Question 1.14: NOISE IN AN RLC CIRCUIT Most radio receivers have a tuned pa...
NOISE IN AN RLC CIRCUIT Most radio receivers have a tuned paralle resonant circuit, which consists of an inductor L, capacitor C, and resistance R in parallel. Suppose L is 100 μH; C is 100 pF; and R, the equivalent resistance due to the input resistance of the amplifier and to the loss in the coil (coil resistance plus ferrite losses), is about 200 kΩ. What is the minimum rms radio signal that can be detected?
Consider the bandwidth of this tuned RLC circuit, which can be found in any electrical engineering textbook:
B=\frac{f_{o}}{Q}
where f_{o}=1 /[2 \pi \sqrt{L C}] is the resonant frequency and Q=2 \pi f_{o} C R is the quality factor. Substituting for L, C, {\text and} R, we get f_{o}=10^{7} / 2 \pi=1.6 \times 10^{6} Hz and Q = 200, which gives
B=10^{7} /[2 \pi(200)] Hz, or 8 kHz. The rms noise voltage is
v_{ rms } =[4 k T R B]^{1 / 2}=\left[4\left(1.38 \times 10^{-23} J K ^{-1}\right)(300 K )\left(200 \times 10^{3} \Omega\right)\left(8 \times 10^{3} Hz \right)\right]^{1 / 2}
=5.1 \times 10^{-6} V \quad \text { or } \quad 5.1 \mu V
This rms voltage is within a bandwidth of 8 kHz centered at 1.6 MHz. This last information is totally absent in Equation 1.38. If we attempt to use
v_{ rms }=\left[\frac{k T}{C}\right]^{1 / 2}
we get
v_{\text {rms }}=\left[\frac{\left(1.38 \times 10^{-23} J K ^{-1}\right)(300 K )}{100 \times 10^{-12} F }\right]^{1 / 2}=6.4 \mu V
However, Equation 1.37 was derived using the RC circuit in Figure 1.28, whereas we now have an LCR circuit. The correct approach uses Equation 1.38, which is generally valid, and the appropriate bandwidth B.
\overline{v(t)^{2}}=\frac{k T}{C} [1.37]
v_{ rms }=\sqrt{4 k T R B} [1.38]
