# BJT amplifier

Discussion in 'Homework Help' started by full, Oct 18, 2014.

1. ### full Thread Starter Member

May 3, 2014
225
2
hello everyone,

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

me solution true? or there are some wrong?

thanks

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Last edited: Oct 18, 2014
2. ### full Thread Starter Member

May 3, 2014
225
2
why some time in Bipolar junction transistor the Ie=(1+B)Ib & Ie=BIb ?

thanks,

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

4. ### full Thread Starter Member

May 3, 2014
225
2
thanks sir,
when current collector equal current emitter (Ic=Ie)?

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
601
I guess when transistor is in Cutoff region.

6. ### full Thread Starter Member

May 3, 2014
225
2
thanks sir,
is mean there isn't resistor and source in emitter region?

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
601
In Cutoff Ic=0, Ib=0, therefore Ie=Ic+Ib=0. Therefore Ic=Ie=0.

8. ### full Thread Starter Member

May 3, 2014
225
2
obs

see this ie=ic why?
thanks sir,

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Last edited: Oct 18, 2014
9. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
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Ib can be VERY small. So. It becomes a question of precision. How precise do you want to be?

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10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
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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

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11. ### full Thread Starter Member

May 3, 2014
225
2
I do another problem ,but this is difficult

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12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Your DC current calculation are totally wrong.
Vth is ok and Rth also. But your KVL loop (Vth - VRB - Vbe - VRe = 0) is wrong.

13. ### full Thread Starter Member

May 3, 2014
225
2
if I do KVL loop (Vth - VRB - Vbe - VRe +VEE= 0) true?
thanks sir

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
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No, wrong again.

15. ### full Thread Starter Member

May 3, 2014
225
2
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

Last edited: Oct 19, 2014
16. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
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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

Ri ?? And where is RL ?

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17. ### full Thread Starter Member

May 3, 2014
225
2
thanks sir

Ri =(Rth//r-bi)

I forget RL sorry ,

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

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
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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

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19. ### full Thread Starter Member

May 3, 2014
225
2
thanks sir ,you are great man......

20. ### full Thread Starter Member

May 3, 2014
225
2
I do this problem but in common collector :
I think there is wrong in Ac analysis ...

thanks,

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