# Transistor help

Discussion in 'Homework Help' started by Garurumon, Oct 28, 2013.

1. ### Garurumon Thread Starter Member

Mar 17, 2013
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Could someone check if the relations written in attached file are correct?

I was confused, if I have some base current, but I have resistors on collector and emitter of that transistor, will the current through the transistor still be (1+Beta)*Ib, or will it be limited by those transistors, never minding the Ie=(1+Beta)*Ib equation?

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2. ### Garurumon Thread Starter Member

Mar 17, 2013
99
1
I see the mistake on T1 base current

3. ### Garurumon Thread Starter Member

Mar 17, 2013
99
1
and in the voltage divider xD I just wanna know about the last question

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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What is the last question ?
This equation Ie = (1+Beta)*Ib holds until Ie < Vcc/(RC+RE)

5. ### Garurumon Thread Starter Member

Mar 17, 2013
99
1
you just answered it, thank you..

But Ib still flows through Ic if not..?

I have to find Vi1 and Vi2 on the image, and I do not know what currents flow through Rc and Re.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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IB never flows through Ic.
IB current flows from Vcc --->R1 --->RB--->Base-emitter T1--->Base-emitter T2--->RE ---->GND
And first use Kirchhoff law to find Vi2 voltage

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7. ### anhnha Active Member

Apr 19, 2012
776
48
Jony, did you mean that the equation only holds if transistor is in active region?
And, isn't Ie always smaller than Vcc/(RC+RE)?

8. ### Garurumon Thread Starter Member

Mar 17, 2013
99
1
So, through Rc flows Vcc/(Rc+Re), and through Re flows Vcc/(Rc+Re)+Ie1 ?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes.

In real circuit yes. But not always in calculation when you use "active region" equation.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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No, this image explains everything.

Last edited: Oct 28, 2013
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11. ### anhnha Active Member

Apr 19, 2012
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I think you meant that is incorrect as transistor operates in saturation region.
In saturation region both junctions are foward biased but this does not make sure that VCE always larger than zero.
Is that right?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I'm not quite sure what you mean by writing "but this does not make sure that VCE always larger than zero"
In saturation Vce is always smaller than < 0.6 but larger then 0V

13. ### Garurumon Thread Starter Member

Mar 17, 2013
99
1
yes, but i have limiting resistors, that's what confuses me! if I take that Ic=beta*Ib, then I get Vi2 voltage about 1kV!

14. ### anhnha Active Member

Apr 19, 2012
776
48
If so, I am confused with this.

Is Ie always smaller than Vcc/(Rc +Re)?

Ie = (Vcc - Vce)/(Rc + Re) and Vce always larger than zero => Ie < Vcc/(Rc +Re) always, right?

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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For what values??

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Try to solve this circuit

For Vcc = 10V ; Rb =10K; Rc = 1K; Re = 100Ω ; and Vbe = 0.6V and Hfe = 100

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17. ### Garurumon Thread Starter Member

Mar 17, 2013
99
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so, what does Ic equals to here? And what is the voltage above Re, and bellow Rc?

18. ### anhnha Active Member

Apr 19, 2012
776
48
Vcc = Ib*Rb + Vbe + Ie*Re

=> Ib = (Vcc - Vbe)/( Rb + (Hfe +1)Re) = (10 - 0.6)/( 10K + (100 + 1)0.1K) = 0.468mA

Ie = (Hfe + 1)Ib = 101*0.468mA = 47.268mA
Ic = Hfe*Ib = 100*0.468mA = 46.8mA
Ve = Ie*Re = 47.268mA*0.1K = 4.7268V
Vce = Vcc - Ic*Rc - Ie*Re = 10V - 46.8mA*1K - 47.268mA*0.1K = -41.5268V

If Vce < 0 how current flow from C to E?

19. ### Jony130 AAC Fanatic!

Feb 17, 2009
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This result (Vce = -41.5V) give as one important hint.
Our BJT is saturated and Ic = Hfe*Ib don't hold anymore.
Because if Ic = Hfe*Ib = 46.8mA >> Ic_max = Vcc/(Rc+RE) =10V/1.1K = 9mA transistor is in saturation.
So to solve tis circuit we need to assume Vce_sat voltage and use this equation
Ie = Ib + Ic.
Let as assume Vce_sat = 0.1V and Vbe = 0.7V.

Ie = Ve/Re
(1)

Ib = (Vcc - Vbe - Ve)/RB (2)

Ic = (Vcc - Vce_sat - Ve)/Rc
(3)

And now if we solve this for VE we have this

$\Large Ve = (\frac{Vcc - Vbe}{RB} + \frac{Vcc - Vce_{sat}}{RC}) * RB||RC||RE = 0.975676V$

So

Ve = 0.975676V

Vb = Ve + Vbe = 0.975676V + 0.7V = 1.675676V

Vc = Ve + Vce_sat = 1.075676V

And the currents

Ie = Ve/Re = 9.75676mA
Ic = (Vcc - Vc)/Rc = 8.924324mA
Ib = (Vcc - Vb)/RB = 0.8324324mA

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20. ### anhnha Active Member

Apr 19, 2012
776
48
Wow, that makes lot of sense now.
Just one more question.
In saturation Vce is vey small but is it constant in the region?