiode in >>>>>> the conventional sense. It is a custom made photoelectromotive force >>>>>> detector for use in a laser ultrasonics application. This device >>>>>> cannot be reverse biased like a PIN diode. >> >>>>>> A major concern about the low series resistance is that it will create >>>>>> a high gain noninverting amplifier with the feedback resistor for the >>>>>> equivalent input noise on the inverting input. This gain will also >>>>>> reduce the bandwidth of the opamp circuit. >> >>>>>> The zero the capacitance will make is another reason I am looking to >>>>>> bootstrap this. >> >>>>> Bootstraps have the same noise multiplication problem as TIAs, for the >>>>> same reason: they put their own noise voltage across the PD capacitance. >>>>> With equivalent devices, you can get a 3 dB improvement by using both, >>>>> but bootstrapping is not a slam dunk. One good thing about it is that >>>>> you can AC-couple the bootstrap, which means it can be single-ended >>>>> rather than differential. >> >>>>> You can get the same 3 dB improvement by putting a TIA on each end of >>>>> the PD. >> >>>>> If it's a photoacoustic measurement, you may not need DC-10 MHz. What's >>>>> the actual measurement bandwidth? >> >>>>> Cheers >> >>>>> Phil Hobbs >> >>>> The low end of the bandwidth is 100kHz. >> >>>> A bootstrap does not necessarily add noise to the circuit. If the >>>> bootstrap amplifier has less noise than the op amp its noise is >>>> swapped for that of the op amp. This measured reduction in noise is >>>> documented in figures 4b, 5b, and 6b on pages 17 and 18 of: >>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >> >>>> Suppose I returned the photoelectromotive detector to ground. Since >>>> the expected detector current is 1uA, to get some reasonable output >>>> from this first TIA stage I would need a feedback resistor on the >>>> order of 1Mohm. Since the series resistance of the detector is 1kOhm >>>> there is a noise gain of 1001 before the zero from the 30pF >>>> capacitance kicks in at 5.3MHz. Taking into account the DC noise gain, >>>> the capacitance on the virtual ground (dominated by detector >>>> capacitance) and the needed bandwidth of the TIA (10MHz) I would need >>>> an opamp with an 18.8THz GBW. If it weren't for the noise gain I would >>>> have needed only 18.8MHz. The only way this is going to be done is by >>>> means of bootstrapping to reduce the effective impedance of the detector. >> >>> In general you can make the TIA out of the same type of device as the >>> bootstrap, which is what I'm assuming. If you have a noise problem, >>> then reducing the TIA noise is job 1, followed by bootstrapping. Using >>> the same device type in the TIA and the bootstrap gives a noise level >>> about 3 dB better than either one alone. >> >>> A quiet bootstrap plus a noisy TIA is a lot better than just the noisy >>> TIA barefoot, but not as good as a quiet bootstrap and a quiet TIA. >> >>> The resistance of the detector is in series with the capacitance, not in >>> parallel, so it looks like a lead-lag network. The capacitance makes a >>> feedback zero, and the resistance puts in a zero. The noise gain may be >>> 1000 at high frequency, but not at low frequency. >> >>> Cheers >> >>> Phil Hobbs >> >> I have been told by the project engineer who has given me this >> assignment the capacitance is a parallel one. It is a custom detector >> device so there is no data sheet I can link to. I have not even been >> given a hard copy of one. So I can only go by what he has told me. Do >> you have knowledge of photoelectromotive force detector physics that >> causes you to disagree? >> >> Right now my main concern is just getting the bandwidth. Once one or >> more ways to do that are established I can be concerned about minimizing >> noise. >> >> I have not until now considered using another op amp instead of an amp >> composed of discrete devices to do the bootstrapping. Doing it this way >> would reduce added capacitance on the virtual ground. >> >> Regarding my earlier posting that was 18.8GHz not 18.8THz. >> >> -- >> If you desire to respond directly remove the "sj." from the domain name >> part of my email address. It is a spam jammer.- Hide quoted text - >> >> - Show quoted text - > > "I have been told by the project engineer who has given me this >> assignment the capacitance is a parallel one." > > That sounds much worse. It's like some photo-resistor.? It will > always be leaking as you bias it. (and leaking as the light shines on > it.) ughh. > > > George H. > > Similar, but unlike the passive photoresistor, the electromotive force detector converts photons into electrical energy. In a TIA circuit it can be modeled as a current source that does not require a bias. The current is proportional to the motion of a beam light across it. It is very good at detecting changes in a laser light's speckled pattern produced when it reflects off any but the most even of surfaces. Right now my plan is get two of the lowest noise JFET or CMOS input op amps I can find that satisfy the bandwidth requirement. One will be used for the feedback resistor and the other will do the bootstrap. -- To reply directly remove the sj. from my email address. This is a spam jammer.
From: Fred Bartoli on 6 Nov 2009 02:37
Artist a �crit : > George Herold wrote: >> On Nov 5, 12:40 pm, Artist<Art...(a)sj.speakeasy.net> wrote: >>> Phil Hobbs wrote: >>>> Artist wrote: >>>>> Phil Hobbs wrote: >>>>>> Artist wrote: >>>>>>> Phil Hobbs wrote: >>>>>>>> Artist wrote: >>>>>>>>> I need to bootstrap a photodiode in a TIA circuit similar to >>>>>>>>> the way >>>>>>>>> it is done as shown on page 18 of: >>>>>>>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >>>>>>>>> This example is much too limited in bandwidth. I need a 10MHz >>>>>>>>> bandwidth. >>> >>>>>>>>> The bootstrapping is needed because of the low impedance of the >>>>>>>>> photodiode. This is 150pF in parallel with 1 Kohm. The problem >>>>>>>>> is one >>>>>>>>> of designing a 10MHz unity gain amplifier with high impedance >>>>>>>>> input, >>>>>>>>> low noise, negligible phase change, and unity gain. >>> >>>>>>>>> Does anyone have any ideas? I am not sure it can be done. >>> >>>>>>>> One method is to connect the PD directly to the input of a nice >>>>>>>> quiet >>>>>>>> 50-ohm amplifier. If you have at least 200 uA of photocurrent, this >>>>>>>> will >>>>>>>> work very well--you can get to the shot noise limit that way. >>> >>>>>>>> At lower photocurrents, life gets a bit harder. Your particular >>>>>>>> problem >>>>>>>> gets quite difficult below about 20 uA--at that point you have to >>>>>>>> start >>>>>>>> trading away SNR or reducing that capacitance. The best Si PIN >>>>>>>> diodes >>>>>>>> have a capacitance of 40-100 pF/cm**2 when reverse biased, so if >>>>>>>> your PD >>>>>>>> isn't at least a half inch square, you can reduce the >>>>>>>> capacitance by >>>>>>>> choosing a different PD and/or reverse biasing. >>> >>>>>>>> So how big a photocurrent are you expecting, and what's your SNR >>>>>>>> target? >>> >>>>>>>> Cheers >>> >>>>>>>> Phil Hobbs >>> >>>>>>> The peak current is expected to be 1 uA. >>> >>>>>> If there's a way to make that 10 uA, your life will be much easier. >>> >>>>>>> The latest value for the capacitance I have is now 30pF. >>> >>>>>>> I do not have a choice on photodiodes. The detector I have been >>>>>>> assigned to make work for this project is not actually a >>>>>>> photodiode in >>>>>>> the conventional sense. It is a custom made photoelectromotive force >>>>>>> detector for use in a laser ultrasonics application. This device >>>>>>> cannot be reverse biased like a PIN diode. >>> >>>>>>> A major concern about the low series resistance is that it will >>>>>>> create >>>>>>> a high gain noninverting amplifier with the feedback resistor for >>>>>>> the >>>>>>> equivalent input noise on the inverting input. This gain will also >>>>>>> reduce the bandwidth of the opamp circuit. >>> >>>>>>> The zero the capacitance will make is another reason I am looking to >>>>>>> bootstrap this. >>> >>>>>> Bootstraps have the same noise multiplication problem as TIAs, for >>>>>> the >>>>>> same reason: they put their own noise voltage across the PD >>>>>> capacitance. >>>>>> With equivalent devices, you can get a 3 dB improvement by using >>>>>> both, >>>>>> but bootstrapping is not a slam dunk. One good thing about it is that >>>>>> you can AC-couple the bootstrap, which means it can be single-ended >>>>>> rather than differential. >>> >>>>>> You can get the same 3 dB improvement by putting a TIA on each end of >>>>>> the PD. >>> >>>>>> If it's a photoacoustic measurement, you may not need DC-10 MHz. >>>>>> What's >>>>>> the actual measurement bandwidth? >>> >>>>>> Cheers >>> >>>>>> Phil Hobbs >>> >>>>> The low end of the bandwidth is 100kHz. >>> >>>>> A bootstrap does not necessarily add noise to the circuit. If the >>>>> bootstrap amplifier has less noise than the op amp its noise is >>>>> swapped for that of the op amp. This measured reduction in noise is >>>>> documented in figures 4b, 5b, and 6b on pages 17 and 18 of: >>>>> http://cds.linear.com/docs/Datasheet/6244fa.pdf >>> >>>>> Suppose I returned the photoelectromotive detector to ground. Since >>>>> the expected detector current is 1uA, to get some reasonable output >>>>> from this first TIA stage I would need a feedback resistor on the >>>>> order of 1Mohm. Since the series resistance of the detector is 1kOhm >>>>> there is a noise gain of 1001 before the zero from the 30pF >>>>> capacitance kicks in at 5.3MHz. Taking into account the DC noise gain, >>>>> the capacitance on the virtual ground (dominated by detector >>>>> capacitance) and the needed bandwidth of the TIA (10MHz) I would need >>>>> an opamp with an 18.8THz GBW. If it weren't for the noise gain I would >>>>> have needed only 18.8MHz. The only way this is going to be done is by >>>>> means of bootstrapping to reduce the effective impedance of the >>>>> detector. >>> >>>> In general you can make the TIA out of the same type of device as the >>>> bootstrap, which is what I'm assuming. If you have a noise problem, >>>> then reducing the TIA noise is job 1, followed by bootstrapping. Using >>>> the same device type in the TIA and the bootstrap gives a noise level >>>> about 3 dB better than either one alone. >>> >>>> A quiet bootstrap plus a noisy TIA is a lot better than just the noisy >>>> TIA barefoot, but not as good as a quiet bootstrap and a quiet TIA. >>> >>>> The resistance of the detector is in series with the capacitance, >>>> not in >>>> parallel, so it looks like a lead-lag network. The capacitance makes a >>>> feedback zero, and the resistance puts in a zero. The noise gain >>>> may be >>>> 1000 at high frequency, but not at low frequency. >>> >>>> Cheers >>> >>>> Phil Hobbs >>> >>> I have been told by the project engineer who has given me this >>> assignment the capacitance is a parallel one. It is a custom detector >>> device so there is no data sheet I can link to. I have not even been >>> given a hard copy of one. So I can only go by what he has told me. Do >>> you have knowledge of photoelectromotive force detector physics that >>> causes you to disagree? >>> >>> Right now my main concern is just getting the bandwidth. Once one or >>> more ways to do that are established I can be concerned about minimizing >>> noise. >>> >>> I have not until now considered using another op amp instead of an amp >>> composed of discrete devices to do the bootstrapping. Doing it this way >>> would reduce added capacitance on the virtual ground. >>> >>> Regarding my earlier posting that was 18.8GHz not 18.8THz. >>> >>> -- >>> If you desire to respond directly remove the "sj." from the domain name >>> part of my email address. It is a spam jammer.- Hide quoted text - >>> >>> - Show quoted text - >> >> "I have been told by the project engineer who has given me this >>> assignment the capacitance is a parallel one." >> >> That sounds much worse. It's like some photo-resistor.? It will >> always be leaking as you bias it. (and leaking as the light shines on >> it.) ughh. >> >> >> George H. >> >> > > Similar, but unlike the passive photoresistor, the electromotive force > detector converts photons into electrical energy. In a TIA circuit it > can be modeled as a current source that does not require a bias. The > current is proportional to the motion of a beam light across it. It is > very good at detecting changes in a laser light's speckled pattern > produced when it reflects off any but the most even of surfaces. > > Right now my plan is get two of the lowest noise JFET or CMOS input op > amps I can find that satisfy the bandwidth requirement. One will be used > for the feedback resistor and the other will do the bootstrap. > I just had the exact same requirement for an IR CO2 sensor: Cust: we've made something, but it's way too noisy and we don't understand why (I was said it has 40R leak resistance). Can you improve on that? And no, we don't have the sensor spec... Oh, and you'd get paid only if the outcome please us. Oh, and we want you cheap... Me: OK, no way for me to work with you. Good luck :-) They've finally found someone accepting that. And no, it isn't really a small firm. Rather, a very big one... -- Thanks, Fred. |