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Two main steps. Set up the DC circuit and the AC circuit. See below.
This is not the only way to build an Amp.
Have fun with Electronic!
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1st DC Circuit:
==> Chosen: IC ~= 1mA
==> Calculated: IB ~= IC/Beta = 1mA/100 = 10uA
==> Chosen: Current through the voltage divider is chosen to be 10*IB = 10*10uA = 100uA
==> Chosen: VE ~= Vcc - 1V = 12V - 1V = 11V
==> Calculated: VB ~= VE - Vbe = 11V - 0.7V = 10.3V
==> Calculated: R1 = (Vcc - VB)/(10*IB) = (12V - 10.3V)/100uA = 17kohm
==> Calculated: R2 = (VB - Vgnd)/(10*IB) = (10.3V - 0V)/100uA = 103kOhm
==> Calculated: RE = (Vcc - VE)/IE = (12V - 11V)/1mA = 1kOhm
==> Chosen: VC ~= Vcc/2 = 12V/2 = 6V
==> Calculated: RC = (Vcc/2 - Vgnd)/IC = (6V - 0V)/1mA = 6kOhm
*Assuming one may be using 5% metal-film resistors, we'll round these to the nearest available value.
R1 = 17kOhm ==> 16kOhm
R2 = 103kOhm ==> 100kOhm
RE = 1kOhm
RC = 6kOhm ==> 6.2kOhm
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2nd AC Circuit:
==> Chosen: Gain = 10
==> Chosen: Rload ~= 10*RC = 10*6kOhm = 60kOhm
==> Calculated: Re = V_thermal/IE = 25mV/1mA = 25Ohm
==> Since we want Gain = 10 and Gain ~= RC/(RE_un-bypassed + Re) = 6kOhm/(RE_un-bypassed + 25Ohm)
==> Solving for RE_un-bypassed = (RC - 10*Re)/10 = (6kOhm - 10*25Ohm)/10 = 575Ohm
==> In order to NOT disturb the DC bias we carefully set up, and we already know what we wanted RE to be for IE = 1mA, the new RE is split into two resistors, one is AC bypassed, the other is not, but their sum is 1kOhm. Therefore, RE_bypassed = RE_DC_BIAS - RE_un-bypassed = 1kOhm - 575Ohm = 425Ohm
Rload = 60kOhm ==> 62kOhm
RE_bypassed = 425Ohm ==> 430Ohm
RE_un-bypassed 575Ohm ==> 560Ohm
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Equation
==> Rin = R1||R2||RE', where RE' = RE*(Beta + 1)
==> Vo = (Vi')*Gain, where Vi' = Vi*(Rin/(Rin + RS)) and RS = 50Ohm in this case, and Gain = R_load-equivalent/(RE + Re), where R_load-equivalent = RC||Rload
==> Putting it all together:
Vo = Vi*[(R1||R2||RE*(Beta + 1))/(R1||R2||RE*(Beta + 1) + RS)]*[R_load-equivalent/(RE + Re)] = Vi*0.995511*9.63481.
Let Vi = 500mV, we have:
Vo = 500mV*0.995511*9.63481= 4.79578V ~= 4.8V
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