The gain, A_N, and impedance, Z, values for amplifiers AMP_1 and AMP_2 in Fig. 34-8 are
V_G = 50 mVR_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
Chapter 34
Q. 34.6
Step-by-Step
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 mVThe 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 VThe 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)=1.11VNote: Without the voltage divider effect, the total gain would be
A_T=A_1A_2=15(4) = 60With 50 mV in, the voltage on the load would be
V_{RL}= A_TV_G =60(50 mV) = 3 VAs 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.