Question

48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers have voltage gain, input resistance, and output resistance as follows: for A, 100 V/V, 100 k,10k, respectively; for B, 10 V/V, 10 k,1 k, respectively. Your problem is to decide how the amplifiers should be connected. To proceed, evaluate the two possible connections between source S and load L, namely, SABL and SBAL. Find the voltage gain for each both as a ratio and in decibels. Which amplifier arrangement is best?

Answer 1

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

$Fp=\frac{R_{in}}{R_{in}+R_{out}}$

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

$Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db$

Analyzing SBAL:

$Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db$

So the best amplifier arrangement is SABL.