Question

Z; = (1 + A)R. Give numerical values. 7. Repeat the calculations in Example 6.2 when the input device is a microphone with vs = 10 mV (rms) and source impedance Rs = 50 k 2. Draw a conclusion regarding the effectiveness of a buffer. 741

number 7

Answer 1

Given , microphone cophone load Cchalt has has y to my crms) y to my coms) and Rs= 50kr a recorder has impedance of soon and Solenie 24 3 he chost i 500h recorder loma (()! censi I Microphone

Using Voltage division rule the voltage available at the output is

$V_o=\frac{500}{5k+500}10mV=\frac{5}{5.5k}V=0.91mV(rms)$

and the power delivered to the load is

PL = V? (rms) RL (0.91mV)2 L = 1.66nW 50012

When a buffer is added, the total voltage is available at the load as it has infinite input resistance, zero output resistance and unity gain as shown below.

​​​​​From circuit, we observe that the voltage drop across Rs=0V as no current flows through it. So,

。。 ༥ so པ་མ ༴ སྤྱ° k vo Us ( ) ( Buffer has Reu aos, Roao Gain = 1)

$\\V_o'=V_i=V_s=10mV(rms)\\\\ P_L'=\frac{V_0^2(rms)}{R_L}=\frac{(10mV)^2}{500\Omega}=200nW$

Due to the buffer there is no loss in the voltage and power delivered from source to the load. It enables the matching of input and load.

$\\Voltage\ gain= \frac{V_o'}{V_o}=\frac{10mV}{0.91mV}=10.99\\\\Power \ gain= \frac{P_L'}{P_L}=\frac{200nW}{1.66nW}=120.48$