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?

 

I see the mistake on T1 base current

 

and in the voltage divider xD I just wanna know about the last question

 

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

 

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.

 

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

 

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

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)?

 

So, through Rc flows Vcc/(Rc+Re), and through Re flows Vcc/(Rc+Re)+Ie1 ?

 

Jony, did you mean that the equation only holds if transistor is in active region?

Yes.

And, isn't Ie always smaller than Vcc/(RC+RE)?

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

 

So, through Rc flows Vcc/(Rc+Re), and through Re flows Vcc/(Rc+Re)+Ie1 ?

No, this image explains everything.

 

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

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?

 

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?

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

 

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!

 

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

If so, I am confused with this.

But not always in calculation when you use "active region" equation.

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?

 

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!

For what values??

 

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?

Try to solve this circuit

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

 

Try to solve this circuit

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

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

 

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?

 

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

 

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