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| Artist... |
 Posted: Mon Nov 02, 2009 8:39 pm |
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Guest
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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.
--
If you desire to respond directly remove the "sj." from the domain name
part of my email address. It is a spam jammer. |
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| George Herold... |
| Posted: Mon Nov 02, 2009 8:39 pm |
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Guest
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On Nov 2, 8:39 pm, Artist <Art... at (no spam) sj.speakeasy.net> wrote:
[quote]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.
--
If you desire to respond directly remove the "sj." from the domain name
part of my email address. It is a spam jammer.
[/quote]
Cool circuit, thanks for the link. (I don't quite understand
bootstrapping.... something for me to work on.)
The gain is due, in part, to the changing of input C from 3nF of the
PD to 10pF of the JFET+opamp+stray. Bootstrapping a 150pF PD will
give you less improvement. But still perhaps enough. Do you have
enough photocurrent to reduce the TIA resistor from 1 Meg to 1k? I
would then 'naively' expect a bandwidth improvement of sqrt(R) so a
factor of 30... X 350kHz... something near 10MHz may not be out of
the question. I've never built PD circuits this fast though......
George H. |
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| John Larkin... |
| Posted: Mon Nov 02, 2009 8:55 pm |
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Guest
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On Mon, 02 Nov 2009 17:39:35 -0800, Artist <Artist at (no spam) sj.speakeasy.net>
wrote:
[quote]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.
[/quote]
Do you have Phil Hobbs' book? That is Step One for issues like this.
That opamp is a little noisy.
John |
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| Jamie... |
| Posted: Mon Nov 02, 2009 9:48 pm |
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Guest
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Artist wrote:
[quote]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.
That's funny. Brings back memories of years ago trying to make[/quote]
a photo receiver for a specialized light wall. It worked for what
I had to do at the time how ever, the next version I made was with a
cluster of 4 small body photo diodes into a 4 channel op-amp. I then
summed the results. That generated a cleaner output..
P.S.
I was only doing 500 khz and it was a digital stream with a little
hysteresis in the circuit. |
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| Phil Hobbs... |
| Posted: Mon Nov 02, 2009 10:22 pm |
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Guest
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Artist wrote:
[quote]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.
[/quote]
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
--
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 |
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| Artist... |
| Posted: Tue Nov 03, 2009 4:36 pm |
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Guest
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John Larkin wrote:
[quote]On Mon, 02 Nov 2009 17:39:35 -0800, Artist<Artist at (no spam) sj.speakeasy.net
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.
Do you have Phil Hobbs' book? That is Step One for issues like this.
That opamp is a little noisy.
John
I don't. A book I do have is by Jerald Graeme, "Photodiode Amplifiers,[/quote]
Opamp Solutions"
http://books.google.com/books?id=sHV0c5hBW4QC&dq=Jerald+G+Graeme&printsec=frontcover&source=an&hl=en&ei=_pzwStuuMYPYsgOh5LX0BQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBMQ6AEwAw#v=onepage&q=&f=false
There is a topology for a boot strap amplifier in it. He begins his
treatment of them in Chapter 4.3 page 71. He does not give any component
values or part numbers for the topology (fig 4.12 page 81). It would
take a lot of research for me to figure out if there are any transistors
and resistors that can make this topology go up to 10MHz.
Is this the book you are referring to?
http://www.amazon.com/gp/pdp/profile/AKFB26K3TYMSS
I will get back all of you on the expected photodiode current.
--
To reply directly remove the sj. from my email address. This is a spam
jammer. |
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| George Herold... |
| Posted: Tue Nov 03, 2009 4:43 pm |
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Guest
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On Nov 3, 8:32 pm, Phil Hobbs <pcdhSpamMeSensel... at (no spam) electrooptical.net>
wrote:
[quote]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
--
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 -- Hide quoted text -
- Show quoted text -
[/quote]
"> You can get the same 3 dB improvement by putting a TIA on each end
of
[quote]the PD."
[/quote]
Hey, I remember that idea! No one’s ever tried it though, have
they?
Say, will your cascode circuit work for photo detectors other than
photodiodes? I have this ‘silly’ idea that electrons from photodiodes
are born at half a volt or so, and are thus able to do a bit of work
before they are detected.
George H. |
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| John Larkin... |
| Posted: Tue Nov 03, 2009 7:19 pm |
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Guest
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| Artist... |
| Posted: Tue Nov 03, 2009 8:13 pm |
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Guest
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Phil Hobbs wrote:
[quote]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.[/quote]
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.
--
To reply directly remove the sj. from my email address. This is a spam
jammer. |
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| Phil Hobbs... |
| Posted: Tue Nov 03, 2009 8:32 pm |
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Guest
|
Artist wrote:
[quote]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.
[/quote]
If there's a way to make that 10 uA, your life will be much easier.
[quote]
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.
[/quote]
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
--
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 |
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| Artist... |
| Posted: Tue Nov 03, 2009 10:12 pm |
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Guest
|
Phil Hobbs wrote:
[quote]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
[/quote]
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.
--
To reply directly remove the sj. from my email address. This is a spam
jammer. |
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| miso at (no spam) sushi.com... |
| Posted: Tue Nov 03, 2009 11:16 pm |
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Guest
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On Nov 3, 10:39 pm, Phil Hobbs
<pcdhSpamMeSensel... at (no spam) electrooptical.net> wrote:
[quote]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
[/quote]
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)? |
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| Phil Hobbs... |
| Posted: Wed Nov 04, 2009 1:37 am |
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Guest
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George Herold wrote:
[quote]On Nov 3, 8:32 pm, Phil Hobbs <pcdhSpamMeSensel... at (no spam) electrooptical.net
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
--
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 -- Hide quoted text -
- Show quoted text -
"> You can get the same 3 dB improvement by putting a TIA on each end
of
the PD."
Hey, I remember that idea! No one’s ever tried it though, have
they?
Say, will your cascode circuit work for photo detectors other than
photodiodes? I have this ‘silly’ idea that electrons from photodiodes
are born at half a volt or so, and are thus able to do a bit of work
before they are detected.
George H.
[/quote]
Sure, but it helps the most with PDs. Most other detectors are either
intrinsically slow and noisy, like many photoconductors, or else have
low capacitance anyway, like PMTs. Of course there are lots of
non-optical transducers, but not too many of them have the difficult
feature of PDs, namely high speed with predominantly capacitive
impedance and wide dynamic range.
Cheers
Phil
--
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 |
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| Phil Hobbs... |
| Posted: Wed Nov 04, 2009 1:39 am |
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Guest
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Artist wrote:
[quote]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.
[/quote]
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 net
http://electrooptical.net |
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| George Herold... |
| Posted: Wed Nov 04, 2009 6:30 am |
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Guest
|
On Nov 4, 7:49 am, Fred Bartoli <" "> wrote:
[quote]m... at (no spam) sushi.com a écrit :
On Nov 3, 10:39 pm, Phil Hobbs
pcdhSpamMeSensel... at (no spam) 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 -
[/quote]
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. |
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