COMPUTING

4.65 – V B = 0.1V B + 0.2V B + 2.1Solving for V B ,2.55V B = --------- = 1.96 V1.3Then1.96I B = ------------- = 0.196 mA10 KΩSolutions to End−of−Chapter ExercisesAlso,I C sat = 0.2V B + 2.1 ≈ 3.03 mAβ satI---------- C satI B= =------------3.03≈ 15.50.19618. We recall thatv 1 = h 11 i 1 + h 12 v 2i 2 = h 21 i 1 + h 22 v 2(1)(2)With the voltage source = cos ωt mVin series with 800 Ω connected at the input and av 15 KΩ load connected at the output the network is as shown below.800 Ω 1200 Ω1∠0° mV+i 1i 2v 1 +−−2 × 10 – 4 v 250i – 61 50 10×Ω – 15000 Ω+v 2−The network above is described by the equationsor( 800 + 1200)i 1 + 2 × 10 – 4v 2 = 10 – 350i 1 50 10 – 6– v+ × v 2 = i 22 = -----------50002 × 10 3 i 1 + 2×10 4v 2 = 10 350i 1 + 250 × 10 6= 0– v 2We write the two equations above in matrix form and use MATLAB for the solution.A=[2*10^3 2*10^(−4); 50 250*10^(−6)]; B=[10^(−3) 0]'; X=A\B;...fprintf(' \n'); fprintf('i1 = %5.2e A \t',X(1)); fprintf('v2 = %5.2e V',X(2))Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications3−113