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## Q. 34.6

The gain, $A_N$, and impedance, Z, values for amplifiers $AMP_1 and AMP_2$ in Fig. 34-8 are

$V_G = 50 mV$

$R_G = 600 \Omega$

$Z_{in1}= 1200\Omega$

$A_1 = 15$

$Z_{out1}= 4 k\Omega$

$Z_{in2}= 20 k\Omega$

$A_2 = 4$

$Z_{out2}= 100 \Omega$

$R_L = 200 \Omega$ ## Verified Solution

The actual input voltage, $v_{in_1} , to AMP_1$ is

$v_{in1}=V_G(\frac{Z_{in_1} }{Z_{in}+R_G } )$

$=50 mV(\frac{1200}{1200+600} )=33.33 mV$

Amplifier $AMP_1$amplifies this by 15 to produce a$V_{out_1}$ of

$V_{out1}= A_1(v_{in_1}) = 15(33.33 mV) = 500 mV$

The input voltage to $AMP_2$ is

$v_{in_2}=V_{out_1}(\frac{Z_{in_2} }{Z_{in_2}+Z_{out_1} } )$

$=500 mV(\frac{20 k\Omega}{20k\Omega+4k\Omega} )=416.7 mV$

Using AMP$_2$ amplifies this to

$V_{out_2} = A_2(V_{in_2}) \\ = 4(416.7 mV) = 1666.7 mV or 1.67 V$

The voltage across the load is

$V_{RL}=V_{out_2} \left(\frac{R_L}{R_L+Z_{out_2}} \right) \\ =1.67V\left(\frac{200}{200+100}\right)=1.11V$

Note: Without the voltage divider effect, the total gain would be

$A_T=A_1A_2=15(4) = 60$

With 50 mV in, the voltage on the load would be

$V_{RL}= A_TV_G =60(50 mV) = 3 V$

As you can see, this load effect of one amplifier by another has a significant effect. That is why most voltage amplifiers are designed with high input impedance and low output impedance to minimize this effect.