Hi

Basically I have been toying around with my electronics homework (which is one of my weaker subjects) and I have become stumped, I'l post a link of the circuit I am looking at further down.

Basically the question asks: Calculate the Power Loss in Transistor T

Information given:
R1 = 400 ohms
R2 = 500 ohms
R3 = 50 ohms
E = 10v
Vbe = 0.6v
Hfe = 10




A formula I have worked out is(although not sure if correct):
E= (Ir1xR1) + Vbe + (IeR3)..... where Ir1 = (Ib + Ir2)

Basically I have been struggling with the initial phase of calculations this, ie.... E, Ib, Ir2, Ir1,
I think if i had a formula to work out those then maybe I could get it.

Thanks for taking the time to read this!

 

Solve this
E-I1*R1-I2*R2=0
I1=Ib+I2
I2*R2-Vbe-Ie*R3=0
Ie=(hfe+1)*Ib

Or use the Thévenin
Et=E*R2/(R1+R2)=5.5V
Rth=R1||R2=222Ω
And then
Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.9V/70.18Ω=69.81mA

 

Solve this
E-I1*R1-I2*R2=0
I1=Ib+I2
I2*R2-Vbe-Ie*R3=0
Ie=(hfe+1)*Ib

Or use the Thévenin
Et=E*R2/(R1+R2)=5.5V
Rth=R1||R2=222Ω
And then
Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.9V/70.18Ω=69.81mA

Soo from what I can get from this I can use thevenins to work out Ie, and from there Ib can be solved using ib= (hfe+1)/Ie and from there I should be able to work my way through it?

slightly disappointed I didnt spot the thevenins way

Thanks for the help

 

To determine the power dissipated in bjt we usually ignore Ib*Vbe losses.
So Ptot=Ic*Vce.

 

Solve this
E-I1*R1-I2*R2=0
I1=Ib+I2
I2*R2-Vbe-Ie*R3=0
Ie=(hfe+1)*Ib

Or use the Thévenin
Et=E*R2/(R1+R2)=5.5V
Rth=R1||R2=222Ω
And then
Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.9V/70.18Ω=69.81mA

Hi, the teacher has asked for us to use the formula:

Ptot = Pc + Pb

therefore I have used:
Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.95V/70.18Ω=70.5mA

I then used:

Ib = Ie / (hfe +1) = 6.5mA

Therefore Ic = Ie - Ib = 64mA

Vce = E - Ib(hfe+1)R3 = 6.425v

Pb = Ib*Vbe = 3.9mW
Pc = Ic*Vce = 411.2mW

Therefore Ptot = 415.1mW

Could you check through this working and see what you think please?