From: Winfield Hill
Subject: Re: photodetector circuit, high speed with good ambient light rejection
Date: 28 Nov 2002 10:12:29 -0800
Organization: Rowland Institute
X-Newsreader: Direct Read News 4.11
> 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.
For example, a 4k resistor has a current-noise density of 2pA/Hz^1/2,
yielding a Dc-to-90MHz-bandwidth noise of 20nA rms. (We'll ignore
e_n-Cin noise.) If Cin was 5pF, you'd only get an 8MHz bandwidth with
4k connected to ground, so a transimpedence amplifier is required, with
an opamp f_T >= 1000Mz. Depending on the opamp's f_T and any desired
response peaking, about 0.44pF of feedback capacitance is also required,
including the resistor's self capacitance.
As for using an inductor to eliminate DC light currents, this may be
reasonable for lower reistance values, and can be used in parallel with
either R to ground or with the Rf transimpedence feedback resistor.
However, either way be careful of the inductor's self capacitance. In
the 4k example above the inductor would have to be say 22uH for a 30MHz
low-frequency cutoff, and have a self-capacitance below 0.4pF, which is
a self-resonate frequency of over 53MHz. I don't think such a beast is
available. A typical 22uH has SRF = 15MHz, so an inductor with Rf = 4k
would force you into a narrowband resonate system. On the other hand,
if Rf = 1k or less you might have success.
Tell us more about what you're working on.