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

 

Simply post the diagram

 

I redrawn the circuit for you.
Now if you look at circuit B, do you know how to solve it ?

 

I redrawn the circuit for you.
Now if you look at circuit B, do you 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

 

U R a genius MAAN
thank u very very much and happy new year

 

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.