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Optical Sources, Detectors, and Systems by Robert H. Kingston

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4.2 The Detector Circuit: Resistor and Amplifier Noise
79
4.2 The Detector Circuit: Resistor and Amplifier Noise
i
= riqP/hv
jo
'•" = 2<?iB
|() \ ^
Figure 43 Eqiiivalent circuit for photodetection.
To
measure
or process the information from the detector, we need a
load resistance R followed by an amplifier with gain A to raise the
signal to a level suitable for processing or for transmission to another
location. The equivalent circuit in Figure 4.3 shows this arrangement
with the expressions for the signal and shot noise currents. The resist-
ance R includes the effective input resistance of the amplifier and must
be set to a value that, combined with the ever-present detector cap-
acitance, yields the required bandwidth B.
Just as in the case of blackbody radiation from an absorbing surface,
the resistor R exhibits an electromagnetic wave emission capability.
Actually it only emits power into one mode of
s^pace,
that associated
with a single-pair transmission line, as shown in Figure 4.4. Here the
transmission line is terminated at each end with matched resistors R and
the mode density in the frequency domain (see Problem 1.2) may be
shown to be dN =
(2L/c)df,
which is the one-dimensional version of Eq.
(1.11).
Since we are concerned with the frequency of electrical circuit
R
dP-\-
=
cdu+ ^
-^ dP-
=
cdu-
R
Figure 4.4 Thermal energy and power flow on transmission line.
80 Chapter
4
The Ideal Detector and Noise Limitations
currents, we shall again use/to distinguish it from optical or infrared
frequencies. The energy per mode from Eq. (1.8) becomes kT since hf
« kT, and the energy per unit length becomes
du = kTdN IL = (2kT/c)df (4.7)
One-half of this energy is flowing to the right and the other half to the
left at a velocity c, so that the respective power flows, dP+ and dP-, are
each given by
cdu-^
-
kTdf.
If we denote B as the effective electrical
bandwidth of a circuit, then the resistor must be emitting and absorbing
a total thermal power of
kTB.
Now this power is not constant, since we
know from Eq. (1.1) that the probability distribution of expected energies
on the transmission line obeys the Boltzmann factor. Since the prob-
ability of observing a specific energy is Boltzmann distributed so also is
the instantaneous power P. We may thus write
V(P) =
le-''l^
(4.8)
P
and the mean square fluctuation of P, from Eq. (4.3), becomes
oo
oo
—PIP
*^
p^
=
j p^p(P)dP =JP^ ^-^=^
dP =
(P)^
J x^e'^'dx
=
2(W (4.9)
0 0 0
:.AP'^
=(P)^
Therefore the rms power fluctuation is equal to the mean or kTB. This
fluctuation is the available noise power from the resistor, and it is a
simple circuit exercise to show that the equivalent shunt noise current is
given by
.1 = ^ (4.10)
The exponential distribution of expected power corresponds to a
Gaussian distribution of expected current and the thermal noise current
4.2 The Detector Circuit: Resistor and Amplifier Noise
81
behaves exactly the same as the shot noise current in Eq. (4.6), but with i
= 0.
An alternative way of deriving the Johnson or thermal noise pro-
duced by a resistor uses the detailed statistics of the electron motion in
the resistor material. Following the treatment of (Yariv, 1991), we
consider the rectangular resistor structure shown in Figure 4.5(a), which
contains an electron density n and has dimensions as shown. Each
electron moves randomly and has an average collision time, t^, and
(a)
i(t)k
-^r \^
(b)
Figure
4.5.
(a) Model of simple resistor showing the a:-component of electron velocity.
(b) Ciirrent waveform in external circuit due to single electron, ris a particular value of
the collision time which has average value, t .

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