Today I don't have the time to check your solution. But here is my solution
Ie = ( 9 - 0.7)/(5K/181 + 0.5K) = 15.73mA

Av = Rin/(Rs + Rin) * (RE||RL)/(re + RE||RL) = 4.36/(1 +4.36) * 187.5/(1.652 + 187.5) = 0.806V/V

Rin = R1||R2||Rib = 4.36K ohm

Rib = (Hfe +1) * (re +RE||RL) = 34.23k ohm

re = 26mV/Ie = 1.65 ohm

 

Today I don't have the time to check your solution. But here is my solution
Ie = ( 9 - 0.7)/(5K/181 + 0.5K) = 15.73mA

Av = Rin/(Rs + Rin) * (RE||RL)/(re + RE||RL) = 4.36/(1 +4.36) * 187.5/(1.652 + 187.5) = 0.806V/V

Rin = R1||R2||Rib = 4.36K ohm

Rib = (Hfe +1) * (re +RE||RL) = 34.23k ohm

re = 26mV/Ie = 1.65 ohm

thanks sir , I know you are not have time and your time are very important.......

 

Well you made a mistake R1 is 250k not 225K. Because Vth = 1.1538V not 1.25V.
Ib = 3.48329μA; Ic = 417.995μA; Ie = 421.478μA.

As for the gain

Av = Rc/(re + RE) * (Rib||Rth)/(Rs + Rib||Rth) = 8.46 * 0.985 =8.3V/V
re = 26mV/Ic
Rib = ( β+1 )*(re + RE) = 80kΩ
Rth = R1||R2 = 57.7kΩ

And for BJT we have Ie = Ib + Ic but as you know Ic = Ib*β. So we have Ie = Ib + β*Ib = (β +1)*Ib.

I think the Av in negative true?
Av=(Vo/Vs) = (-B*Rc/r-pi+(1+B)*Re)(Ri/Ri+Rs)
Ri = Rth || Rib
thanks

 

the Vth is zero?
or Vth=(v2/V1+V2)*{ (Vee)-(vcc) }?
Vth=(250k/175k+250k)*{ (9)-(-9) }
Vth= 10.6v
thanks

 

I do this problem
find Q-point
I very sorry if the image size is big
thanks for help me

 

I think the Av in negative true?
Av=(Vo/Vs) = (-B*Rc/r-pi+(1+B)*Re)(Ri/Ri+Rs)
Ri = Rth || Rib
thanks

Yes, the voltage gain for CE amplifier is negative. But we usually forget about this "minus" sign. Because this "minus" only inform us that our amplifier output is 180° out of phase with its input nothing more.

View attachment 74666
the Vth is zero?
or Vth=(v2/V1+V2)*{ (Vee)-(vcc) }?
Vth=(250k/175k+250k)*{ (9)-(-9) }
Vth= 10.6v
thanks

You have Vcc upside down in full2 pic.
And Vth = 1.6V or 10.6V see this two diagrams

 

Yes, the voltage gain for CE amplifier is negative. But we usually forget about this "minus" sign. Because this "minus" only inform us that our amplifier output is 180° out of phase with its input nothing more.


You have Vcc upside down in full2 pic.
And Vth = 1.6V or 10.6V see this two diagrams

View attachment 74673

is mean the analysis for Vth=10.6

Vee - VRe - Vbe - VRth - Vth =0 ( emitter-base )
(Vee=18)

Vee - VRe - Vce - VRc=0 (emitter-Collier) true?

please see this :
http://forum.allaboutcircuits.com/threads/analyses-for-bipolar-junction-transistor.102779/

thanks sir for help me

 

is mean the analysis for Vth=10.6

Vee - VRe - Vbe - VRth - Vth =0 ( emitter-base )
(Vee=18)

Vee - VRe - Vce - VRc=0 (emitter-Collier) true?

Looks good to me. But only if Vee = 18V

 

V1=9 and V2=-9
if I do analysis for circuit :
V1 + V2= 9 + (-9 )
=0 v true?

thanks sir

 

Wrong, the voltage is potential difference. So V+ = +9 and V- = -9V
So the voltage between V+ and V- is equal to +9V - (-9V) = 18V

 

hello again
I did new problem
from common emitter
please see my solution true?