20mW output two stage amp

How it was built, below. Output stage: Required: Rload = 5.1k (light bulb for fun), Pload = 20mW Iload = sqrt(Pload/Rload) = sqrt(20mW/5.1k) = 1.98mA[RMS] Vload = Pload/Iload = 20mW/1.98mA = 10.101V[RMS] Vload = sqrt(2)*Vload[RMS] = sqrt(2)*10.101V = 14.3V[PEAK] IE2 ~= 10*Iload[RMS] = 10*1.98mA ~= 20mA RE2 = (Vcc/2)/IE2 = 20V/20mA = 1k Rin2 = (RE2||Rload)*(Beta + 1) = [(1k*5.1k)/(1k + 5.1k)]*(301) = 251.7k __________________________________ Input stage: Let IC1 ~= 1mA RC1 = (Vcc - (Vcc/2 - VBE2))/(IC1) = (40V - (20V - 0.7V))/1mA = 20.7k Let Gain ~= 10V/V RC1_eq = RC1||Rin2 = (20.7k*251.7k)/(20.7k + 251.7k) = 19.31k Since Gain ~= RC1_eq/(RE1 + Re1) ==> RE1 = (19.31k - 10.5*(25mV/1mA))/10.5 = 1796.9 ~= 1.8k VE1 ~= IE1*RE1 = 1mA*1.8k = 1.8V VB1 = VE1 + VBE1 = 1.8V + 0.7V = 2.5V R1 = (Vcc - VB1)/1mA = (40V - 2.5V)/1mA = 37.5k R2 = VB1/1mA = 2.5V/1mA = 2.5k __________________________________ Equation: Vo = Vi*(A_attenuation)*(A_gain) Vi = 1.43V[PEAK] A_attenuation = Rin1/(Rin1 + Rs) = 2.34k/(2.34k + 50) = 0.979V/V A_gain = RC1_eq/(RE1 + Re1) = 19.13k/(1.8k + 25) = 10.34V/V Vo = 1.43*(0.979V/V)*(10.4V/V) = 14.48V Vo[RMS] = Vo/sqrt(2) = 14.48V/sqrt(2) = 10.24V[RMS] Pload = Vo[RMS]^2/Rload = (10.24V)^2/5.1k = 20.56mW Done.

published 6 years ago