Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
Subject: Re: photodetector circuit, high speed with good ambient light rejection
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Date: Thu, 28 Nov 2002 19:25:32 -0000
NNTP-Posting-Date: Thu, 28 Nov 2002 19:25:36 GMT
Winfield Hill wrote:
>> Dr C.I. Swift wrote:
>>> Need to make such a circuit good for up to 90MHz. will have a
>>> considerable DC light component so want too reject this.
>>> Is it possible to reverse bias the photodiode, running it to ground
>>> via an inductor in series with the resistor? The idea being the ac
>>> component will seee a higher impedence and therefore produce a
>>> higher voltage.
>>> Or should I stick to the transimpedence amplifier with a feedback
>>> resistor small enough so that the system does not saturate and then
>>> just ac couple the signal to further amplification stages.
> Rene Tschaggelar answered...
>> Provided sufficient light, eg some mW, a reverse biased photodiode
>> onto a coil or 100 Ohm resistor gives sufficient signal for an AC
>> coupled Minicircuits MAR-6 amplifier.
> The answer to whether to use a transimpedence amplifier or a
> resistive load to ground, etc., depends on several things, the total
> capacitance (photodiode + cable + amplifier), your photodiode signal
> current, and your signal-to-noise ratio requirements.
> The Johnson current-noise density of a resistor (connected to ground
> or used as a feedback element) is sqrt(4kT/R), which means that low
> resistor values create high current noise. If you have high light
> currents, you may escape this issue, otherwise this formula will
> determine the minimum resistor value necessary. Once your minimum R
> value is known, evaluate your total capacitance Cin and see if your
> needed bandwidth is possible using just a resistor to ground, i.e.
> BW = 1 / 2pi R Cin.
> If your capacitance Cin is too high for enough bandwidth using
> resistor R to ground, you'll need a transimpedence amplifier. It
> can use the same value R, or an even higher Rf value for more
> signal, if you have enough amplifier bandwidth. The required opamp
> GBW is f_T > fc^2 / (2pi Rf Cin), where fc is your required system
> bandwidth. You'll also need a parallel feedback capacitor limiting
> the bandwidth to fc, so Cf = 1 / 2pi Rf fc.
Ahmmm...Well, I use a little different logic than this, but this is
typically for the cases of low/zero dc bias which keeps the noise floor
low. It does not matter if the initial BW is too low with regard to the
choice to go from shunt resistor to a transinpedance amplifier. The key
issue is the final S/N ratio at the required BW. Indeed, if the BW
required is too high, it can demand that one must use the shunt!
If the *lowest* noise possible is required, then a very large resistor
must be used. The problem here, in a feedback transinpedance amp, is
that to get an adequate C.Rf/A, it might require an effective 100Ghz+
GBW, which is not actually realisable.
In the shunt case the initial 3db BW is, of course, quite low. However,
this is not important as the signal and noise of the resistor go hand in
hand to keep a constant S/N in the roll off portion. The idea here is to
buffer the signal with a low noise fet, then voltage amplifier the
signal, then differentiate it to the recover the flat response. This is
a technique that is used in very low noise fibre communication systems.
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