Transistor Biasing: Calculating Ic, Ib, and Power Dissipation

[shaunakde]

Problem Statement:
1. Given that the beta of a transistor is 36, Vcc=5v, Vbb= 3.6V, Rc=440 ohm, Rb=640 ohm, find Ic, and Ib. Also find power dissipated across transistor.
Figure: http://yfrog.com/7hdiagram1j
Assume silicon diode. [Ie vbe = 0.7]

The attempt at a solution
I applied voltage equations on the collector and base loop. My equations are:
5 - 440Ic - Vce = 0 ------(1)
and
3.6 -40Ib - 0.7 =0 ------(2)

I got the value of Ib from equation 2.

Now I can't find Ic.
I tried using the following relation:
Ic = beta Ib.
But then Ic comes to 3.625 A which is wrong.

What am I doing wrong? How do I solve this?

 

[Mr.Green]

3.6 -40Ib - 0.7 =0 ------(2)

This has error in number.

3.6 - 640 Ib - 0.7 = 0

 

[vk6kro]

Test for what collector current is possible through the 440 ohm resistor. Even if the transistor was shorted, how much current could flow?

Now, work out the base current and multiply it by BETA.

See why the equations didn't work?

In this case, the base current is comparable with the collector current, so you would even have to allow for the power due to the base current.

Just work it out one bit at a time, without equations.
Voltage across 640 ohm resistor=
so current into base =
Voltage across 440 ohm resistor =
so current into collector =

 

[shaunakde]

Thanks for the hints. I got one step closer to the right solution.

Since Ic and Ib are comparable [the ratio is less than beta] the transistor is in saturation. ie Vce = 0.25 [standard value]

Using this I was able to figure out the correct Ic and Ib.

I still can't figure out the power dissipated across the transistor.
My guess is that P should be Vce * total currect through transistor(Ic +Ib). But the answer is not matching.

 

[vk6kro]

The base current would flow into the base which would have about 0.6 volts across it.

So, you would have to work this power out and add it to the collector power.
The actual power depends on the assumptions about voltage. With a big base current the voltage could be higher than 0.6 volts.

Yes, the transistor is very saturated in this problem, mainly because of the excess base current, so all the equations fall in a heap.