TIA Photodiode Bootstrap at 10MHz
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From: George Herold on 4 Nov 2009 11:30 On Nov 4, 7:49 am, Fred Bartoli <" "> wrote: > m...(a)sushi.com a écrit : > > > > > > > On Nov 3, 10:39 pm, Phil Hobbs > > <pcdhSpamMeSensel...(a)electrooptical.net> 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 > >>> How can a TIA on each end accomplish the same thing as bootstrapping? > >>> The idea of bootstrapping is to increase the effective impedance of the > >>> photodetector by making the same virtual ground voltage appear on both > >>> ends of it. In a dual TIA arrangement the current going into the virtual > >>> ground of one TIA comes out of the virtual ground of the other. That > >>> means the outputs of the TIA will be equal and opposite with the > >>> consequence that the voltages on the virtual grounds will also be equal > >>> and opposite (neglecting part tolerances.) This would halve the > >>> effective photodiode impedance rather than increase it. > >> I didn't say it did the same thing, only that it gets the same SNR > >> improvement. Putting a TIA on one end of a PD requires bypassing the > >> other end to ground, so the bypassing might as well be done by the > >> summing junction of another TIA! > > >> 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 nethttp://electrooptical.net > > > I see there are papers and patents on fully differential TIAs, but no > > real products. > > > But isn't this all a wash since you added another noise source (2nd > > TIA)? > > Translate one TIA input noise to the other TIA. > Being uncorrelated their power add up, so it's +3db noise. > The signal is +6dB level for a net +3db S/N ratio... > > -- > Thanks, > Fred.- Hide quoted text - > > - Show quoted text - Exactly, I came to think of differential TIA's from correlation noise measurement circuits. (You look at the same noise signal (say a resistor) with two identical amps. You then look for the correlations in the noise in each channel.) They do this by multiplying the two signals. The amplifer noise averages to zero and you are left with the 'signal' from the noise source. So if you have a differential TIA would you be better to multiply the two signals rather than summing them? George H.
From: Phil Hobbs on 4 Nov 2009 11:33 miso(a)sushi.com wrote: > On Nov 3, 10:39 pm, Phil Hobbs > <pcdhSpamMeSensel...(a)electrooptical.net> 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 >>> How can a TIA on each end accomplish the same thing as bootstrapping? >>> The idea of bootstrapping is to increase the effective impedance of the >>> photodetector by making the same virtual ground voltage appear on both >>> ends of it. In a dual TIA arrangement the current going into the virtual >>> ground of one TIA comes out of the virtual ground of the other. That >>> means the outputs of the TIA will be equal and opposite with the >>> consequence that the voltages on the virtual grounds will also be equal >>> and opposite (neglecting part tolerances.) This would halve the >>> effective photodiode impedance rather than increase it. >> I didn't say it did the same thing, only that it gets the same SNR >> improvement. Putting a TIA on one end of a PD requires bypassing the >> other end to ground, so the bypassing might as well be done by the >> summing junction of another TIA! >> >> Cheers >> >> Phil Hobbs > > I see there are papers and patents on fully differential TIAs, but no > real products. > > But isn't this all a wash since you added another noise source (2nd > TIA)? No, because the noise adds in power and the signal adds in amplitude. That's why it's 3 dB not 6 dB improvement. 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: Phil Hobbs on 4 Nov 2009 11:36 George Herold wrote: > On Nov 4, 7:49 am, Fred Bartoli <" "> wrote: >> m...(a)sushi.com a �crit : >> >> >> >> >> >>> On Nov 3, 10:39 pm, Phil Hobbs >>> <pcdhSpamMeSensel...(a)electrooptical.net> 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 >>>>> How can a TIA on each end accomplish the same thing as bootstrapping? >>>>> The idea of bootstrapping is to increase the effective impedance of the >>>>> photodetector by making the same virtual ground voltage appear on both >>>>> ends of it. In a dual TIA arrangement the current going into the virtual >>>>> ground of one TIA comes out of the virtual ground of the other. That >>>>> means the outputs of the TIA will be equal and opposite with the >>>>> consequence that the voltages on the virtual grounds will also be equal >>>>> and opposite (neglecting part tolerances.) This would halve the >>>>> effective photodiode impedance rather than increase it. >>>> I didn't say it did the same thing, only that it gets the same SNR >>>> improvement. Putting a TIA on one end of a PD requires bypassing the >>>> other end to ground, so the bypassing might as well be done by the >>>> summing junction of another TIA! >>>> Cheers >>>> Phil Hobbs >>> I see there are papers and patents on fully differential TIAs, but no >>> real products. >>> But isn't this all a wash since you added another noise source (2nd >>> TIA)? >> Translate one TIA input noise to the other TIA. >> Being uncorrelated their power add up, so it's +3db noise. >> The signal is +6dB level for a net +3db S/N ratio... >> >> -- >> Thanks, >> Fred.- Hide quoted text - >> >> - Show quoted text - > > Exactly, I came to think of differential TIA's from correlation noise > measurement circuits. (You look at the same noise signal (say a > resistor) with two identical amps. You then look for the correlations > in the noise in each channel.) They do this by multiplying the two > signals. The amplifer noise averages to zero and you are left with > the 'signal' from the noise source. So if you have a differential TIA > would you be better to multiply the two signals rather than summing > them? > > George H. Using correlation will help suppress the amplifier voltage noise, but won't affect the amplifier current noise or the shot noise, because both of those are real currents that flow into both amplifier inputs. (It also won't help the pickup and power supply feedthrough much.) 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: George Herold on 4 Nov 2009 12:35 On Nov 4, 11:36 am, Phil Hobbs <pcdhSpamMeSensel...(a)electrooptical.net> wrote: > George Herold wrote: > > On Nov 4, 7:49 am, Fred Bartoli <" "> wrote: > >> m...(a)sushi.com a écrit : > > >>> On Nov 3, 10:39 pm, Phil Hobbs > >>> <pcdhSpamMeSensel...(a)electrooptical.net> 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 > >>>>> How can a TIA on each end accomplish the same thing as bootstrapping? > >>>>> The idea of bootstrapping is to increase the effective impedance of the > >>>>> photodetector by making the same virtual ground voltage appear on both > >>>>> ends of it. In a dual TIA arrangement the current going into the virtual > >>>>> ground of one TIA comes out of the virtual ground of the other. That > >>>>> means the outputs of the TIA will be equal and opposite with the > >>>>> consequence that the voltages on the virtual grounds will also be equal > >>>>> and opposite (neglecting part tolerances.) This would halve the > >>>>> effective photodiode impedance rather than increase it. > >>>> I didn't say it did the same thing, only that it gets the same SNR > >>>> improvement. Putting a TIA on one end of a PD requires bypassing the > >>>> other end to ground, so the bypassing might as well be done by the > >>>> summing junction of another TIA! > >>>> Cheers > >>>> Phil Hobbs > >>> I see there are papers and patents on fully differential TIAs, but no > >>> real products. > >>> But isn't this all a wash since you added another noise source (2nd > >>> TIA)? > >> Translate one TIA input noise to the other TIA. > >> Being uncorrelated their power add up, so it's +3db noise. > >> The signal is +6dB level for a net +3db S/N ratio... > > >> -- > >> Thanks, > >> Fred.- Hide quoted text - > > >> - Show quoted text - > > > Exactly, I came to think of differential TIA's from correlation noise > > measurement circuits. (You look at the same noise signal (say a > > resistor) with two identical amps. You then look for the correlations > > in the noise in each channel.) They do this by multiplying the two > > signals. The amplifer noise averages to zero and you are left with > > the 'signal' from the noise source. So if you have a differential TIA > > would you be better to multiply the two signals rather than summing > > them? > > > George H. > > Using correlation will help suppress the amplifier voltage noise, but > won't affect the amplifier current noise or the shot noise, because both > of those are real currents that flow into both amplifier inputs. (It > also won't help the pickup and power supply feedthrough much.) > > 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 nethttp://electrooptical.net- Hide quoted text - > > - Show quoted text - Sure, But it's often the case that the amp voltage noise is the dominant noise source. Mind you Ive never had an occasion to worry that much about 3dB of improvement. Its usually a lot easier to increase the signal upstream somewhere... more light, better focus, etc.. George H.
From: Artist on 4 Nov 2009 17:11
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. -- To reply directly remove the sj. from my email address. This is a spam jammer. |