transimpeadance vs voltage amp for photodiode

B

Bill

Guest
I am working on an amplifier circuit for a avalanche photodiode that
is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8 amps
looking at the photodiode signal across 50 ohms. I am pressuring a
transimpeadance amplifier using the OPA695 with a 1k Rf. The
photodiode current is ~300ua. On a general basis what are the pluses
and minuses considering noise, bandwidth, etc. of using a voltage vs.
a transimpeadance amp in this application.
 
In article <27c04ef4.0411241340.5b73f35@posting.google.com>,
Bill <r.u@cox.net> wrote:
I am working on an amplifier circuit for a avalanche photodiode that
is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8 amps
looking at the photodiode signal across 50 ohms. I am pressuring a
transimpeadance amplifier using the OPA695 with a 1k Rf. The
photodiode current is ~300ua. On a general basis what are the pluses
and minuses considering noise, bandwidth, etc. of using a voltage vs.
a transimpeadance amp in this application.
Click to expand...
If you have to put a load resistance on the photodiode to make the
frequency responce flat, the transimpedance amplifier will produce a lower
noise The "noise power" of a resistor at a given temperature is a
constant. As R goes down noise current goes up. The transimpedance
design gets a low impedance with a higher value resistor and hence les
noise current.

If you use the capacitance of the photocell to integrate a small signal,
the transimpedance and the voltage amplifier method end up with the same
noise.

Until you get to frequencies that require silly values of resistance, the
transimpedance amplifier is the best option from a bandwidth point of
view. At really high frequencies, you want to match the amplifier
transistor to the photodiode's impedance.


--
--
kensmith@rahul.net forging knowledge
 
On 24 Nov 2004 13:40:51 -0800, r.u@cox.net (Bill) wrote:

I am working on an amplifier circuit for a avalanche photodiode that
is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8 amps
looking at the photodiode signal across 50 ohms. I am pressuring a
transimpeadance amplifier using the OPA695 with a 1k Rf. The
photodiode current is ~300ua. On a general basis what are the pluses
and minuses considering noise, bandwidth, etc. of using a voltage vs.
a transimpeadance amp in this application.
Click to expand...

Dumping your signal into 50 ohms is a waste, noise-wise. The place for
signal to go is into the silicon.

I just did a PIN diode amp using two AD8014's, current-mode opamps.
The first stage was a TIA with (I recall) 680 ohms feedback || 1 pF,
and the second stage was an inverting gain-of-3 or so. Risetime (from
a vcsel laser pulse) was just under 2 ns overall with about a 1.5 pF
silicon pin diode. I had a similar amount of signal as you do, a few
hundred uA, so s/n was pretty good and edge jitter was around 5 ps
RMS, not bad. I think (data's at work, I'm not) I'm getting just about
1 volt out per optical milliwatt in at 850 nm.

Phil, avowed enemy of bad TIA designs, will probably rip me up for
this one.

John
 
John Larkin wrote...
On 24 Nov 2004 13:40:51 -0800, r.u@cox.net (Bill) wrote:

I am working on an amplifier circuit for a avalanche photodiode
that is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8
amps looking at the photodiode signal across 50 ohms. I am
pressuring a transimpeadance amplifier using the OPA695 with a 1k Rf.
The photodiode current is ~300ua. On a general basis what are the
pluses and minuses considering noise, bandwidth, etc. of using a
voltage vs. a transimpeadance amp in this application.


Dumping your signal into 50 ohms is a waste, noise-wise. The place
for signal to go is into the silicon.

I just did a PIN diode amp using two AD8014's, current-mode opamps.
The first stage was a TIA with (I recall) 680 ohms feedback || 1 pF,
and the second stage was an inverting gain-of-3 or so. Risetime (from
a vcsel laser pulse) was just under 2 ns overall with about a 1.5 pF
silicon pin diode. I had a similar amount of signal as you do, a few
hundred uA, so s/n was pretty good and edge jitter was around 5 ps
RMS, not bad. I think (data's at work, I'm not) I'm getting just
about 1 volt out per optical milliwatt in at 850 nm.

Phil, avowed enemy of bad TIA designs, will probably rip me up for
this one.
Click to expand...
He won't be happy unless we're using a biased common-base input stage.


--
Thanks,
- Win
 
On 25 Nov 2004 06:02:03 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

John Larkin wrote...

On 24 Nov 2004 13:40:51 -0800, r.u@cox.net (Bill) wrote:

I am working on an amplifier circuit for a avalanche photodiode
that is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8
amps looking at the photodiode signal across 50 ohms. I am
pressuring a transimpeadance amplifier using the OPA695 with a 1k Rf.
The photodiode current is ~300ua. On a general basis what are the
pluses and minuses considering noise, bandwidth, etc. of using a
voltage vs. a transimpeadance amp in this application.


Dumping your signal into 50 ohms is a waste, noise-wise. The place
for signal to go is into the silicon.

I just did a PIN diode amp using two AD8014's, current-mode opamps.
The first stage was a TIA with (I recall) 680 ohms feedback || 1 pF,
and the second stage was an inverting gain-of-3 or so. Risetime (from
a vcsel laser pulse) was just under 2 ns overall with about a 1.5 pF
silicon pin diode. I had a similar amount of signal as you do, a few
hundred uA, so s/n was pretty good and edge jitter was around 5 ps
RMS, not bad. I think (data's at work, I'm not) I'm getting just
about 1 volt out per optical milliwatt in at 850 nm.

Phil, avowed enemy of bad TIA designs, will probably rip me up for
this one.

He won't be happy unless we're using a biased common-base input stage.
Click to expand...

Well, that lets me off the hook. The inverting input of a current-mode
opamp *is* a biased common-base stage.

John
 
On Thu, 25 Nov 2004 12:54:12 -0500, Phil Hobbs
<pcdhSpamMeSenseless@us.ibm.com> wrote:

John Larkin wrote:
On 25 Nov 2004 06:02:03 -0800, Winfield Hill
hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:


John Larkin wrote...

On 24 Nov 2004 13:40:51 -0800, r.u@cox.net (Bill) wrote:


I am working on an amplifier circuit for a avalanche photodiode
that is looking at very high speed 2 nsec rise 10nsec wide pulses.
Previous work had been done on a circuit that uses 2 TI OPA695 X8
amps looking at the photodiode signal across 50 ohms. I am
pressuring a transimpeadance amplifier using the OPA695 with a 1k Rf.
The photodiode current is ~300ua. On a general basis what are the
pluses and minuses considering noise, bandwidth, etc. of using a
voltage vs. a transimpeadance amp in this application.


Dumping your signal into 50 ohms is a waste, noise-wise. The place
for signal to go is into the silicon.

I just did a PIN diode amp using two AD8014's, current-mode opamps.
The first stage was a TIA with (I recall) 680 ohms feedback || 1 pF,
and the second stage was an inverting gain-of-3 or so. Risetime (from
a vcsel laser pulse) was just under 2 ns overall with about a 1.5 pF
silicon pin diode. I had a similar amount of signal as you do, a few
hundred uA, so s/n was pretty good and edge jitter was around 5 ps
RMS, not bad. I think (data's at work, I'm not) I'm getting just
about 1 volt out per optical milliwatt in at 850 nm.

Phil, avowed enemy of bad TIA designs, will probably rip me up for
this one.

He won't be happy unless we're using a biased common-base input stage.



Well, that lets me off the hook. The inverting input of a current-mode
opamp *is* a biased common-base stage.

John

I am of course The TIA Avenger, but that's because I have this deep committed
long-term faithful monogamous relationship with my signal-to-noise ratio.

Circuit details are much less important, and in this case 50 ohms is probably
fine. This is an APD we're talking about, after all, so assuming that it is
running at a gain of 100 (pretty typical), the noise current is sqrt(100)=10
times larger than that of a PIN diode running at the same output current.

With a current having full shot noise, such as a photocurrent from a PIN
diode, shot noise will dominate Johnson noise if the current drops at least
50 mV across the load resistor.

That means that the APD's output current, running at 20 dB above full shot
noise, will be quantum limited if it drops at least 500 uV across the load.

M sqrt(2q I_primary) = sqrt(4 kT/R_L)

makes amplified shot noise equal to Johnson noise, where I_primary is the
photocurrent _before_ multiplication, M is the APD gain, and everything else
is as usual.

Rearranging this a bit, the voltage drop required

M**2 I_primary R_L = 2 kT/q (50 mV at room temp)

so since amplified photocurrent I_photo=M I_primary,

I_photo R_L = (50 mV)/M

puts you in the quantum limit.

Thus for a 50-ohm R_L, you need only (50 mV)/((50 ohms)*100) or 10 uA of
amplified photocurrent to get into the quantum limit.

With a 300-uA pulse, 50 ohms is better than good enough.

Cheers,

Phil "TIA Avenger--I like that" Hobbs
Click to expand...

OK, given all that, a standard MMIC should be a great, simple,
superfast amp for an APD, unless DC performance matters. MMICS are
crap dc-wise. But cheap to replace every time the APD blows one out.

John
 
Phil Hobbs wrote:
Thus for a 50-ohm R_L, you need only (50 mV)/((50 ohms)*100) or 10 uA
of amplified photocurrent to get into the quantum limit.

With a 300-uA pulse, 50 ohms is better than good enough.
Click to expand...
Apologies for the faux pas of replying to my own post:

The noise figure of the amplifier comes in here as well. Since we're not
conjugate-matching the input of the amp, we get the full benefit of the low
noise temperature of the amplifier. A good amp can have a 35K noise
temperature, and that's the number to put in to the sqrt(4kT/R) Johnson noise
formula. Therefore with a really good amp, you could be in the quantum limit
with an 8x smaller voltage drop--only 60 uV or thereabouts. It's the dc
current times the ac resistance that matters.

Cheers,

Phil Hobbs
 
John Larkin <jjlarkin@highSNIPlandTHIStechPLEASEnology.com> writes:

<SNIP>

Since shot noise is dominant, does that automatically mean you can
resolve photons?
Click to expand...
I would think it means that your signal to noise is being limited by
the fact that light comes in photons rather than
continuously. However, this in itself is not the same thing as being
able to count the photons individually!


--

John Devereux
 
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