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+//Tested on Windows 7 Ultimate 32-bit

+//Chapter 8 Power Amplifiers Pg no. 277,278 and 279

+clear;

+clc;

+

+//Given Data

+//Figure 8.13

+

+VCC=20;//collector supply voltage in volts

+RC=270;//collector resistance in ohms

+RE=150;//emitter resistance in ohms

+R1=3.3D3;//divider network resistance R1 in ohms

+R2=1.5D3;//divider network resistance R2 in ohms

+VBE=0.7;//forward voltage drop of emitter diode in volts

+B=100;//DC CE current gain beta

+RL=470;//load resistance in ohms

+C1=15D-6;//input coupling capacitance in farads

+C2=15D-6;//output coupling capacitance in farads

+

+//Solution

+

+VB=VCC*R2/(R1+R2);//base to ground voltage in volts

+VE=VB-VBE;//emitter to ground voltage in volts

+IE=VE/RE;//emitter current in amperes

+ICQ=IE;//neglecting base current, collector current is equal to emitter current in amperes

+VC=VCC-ICQ*RC;//collector to ground voltage in volts

+VCEQ=VC-VE;//collector to emitter quiscent voltage in volts

+PD=VCEQ*ICQ;//power dissipation in watts

+RL_dash=RC*RL/(RC+RL);//equivalent a.c. load resistance in ohms

+IC_sat=ICQ+VCEQ/RL_dash;//saturation collector current in amperes

+VCE_cutoff=VCEQ+ICQ*RL_dash;//cutoff collector to emitter voltage in volts

+Pout=0.5*ICQ^2*RL_dash;//output a.c. power in watts

+e=Pout/VCC/ICQ;//efficiency of circuit = Pout/Pin(dc)

+printf("(a) The minimum transistor power rating required PD = %.3f Watts\n ",PD);

+printf("(b) AC output power Pout = %d milli-Watts\n ",Pout*10^3);

+printf("(c) Efficiency of the amplifier η = %.2f\n ",e);

+

+//decimal approximation taken here in efficiency