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Chapter 34

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 aV_{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.