hello everyone,

I do this problem in basic common emitter Bipolar junction transistor (BJT) amplifier ........

me solution true? or there are some wrong?

thanks

 

why some time in Bipolar junction transistor the Ie=(1+B)Ib & Ie=BIb ?

thanks,

 

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.

 

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.

thanks sir,
when current collector equal current emitter (Ic=Ie)?

 

thanks sir,
when current collector equal current emitter (Ic=Ie)?

I guess when transistor is in Cutoff region.

 

I guess when transistor is in Cutoff region.

thanks sir,
is mean there isn't resistor and source in emitter region?

 

thanks sir,
is mean there isn't resistor and source in emitter region?

In Cutoff Ic=0, Ib=0, therefore Ie=Ic+Ib=0. Therefore Ic=Ie=0.

 

In Cutoff Ic=0, Ib=0, therefore Ie=Ic+Ib=0. Therefore Ic=Ie=0.

obs

see this ie=ic why?
thanks sir,

 

obs

see this ie=ic why?

Ib can be VERY small. So. It becomes a question of precision. How precise do you want to be?

 

The emitter current is always equal to Ib + Ic. But because the beta value is large we can ignore the base current and say that Ie = Ic.
For example if β = 100 and Ib = 10μA we have Ic = 100 * 10μA = 1mA and Ie = Ib + Ic = 1.01mA
So we don't do the big mistake/error if we ignore the Ib current and say that Ie = Ic = 1mA

 

I do another problem ,but this is difficult

 

Your DC current calculation are totally wrong.
Vth is ok and Rth also. But your KVL loop (Vth - VRB - Vbe - VRe = 0) is wrong.
You forget about Vee in your loop.

 

Your DC current calculation are totally wrong.
Vth is ok and Rth also. But your KVL loop (Vth - VRB - Vbe - VRe = 0) is wrong.
You forget about Vee in your loop.

if I do KVL loop (Vth - VRB - Vbe - VRe +VEE= 0) true?
thanks sir

 

No, wrong again.

 

No, wrong again.

ok , If I do KVL loop (-Vth - VRB - Vbe - VRe +VEE= 0)
but the loop flowing from -Vth+ is mean positive in voltage true?

about Av= (B Rc)/(r-bi +(1+B)Re) *Ri true?

thanks sir

 

ok , If I do KVL loop (-Vth - VRB - Vbe - VRe +VEE= 0)
but the loop flowing from -V+ is mean positive in voltage true?

In PNP transistor current flow from +Vee--->Re--->emitter-base junction--->Rth--->Vth--->-Vee
Because for PNP Ie current is also equal to Ie = Ib + Ic



So we start at the bottom of a Vee and we go from - to + so we give + sign.
Vee - VRe - Veb - VRth - Vth = 0V

Vee - Ie * Re - Veb - Ib*Rth - Vth = 0

Ib = Ie/(Hfe +1) therefore

Vee - Ie * Re - Veb - Ie/(Hfe +1) *Rth - Vth = 0

Ie = (Vee - Vht - Veb)/( Rth/(Hfe +1) + Re ) = (12 - 10.2 - 0.7)/(12.75/100 + 0.5) = 1.75mA

about Av= (B Rc)/(r-bi +(1+B)Re) *Ri true?

Ri ?? And where is RL ?

 

In PNP transistor current flow from +Vee--->Re--->emitter-base junction--->Rth--->Vth--->-Vee
Because for PNP Ie current is also equal to Ie = Ib + Ic

View attachment 74355

So we start at the bottom of a Vee and we go from - to + so we give + sign.
Vee - VRe - Veb - VRth - Vth = 0V

Vee - Ie * Re - Veb - Ib*Rth - Vth = 0

Ib = Ie/(Hfe +1) therefore

Vee - Ie * Re - Veb - Ie/(Hfe +1) *Rth - Vth = 0

Ie = (Vee - Vht - Veb)/( Rth/(Hfe +1) + Re ) = (12 - 10.2 - 0.7)/(12.75/100 + 0.5) = 1.75mA


Ri ?? And where is RL ?

thanks sir

Ri =(Rth//r-bi)

I forget RL sorry ,

Av= (B Rc//Rl)/(r-bi +(1+B)Re) *Ri

 

But in this circuit we don't have source resistance? So how can Rht has any effect on voltage gain ?
For me the voltage gain is equal to
Av = Rc||RL/(re + RE) = (β * Rc||RL)/(Rpi + (β+1)*Re) ≈ 2.6V/V

 

But in this circuit we don't have source resistance? So how can Rht has any effect on voltage gain ?
For me the voltage gain is equal to
Av = Rc||RL/(re + RE) = (β * Rc||RL)/(Rpi + (β+1)*Re) ≈ 2.6V/V

thanks sir ,you are great man......

 

I do this problem but in common collector :
I think there is wrong in Ac analysis ...

thanks,