Help with BJT Problems, if you please

[ussfletcher]

These are due in the morning, and I'm kinda stumped as to how to solve for all but A) with Beta(f) = 100

Thanks

 

[esun]

Explain what problems you're having and I'll help. This stuff is very straightforward but I'm not going to do it for you.

 

[ussfletcher]

Alright, well honestly my problem lies with part B at this point, I am having trouble figuring out what model I should be using (Active, Saturation, Cut-off) etc. Every model I have tried to use has been inconsistent so far. i.e Base current was negative for Active, which I'm not sure is ok.

Thanks

 

[esun]

Okay so you ruled out forward active. Why can't it be in saturation or cut-off? Show me what you derived there.

 

[ussfletcher]

Alright, well for saturation, I thought that Vce would not be equal to Vce(sat) because of the 15v source. i.e 15V + .2v is not equal to .2v

For cut-off it requires Vbe > Vbe(on) (from my notes) which would not be true, because it would be atleast equal if not less than.

Edit:
Hm, i think I found a contradiction in my notes here

 

[ussfletcher]

I now believe that they match the conditions for cut-off.. would this be correct?

 

[esun]

Okay you're starting to get. I think the most important thing to remember about the BJT is that it really is just two pn junctions and that cut-off, saturation, and active are just three combinations of those pn junctions being on/off.

So in cut-off both junctions are off. That means very little (or, in approximation, zero) current is being conducted through the junctions. This means I_B and I_C are both approximately zero. How do we verify this is true? We use the assumption that I_B and I_C are zero and solve for V_BE and V_BC and see if they are large enough to turn on their respective junctions.

If I_B = 0, then we can see that V_BE < 0 since 470k < 1M, which satisfies our constraint. If I_C = 0, then V_BC < 0 as well since there will be no drop across the 10k resistor. Therefore the conditions for cut-off are satisfied.

 

[ussfletcher]

So then, would V_ce be 15v? , because I see that that must be > .2v

 

[esun]

Think about it. In cut-off I_C = 0. V_CE comes naturally from that.

I don't see why you think V_CE must be greater than 0.2 V either. Why do you think that?

 

[William Gaatjes]

Did you succeed in your tests ?

If your interested, i always use an ideal version of a transistor to start with.
Then some variables become constants and it is easier to see.

If you take V-be as a constant and assume there is no base current you can simplify things to start with and then afterwards you replace all ideal situations with real life situations and recalculate.
Always write down the b c and e for the base collector and emitter respectively.

Always write down that there is a V-be between base and emitter and with the polarity signs.
Even draw a small diode in parallel with the transistor where anode is B and kathode is E for a NPN BJT (reverse it for a PNP).
When you do that you see you have a lot more information to use.

Ideal :
V-be is always 0,7 Volt seen from the emitter. V-ce (saturation) is 0 Volt.

For example draw up 2 situations :
1, where the transistor is maximal conducting and Ic = V power supply / Rload. Rload = Rc. Rce = 0Ohm. Rce is 0 Ohm because we assumed an ideal transistor with V-ce(saturation) = 0. 0/ I = 0.


See the transistor ce connection as a single resistor of 0 Ohm. And place a virtual volt meter over the rCollector and Rce.
What do you measure ?


2, where no current flows at all. See the transistor ce connection as a resistor with infinite resistance in series with R collector. Then place again a virtual voltmeter over Rcollector and over Rce. What do you measure ?


Sometimes you have to start from the end to get to the beginning.
You know what you want as output. Then when you know this, you calculate back to the input.

And use Ohms law. U = I * R is your best friend here.

I have on purpose left 1 important thing out. ^_^
Why does a transistor conduct a current from collector to emitter ?