hello guys i just wanted to show u this circuit that i found in adel sedra's book and i want to know the know how he got the final values and how he got Ic3 ,, any help please cuz we got it as a homework and i don't know how to solve it
thanks man thats the circuit with the loading effect but the on calculating output DC voltage on 2K is it this way : we take 10k parallel with 2k and then the current that flows into the 1.6K is 5mA ? and how we calculate the B in this circuit ?
To simplify the calculation simply assume Vout=0V And Ic3=Ie3 = 5mA And if we do this, we can simply determine the gain (open loop gain) Ao = K1 * K2 K1 = Rc/2re = ( Rc||( (β+1) * (R1+R2)||RL ) / 2*re2 = 17.876KΩ / 104Ω = 171.885 V/V K2 = (R1+R2)||RL / ( re3 + (R1+R2)||RL ) = 1.66KΩ / (1.66K +5.2Ω) =0.9968 V/V Ao ≈ 171 V/V And open loop Rin is equal Rin_0 = [ (β+1)*2re] + (R1||R2)/(β+1) = 10.504KΩ + 8.91Ω = 10.51KΩ open loop Rout is equal Rout_0 = ( Rc /(β+1) + re3 )||(R1+R2) = 199.172Ω
AS for DC bias point you have to assume β = ∞ So Ic=Ie And then Vout = Vcc - Ic2*Rc - Vbe3 And that all you have to do
Thx, but you better check my equations, because I wrote them by inspection. re = 1/gm = 26mV/Ic And voltage gain of a emitter follower is equal Au = Re/ (re+ Re) And and happy new year for you to.