TIA Photodiode Bootstrap at 10MHz
Artist
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.
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John Larkin
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
Jamie
> 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
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.
George Herold
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.
Phil Hobbs
> 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
--
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
Artist
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.
John Larkin
Artist
> 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 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.
Phil Hobbs
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?
George Herold
wrote:
>
> - 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.
Artist
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.
Phil Hobbs
> they?
>
> Say, will your cascode circuit work for photo detectors other than
> are born at half a volt or so, and are thus able to do a bit of work
> before they are detected.
>
> George H.
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
Phil Hobbs
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!
mi...@sushi.com
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)?
Fred Bartoli
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.
George Herold
> m...@sushi.com a écrit :
>
> - 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.
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.
Phil Hobbs
> On Nov 4, 7:49 am, Fred Bartoli <" "> wrote:
>>> 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.)
George Herold
<pcdhSpamMeSensel...@electrooptical.net> wrote:
> George Herold wrote:
> > On Nov 4, 7:49 am, Fred Bartoli <" "> wrote:
>
> - Show quoted text -
Sure, But it's often the case that the amp voltage noise is the
dominant noise source. Mind you I’ve never had an occasion to worry
that much about 3dB of improvement. It’s usually a lot easier to
increase the signal upstream somewhere... more light, better focus,
etc..
George H.
Artist
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.
Phil Hobbs
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.
Artist
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.
George Herold
>
> - 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.
Artist
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.
Fred Bartoli
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.
George Herold
You might look at the ADA4817 from analog devices. They show a TIA
circuit that almost meets your bandwidth spec with 50k ohm of 'gain'.
At 1uA that would give you 50mV of signal which is starting to get the
shot noise above the johnson noise of the resistor. And perhaps you
can do better with your bootstrap or using a cascode front end ala
Phil H. I've got a sample of one of these in a drawer... whenever I
find the time to try it. There is a nasty gain peak out beyond
100MHz.
George H.
Phil Hobbs
Photodiodes and photoconductors are completely different.
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.
I'd never heard of a photoelectromotive detector, but Google had.
Interesting. From what I gather from the free literature, it's
basically a GaAs 1-D lateral effect photodiode run at zero bias, with
the output taken across the ends of the top layer.
I haven't read any really detailed papers, because they all cost $30,
but it relies on the carrier density responding slowly to changes in the
illumination pattern. It's made of GaAs rather than silicon, which I'm
guessing is because the hole mobility in GaAs is so very low, which
essentially glues the charge density in place. (Silicon would respond
much faster.)
For static illumination, the slow carrier diffusion makes the junction E
field flatten out locally, so the charges all recombine, and the output
current is zero. When the illumination changes, though, you have
photocarriers being generated in regions where the net current is
suddenly nonzero. These carriers produce a displacement current
immediately (not when they arrive at the electrodes), and the carriers
still flowing in the previously illuminated region produce a net
displacement current of the opposite sign. The two sum to zero, but
they're spatially offset from each other, which produces a net current
from the outputs. Like everything else in nature, it's linear for small
displacements.
It's a cute technique, which reminds me a bit of a photorefractive phase
conjugate mirror, in which the beam interference pattern forms a
slowly-varying hologram that corrects for phase and pointing errors.
A few things I'm not clear on:
(1) There has to be a junction in there somewhere, because it works at
zero bias and can produce current in both directions--which in a
symmetric structure requires a third electrode.
(2) Splitting the photocarriers between the ends seems to require a
high-resistance epi layer, which would tend to make the response slow or
else very noisy, like a lateral effect cell--the full Johnson noise
current of the epi layer appears in the output.
(3) The diffusion and recombination currents both have full shot noise,
and the forward-biased regions will have a very low resistance. This
will seemingly contribute a lot of noise to the output--the noise will
be that of the full beam's photocurrent, not merely that of the small
difference current.
So in general this doesn't sound like a low-noise device, although it's
certainly convenient for speckle measurements--the usual alternative
method is TV interferometry, in which you cross-correlate the speckle
patterns before and after some change (e.g. pressurizing a tank). TV
interferometry is slow, since it takes a minimum of two video frame times.
> 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.
It would be worth finding out what the noise level of the device itself
is. If I'm mistaken and it's actually quiet, you'd be better off making
the bootstrap out of a single-ended BF862 JFET rather than an op amp.
Otherwise a couple of op amps is probably fine (one of my favourites is
the ADA4817-1).
George Herold
wrote:
> Artist wrote:
> > George Herold wrote:
> >> On Nov 5, 12:40 pm, Artist<Art...@sj.speakeasy.net> wrote:
> >>> Phil Hobbs wrote:
> >>>> Artist wrote:
> >>>>> Phil Hobbs wrote:
> >>>>>> Artist wrote:
> >>>>>>> Phil Hobbs wrote:
> >>>>>>>> Artist wrote:
<Big snip>
Thanks for the hint Phil, A search for (photo-emf detector) brings up
a nice article from Newport.
George H.
Artist
This project had a break for awhile but is now revived. I am given new
specifications to model the emf detector with. These are 100kOhm with a
parallel capacitance of 1pf. This is a much easier task.
There was discussion in this thread about other methods of noise
reduction. The goal of using a bootstrap was primarily to get the
bandwidth. The low impedance of the emf detector meant bandwidth
challenges because to a first approximation that neglects capacitances
the bandwidth will be the GBW of the op amp divided by the noise gain.
I like the ADA4817-1. Using it may make possible doing this without
bootstrapping.
Given the 1pF parallel capacitance in the model a bootstrap would simply
add as much capacitance as it removes and add noise.
The bootstrap would also increase the equivalent resistance of the emf
detector. But the resistance in the model is high enough now, and the
GBW of the ADA4817-1 is high enough, that I do not see a need.
The manager wants a 10MHz bandwidth and 2 Mohm transimpedance. I am not
going to be able to do this in one stage. Even before considering the
capacitances from the virtual ground to actual ground the parallel
capacitance of 2Mohm to get a 10MHz bandwidth would have to be 8fF. That
is not going to happen.
Whats more, because of the capacitances on the virtual ground the
feedback resistor must have a parallel capacitor for stability.
To reduce inductances of resistors and capacitors I intend to use wide
packages such as 0508 along with wide pcb traces.
Your information about emf detector physics is of much interest to me.
Thanks.
George Herold
>
> - Show quoted text -
"The manager wants a 10MHz bandwidth and 2 Mohm transimpedance. I am
not
going to be able to do this in one stage."
This sounds like a stretch no matter how many stages you use. Sounds
like you may have to educate your manager about capacitance. Even
with only 1pF in the detector you've still got a few pF on the input
to the opamp, plus the PCB trace connecting the two.
"> To reduce inductances of resistors and capacitors I intend to use
wide
> packages such as 0508 along with wide pcb traces."
Well on the photodetector side of the circuit I'd keep traces short
and thin to reduce the capacitance. And be careful of where you put
the ground plane.
(But Phil H. is the expert here. BTW have you bought his book yet?)
George H.
Artist
Wider traces reduce inductance and increase capacitance.
I have not bought his book yet but I will.
John Larkin
wrote:
Trace inductances won't matter in a 10 MHz TIA. They barely matter at
500 MHz.
John