Past exam questions: BJT and Diode

[sphericow]

Homework Statement

[PLAIN]http://dl.dropbox.com/u/14490370/bjt.png

Homework Equations

beta=I_c/I_b
By 'ON' I think the lecturer means the saturation region and by 'OFF' he means cut off.

The Attempt at a Solution

a. My reasoning is I_b= (V_s-V_{base emitter})/R_total(6V-0.6V)/(20,000)=~ 270uAmps

b. Now if beta = 50 then I_c=0.0135A
Therefore collector voltage = 6-(1000*.0135)=-7.5V - This can't be right so the bjt must be saturated instead. It follows that collector voltage = 0V.

Is this correct line of reasoning? It seems like I have answered part c so i can answer part b; or I'm completely wrong.

 

[vk6kro]

The figures look OK.

A transistor will not saturate at 0 volts, though. It is typically 0.2 V to 0.5 volts.

 

[sphericow]

Thanks for the reply i really appreciate it.

d. The diode will be forward biased
e. This is my reasoning
[PLAIN]http://dl.dropbox.com/u/14490370/currents.png
IF that's correct how would I work out I₂=I_b

 

[sphericow]

Double post: Would I_b=0 since current takes path of least resistance?
If so, transistor will be OFF and therefore Vce = 6v, since Ice = 0.

 

[vk6kro]

So, if the bottom 10 K has 0.6 volts at each end of it (relative to ground), what would be the current through the resistor?

So, what is the base current?

 

[sphericow]

So base current = 0. Then I₁=I₃=(6-.6)/10k.
Since bass current = 0 the transistor must be off (cut off region).

 

[vk6kro]

So base current = 0. Then I₁=I₃=(6-.6)/10k.
Since bass current = 0 the transistor must be off (cut off region).

Yes. So the collector voltage equals the supply voltage.

I ran a simulation on this and got 0.544 mA through the diode and 2 nA base current.

In a practical situation, the diode in this circuit would possibly be another transistor which was in saturation, so there would be little chance of any base current.

 

[sphericow]

Thanks so for the help, I really appreciate it. My physical intuition seems to fly straight out the window when I'm analyzing circuits; so its really great to have someone confirm my reasoning.