Please Help- Question on BJT

Thread Starter

devilstormz

Joined Aug 18, 2011
10
Hey,

I am new to this forum and hope to increase my activity on the forums in the future as this forum has alot of help and resources available.

I have decided to read ahead in the summer holidays before uni starts so was hoping for some help on this question I have attempted to solve.

Question: Find the emitter voltage and current in each transistor shown below, for which hfe is 100 and can be
considered to be very high (i.e. ic ~ ie). Remember Vbe ~ 0.6V.

I have managed to get a value (incorrect) the emitter voltage but could not work out how to eliminate/ find the currents in that circuit. I have annotated my working to help you guys follow through my working.

The correct answer is:

ve = 3.6V, ie = 36mA

See the attached pdf for my attempt at solving the question. It includes a circuit diagram aswell.

Thanks in advance.
 

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Jony130

Joined Feb 17, 2009
5,183
Have you not notice that
Vb = I2 * 10K ??

And to get the exact value for Ie current you need take into account Ib current.
Which load the voltage divider.
To do so you need to use Thevenine equivalent circuit or simply write down the KVL expression based on the circuit digram.
And if you do everything right you get this result
Vb = 4.21192V
Ie = 36.119216mA
 

Thread Starter

devilstormz

Joined Aug 18, 2011
10
Thanks for the promt reply.

Yes i did notice it but I dont know how to calculate I2. If you can guide me to calculate that I know how to solve the question.

If you apply KVL to the bottom loop you have I2 and Ie as unknowns?
 

Jony130

Joined Feb 17, 2009
5,183
KCL for Vb node

I1 = I2 + Ib (1)

And KVL

12V - I1*10K - I2*10K = 0
(2)

I2*10K - 0.6V - Ie*100R = 0 (3)

Ie = Ib * (hfe+1) (4)

Or try using Thevenine equivalent, simply replace the voltage divider by his Thevenine equivalent Rth and Vth.
 

hgmjr

Joined Jan 28, 2005
9,029
For us to assist you in a way that helps you to learn how to solve this type of problem on your own, you need to review the material that jony130 has pointed you to along with any other textbook material to which you have access and give the solution another try.

We will be happy to point out where you have made your mistake in your next solution attempt and give you hints as to what you can do to correct your approach if needed.

hgmjr
 

Thread Starter

devilstormz

Joined Aug 18, 2011
10
Thanks jony130, I managed to solve it using Thevenine however can you just explain me equation (2) what loop used to write the kvl equation.

Just confused how the I2*10K bit came.

Also another quick question, when a question relating to transistors has a line "and the supply voltage is 12V" does that mean Vin is 12V? I dont think that is the case because it could relate to another part of the circuit in which a transistor is used.
 
Last edited:

hgmjr

Joined Jan 28, 2005
9,029
Millman's Theorem expression for Ve.

\(V_b\ =\ \left(\frac{\frac{+12}{10000}\ +\ \frac{+0.6}{100*100}}{\frac{1}{10000}\ +\ \frac{1}{10000}\ +\ \frac{1}{10000}} \right)\)

\(V_e\ =\ \left(\frac{\frac{+12}{10000}\ +\ \frac{+0.6}{100*100}}{\frac{1}{10000}\ +\ \frac{1}{10000}\ +\ \frac{1}{10000}}\right)\ -\ 0.6\ =\ 3.6\ volts\)
 

Kermit2

Joined Feb 5, 2010
4,162
Supply V+ and Vin, are usually the same in circuit. Look for obvious cases of difference, such as a voltage regulator, in which case Vin might be the voltage reg. input and Vin of another device will be the regulator output.
 

Jony130

Joined Feb 17, 2009
5,183
can you just explain me equation (2) what loop used to write the kvl equation.
Just confused how the I2*10K bit came.
From this input loop

Vcc - V1 - V2 = 0V
Vcc = I1*R1 - I2*R2








Also another quick question, when a question relating to transistors has a line "and the supply voltage is 12V" does that mean Vin is 12V? I dont think that is the case because it could relate to another part of the circuit in which a transistor is used.
This depends on the circuit diagram.
In our case Vin for voltage divider is equal to Vcc
 

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