8.9. Electronics Noise 517
whereAis a proportionality constant (recall that spectral power density for 1/f
noise is proportional to the inverse of the frequency).
The shot noise due to detector bias voltage can be very well represented by a
current noise source as shown in Fig.8.9.3. Using equation 8.9.10, we write the
spectral current density for detector bias as
i^2 sd=2eId, (8.9.19)whereIdis the current flowing through the detector. Of course here we have assumed
that no current flows through the amplifier/shaper, which is a valid assumption since
for most practical purposes the input impedance of an amplifier can be taken to be
infinite. Also, no current should be flowing through the parallel resistance, which is
again a valid assumption since typical shunt resistances are very large.
We now move to the noise induced by the parallel resistance. Since the parallel
resistance acts as a current source of Johnson noise, therefore we can use equation
8.9.4 to represent its power density.
i^2 jp=4 kBT
Rp. (8.9.20)
Similarly the series resistance induces noise at the amplifier’s input that can be
represented by a voltage noise source. The Johnson noise voltage power density for
this resistance can be written as (cf. equation 8.9.5)
v^2 js=4kBTRs. (8.9.21)Now we are left with the task of determining the total noise power density at the
input of the amplifier/shaper. Since some of the noise components are in terms of
voltage while others are in terms of current therefore we can not simply add them
together. If the voltage amplification factor of the amplifier is knows, it is easiest to
transform all the noise currents into noise voltages using Ohm’s law
v=iR. (8.9.22)Hence the total noise voltage at the input of the amplifier is given by
Vn,in=[(
v^2 as+A
f)
f+2eIdτR^2 s f+4 kBT
Rpf+4kBTRs f] 1 / 2
. (8.9.23)
Here we have multiplied all the noise densities by the bandwidth fto determine
the total noise voltage. If the amplifier’s amplification factor isA, then the output
noise voltage will be given by
Vn,out = AVn,in= A[(
vas^2 +A
f)
f+2eIdτR^2 s f+4 kBT
Rpf+4kBTRs f] 1 / 2
(8.9.24).
Now, if the input signal voltage isVs,in, the signal output voltage will be given by
Vs,out=AVs,in. (8.9.25)