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'[EE] Photodiode interface'
2008\08\27@112440 by Harold Hallikainen

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I'm working on an application where we're sending frequency shift keyed
data on an RF carrier on infrared. There are other carriers on the IR
carrying FM audio. I'm demodulating the RF carrier using an SA639. I
upconvert from my RF frequency to a 10.7MHz IF, use ceramic filters at
10.7MHz, then a ceramic quadrature detector. The trick is the interface
between the photodiode and the SA639.

Right now, I'm using a preamp that's a current to voltage converter based
on an LT application note. It uses a JFET source follower in front of an
op amp. This is then configured as a current to voltage converter with the
photodiode reverse biased about 1V. The current to voltage converter
drives an AC coupled op amp amplifier, then an LC network into the SA639.
The LC provides some RF tuning and voltage gain. The op amps have a GBW
product of about 100MHz.

The performance of this circuit is adequate, but I'd like to do better and
reduce parts count.

Most photodiode interface circuits I've seen assume the diode is a current
source and run it in to a current to voltage converter. It SEEMS, though,
that we'd get maximum power out of the photodiode (allowing us to get over
noise in the rest of the system) if we instead matched the photodiode
impedance. Looking at the datasheet, I divided the open circuit voltage by
the short photo current (with a specified reverse bias) to come up with
the Norton resistance (the resistance across an ideal current source). I
also used the rated capacity at the reverse voltage I'm running to come up
with a model of the photodiode being a current source with 20pF across it
and 400k across it. It SEEMS that we'd get the most power out of this if
we matched this impedance.

I've run some SPICE simulations based on this model and modeled the input
of the SA639 as an 800 ohm resistor with 3.5pF across it. I then, in my
model, use a "tapped capacitor tank circuit" to provide the reverse bias
to the photodiode and impedance match the photodiode to the SA639. My
experimental approach on this was to try adjusting the capacitor across
the SA639 input, then adjust the other capacitor of the tapped capacitor
pair until the peak in the voltage across the SA639 input was at the right
frequency.

We're about to test this against the previously discussed current to
voltage converter approach. However, in all my reading about photodiodes,
I have NEVER seen it suggested that we attempt to match the output
resistance of the photodiode, which is what I'm trying to do here. So...
anyone on the list have photodiode experience and would care to comment?

THANKS!

Harold


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2008\08\27@120440 by Bob Blick

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Hi Harold,

Harold Hallikainen wrote:

> Right now, I'm using a preamp that's a current to voltage converter based
> on an LT application note. It uses a JFET source follower in front of an
> op amp. This is then configured as a current to voltage converter with the
> photodiode reverse biased about 1V. The current to voltage converter
> drives an AC coupled op amp amplifier, then an LC network into the SA639.
> The LC provides some RF tuning and voltage gain. The op amps have a GBW
> product of about 100MHz.

I've used this method too, sometimes with a second FET so I can null out
huge amounts of ambient light.

{Quote hidden}

You're running at such a high frequency(the highest I used was 1 MHz but
the levels were tiny and I needed huge gain) that I somehow doubt this
will work better. The JFET bootstrap method is great because it really
does null out the AC that would develop at the previously highest
acheivable frequency so you can extend it by a few octaves. If you just
try to match impedances, I don't see it extending the frequency range
like the JFET method. You could certainly develop an interesting
oscillator :) Perhaps a careful combination of the impedance matching
and the JFET bootstrap. I was never able to do better than when using a
fast opamp and two JFETs so I guess I can't be much help and only offer
sympathy!


Cheerful regards,

Bob

2008\08\27@120813 by Spehro Pefhany

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Quoting Harold Hallikainen <spam_OUTharoldTakeThisOuTspamhallikainen.org>:

The idea with using the transimpedance amplifier is to avoid voltage
changes across the PD capacitance.

Phil Hobbs has some excellent tips on high performance PD circuitry,
both in his superb book, and there's a quite good paper available
online if you just search for it.

Last one I did, I used an op-amp with a 1.5 or 1.8GHz GBW, if memory serves.


{Quote hidden}

> -

2008\08\27@121954 by Marcel Duchamp

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Harold Hallikainen wrote:
>So...
> anyone on the list have photodiode experience and would care to comment?
>
> THANKS!
>
> Harold
>
>

My experience is only to the extent that I know "it ain't always easy".
 There is usually a tradeoff between speed and gain.  I too, have never
seen any information on using your approach.  So please fill us in on
the results of what you learn.

2008\08\27@154555 by Michael Rigby-Jones

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> -----Original Message-----
> From: .....piclist-bouncesKILLspamspam@spam@mit.edu [piclist-bouncesspamKILLspammit.edu] On
Behalf
> Of Harold Hallikainen
> Sent: 27 August 2008 16:24
> To: Pic List
> Subject: [EE] Photodiode interface
>
>
> Most photodiode interface circuits I've seen assume the diode is a
current
> source and run it in to a current to voltage converter. It SEEMS,
though,
> that we'd get maximum power out of the photodiode (allowing us to get
over
> noise in the rest of the system) if we instead matched the photodiode
> impedance. Looking at the datasheet, I divided the open circuit
voltage by
> the short photo current (with a specified reverse bias) to come up
with
> the Norton resistance (the resistance across an ideal current source).
I
> also used the rated capacity at the reverse voltage I'm running to
come up
> with a model of the photodiode being a current source with 20pF across
it
> and 400k across it. It SEEMS that we'd get the most power out of this
if
> we matched this impedance.

As Sphero has already stated, a transimpedance amplifier allows you to
maintain a constant bias voltage.  This means the PIN's capacitance can
be minimised (by suitable biasing) for the available optical power input
range improving bandwidth and linearity.

I don't deal much with receivers at work, but all PIN and APD receivers
I have seen have used transimpedance amplifier, and that's from
155Mbit/s (in the old days) to 10Gbit/s and higher applications.  That
said, fiber optic links are rarely power constrained these days, if you
need to squeeze every last bit of signal out of the PIN and you can
suffer the loss of bandwidth you might find your method gives you some
advantage.

Regards

Mike

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2008\08\27@164600 by Harold Hallikainen

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>
> As Sphero has already stated, a transimpedance amplifier allows you to
> maintain a constant bias voltage.  This means the PIN's capacitance can
> be minimised (by suitable biasing) for the available optical power input
> range improving bandwidth and linearity.
>
> I don't deal much with receivers at work, but all PIN and APD receivers
> I have seen have used transimpedance amplifier, and that's from
> 155Mbit/s (in the old days) to 10Gbit/s and higher applications.  That
> said, fiber optic links are rarely power constrained these days, if you
> need to squeeze every last bit of signal out of the PIN and you can
> suffer the loss of bandwidth you might find your method gives you some
> advantage.
>
> Regards
>
> Mike


Thanks for the comment! As I said before, all the ones I've seen also use
a transimpedance amplifier (current to voltage converter). This is a free
field IR application, and the output of the photodiode gets pretty low as
we get far away from the emitter. While diode capacity would tend to slow
things down (it takes current to charge and discharge that capacitor), if
we make that capacity part of a resonant circuit, it appears that it
doesn't matter any more. I'll experiment with my method a bit and see what
happens.

Thanks!

Harold


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