# Find the voltages and currents

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#### sulieman

##### New Member
I need to Find the voltages and currents following circuit, where β = 100 Transistor #### microtexan

##### New Member
Do my homework for me.

Do you not have any idea how to do this? Doesn't your textbook have some formulas to calculate the Base current and then from there? Show us you have made some effort #### EN0

##### Member
I'm not an expert at transistors, so hopefully I'm doing this right. If someone see's a mistake, feel free to correct me!

Voltages:

1. Vc=15V
2. Find Base voltage (Vb) according to voltage divider rule. R2/R1+R2 x Vin = 7.5V.
3. Due to the base-emitter drop (-0.7V), Ve = 6.8V.

Currents:

1. Ic = 15V/5k = 3mA
2. Ie = 6.8V/5k = 1.36mA

I'm not sure how to find the current for the base, so I did the following:

R1xR2/R1+R2 = 25k I=E/R so, I=15V/25k = 0.6mA.

I can't build the circuit at the moment, but maybe some else could and let me know if my answers are correct? I'm especially curious If I found the base current correctly...

#### audioguru

##### Well-Known Member
Most Helpful Member
I'm not an expert at transistors, so hopefully I'm doing this right. If someone see's a mistake, feel free to correct me!
What is the definition of beta?
You know the emitter current so calculating the base current from it is easy.
The base current is a lot less than 0.6mA.

#### Miles Prower

##### Member
I'm not an expert at transistors, so hopefully I'm doing this right. If someone see's a mistake, feel free to correct me!

Voltages:

1. Vc=15V
2. Find Base voltage (Vb) according to voltage divider rule. R2/R1+R2 x Vin = 7.5V.
3. Due to the base-emitter drop (-0.7V), Ve = 6.8V.
So far, so good, but everything else is just plain wrong.

Ie= 6.8 / 5E3= 1.36mA= Ic (Close enough if B= 100)
Vc= (15 - 1.36 X 5)= 8.2V
Vce= 8.2 - 6.8= 1.4V
B= Ic / Ib:

Ib = Ic / B= 1.36 / 100= 0.0136mA

Current through divider:

I= 15 / 100E3= 0.15mA (More than 10X Ib, so you're in good shape there.)

#### BrownOut

##### Banned
Acually, he had iE right. He needed to use iC~=iE or else iC = B/(B + 1)*iE. And iB = iC/B. But a pretty good effort!

#### EN0

##### Member
Acually, he had iE right. He needed to use iC~=iE or else iC = B/(B + 1)*iE. And iB = iC/B. But a pretty good effort!
Yeah, I've designed several other circuits and I calculated Ie that way. When I checked on my meter, they were both identical.

#### Jony130

##### Active Member
May I introduce proper method to solve this simple circuit. You can do this in two ways.
The first way, directly form the circuit we can write equation describing the circuit.
If we apply KVL we get two independent mesh.

Input
Vcc=VR1 + VR2 = I1*R1 + I2*R2 = Ie*Re + Vbe + Vcb + Ic*Rc
Just a little bit different
Vcc - I1*R1 - I2*R2 = 0
I1 = IB + I2
I2*R2 - Vbe - Ie*Re = 0
And naturally
Ie = Ib*(β+1); Ib = Ic/β = Ie/(β+1); Ie = Ib + Ic
VB = Vbe + Ie*Re = I2*R2
thus,

Ib=[ (R2*Vcc) - Vbe*(R1+R2) ] / [ Re*(β+1)*(R1+R2) + (R1*R2) ]

Ic = β*Ib

and for the output loop
Vcc = VRc + Vce + VRe = Ic*Rc + Vce + Ie*Re

And a slightly different way.

When we apply Thevenin's Theorem for base voltage divider.
We get Thevenin Equivalent Circuit who looks like this Et = Vcc x RB2 / (RB1+RB2)

Rb = RB1xRB2 / RB1+Rb2

and then
Et = Ib*Rb + Vbe + Ie*Re
and
Ie = Ib*(β+1),

then

Et = Ib*Rb + Vbe + Ib*(β+1)*Re

and finally

Ib=(Et-Vbe) / [Rb+(β+1)*Re]
Ic=β*Ib

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