Stumped on another BJT problem

[jack23]

Hello all,

Given the circuit attached, which will used as a constant current source, I need to find a value of Re that will give I = 2mA.

I am very weak when it comes to BJT amplifiers. So what I did was I used MULTISIM to try and find a value of Re that will satisfy this problem. The value for Re from the simulator was 1.6K ohms.

Now I'm not sure if this is correct, but if so, how can I prove this mathematically??

 

[ljcox]

As a first approximation, you can ignore the base current if the transistor's gain is reasonably high.

Therefore the voltage at the base is approx -5 V, so the voltage at the emitter is -5.7 V.

Thus the voltage across Re is -5.7 - (-10) = 4.3 V

So Re = 4.3/2 = 2.15 k

A 2.2k should be near enough.

If you wish, you can check the approximation by calcuating the base current and then determine how much it will change the base voltage.

For example, if hFE = 200, Ib = 2/200 = 10 uA.

10 uA and 500 Ohm means a voltage of 500 * 10/10^6 = 5 mV. Therefore the error in the assumed base voltage is negligible.

 

[jack23]

Great information. Thank you.

Now, assuming that the current Ib is insignificant. How did you get the voltage at the base to be about -5V??

 

[mneary]

How did you get the voltage at the base to be about -5V??

Thevenin.

The two 1K resistors in a 2:1 divider from -10V. The effective source resistance is 1K/2, or 500.

 

[ljcox]

Yes, have a look at Thevenin's theorm, it is very useful.

You also don't seem to understand how to calculate the voltage from a voltage divider.