Transistors, i can't understand how to solve this problem

1. Jun 30, 2010

kliker

having this transistor how can I find Vo? We know that βdc is too large

and also R1 = 10KΩ and R2 = 30KΩ

i just cant understand what to do, I mean it's pnp, so why does the current goe from top to bottom?

shouldnt it go from bottom to top? since it's pnp and not npn?

also, we have no resistors through the way where the symbol of pnp is, so wouldnt the whole current just go from there? so there would be no current through the resistors, or not?

i please ask someone to explain this in detail, i ve been searching for a month, and found nothing that would help me understand this problem

2. Jun 30, 2010

kliker

I think that I found out the solution

while I was taking a bath 5 minutes ago I thought that maybe the sign next to Vcc shows the current of electrons, which is correct in pnp transistor

we know that b is too large but also we know that b = Ic/Ib

so if b too large then Ib -> 0

hence we are in cut off region where Vce = Vcc

also in cut off region we can assume that the transistor is open hence the whole current will go from the two resistors

that means that

V0 = Vcc - I(R1+R2)

am I correct?

3. Jun 30, 2010

Staff: Mentor

No, the arrow in a current source points in the direction of positive current, not in the direction of electron flow.

So I agree that the transistor looks like it is drawn upside-down. If it really is like that, then you would need to use the reverse Beta number for the transistor in your calculation.

4. Jun 30, 2010

kliker

but how can positive current go like this in pnp? Shouldnt it work like this in npn transistor?

Also, I'm not sure what is reverse beta number, can you please explain it more? thanks a lot :)

5. Jun 30, 2010

Staff: Mentor

As I said, the transistor is drawn upside-down from its usual orientation.

Remember that there is also a PN junction between the collector and the base, and if you draw it, the PN junction points into the base, just like the emitter-base junction. So in a sense, the transistor is symmetrical. However, real BJTs are generally built asymmetrically, with different sizes and doping profiles for the EB and CB junctions. This is done to optimize the forward Beta.

But you can run a BJT backwards (upside-down), you just get a lot less beta (and some breakdown voltages change, I believe). I don't think most BJT datasheets even list a reverse Beta (reverse hfe) number, though...

http://www.fairchildsemi.com/ds/MM/MMBT3906.pdf

.

6. Jun 30, 2010

kliker

thanks a lot i think I understand now

so we have something like this

in our book they use the symbol in a different way but now i can understand that it is the same thing

hence

again the logic is the same I guess

since we have b too large, then Ib -> 0, we are in cut off region

that means all current will go from the two resistors

and V0 = Vcc - I(R1+R2)

since we know that VBE = 0.7

but where is actually BE? If it is upside down, then Vbe will be in the loop that contains R1

but I think Im wrong here, or not?

7. Jun 30, 2010

Staff: Mentor

You're on the right track. You will get about 0.7V drop from C-B, and current flowing out of the base into the junction between the two base resistors. There will be current flowing through the resistors (with Ib extra flowing down through R2), and some current flowing from top to bottom (C-E) through the transistor. Equilibrium will be established with some value of the base voltage. I guess you could assume some low value of Beta for this reverse configuration.... something like 10-20?

8. Jun 30, 2010

kliker

but since it says that beta is too large, cant we assume that it is in cut off region? hence the current will be going only through the two resistors?

9. Jun 30, 2010

Staff: Mentor

I don't know what it means to say "βdc is too large". Especially when the transistor is being used in reverse-Beta mode.

But whatever. If Beta is large, all that means is that you can neglect the Ib compared to all the other currents. So yes, the same current flows in R2 as R1. But when you neglect the base current, I'm not sure how you calculate the transistor current? You need Ib and Beta for that, no? Are you given a value for the current source?

10. Jun 30, 2010

kliker

in my book it says that when we are in cut off region, there wont be any current going through the transistor, because we can assume that it's open, like in an open circuit

so the current will go from the two resistors only

but im not sure if i understood this 100%

11. Jun 30, 2010

Staff: Mentor

While it's true that there is no collector current in cutoff, that does not equate to saying that a BJT with high Beta is always in cutoff.

In the circuit you posted, I believe that the circuit will settle with some transistor current and some current through the two resistors. Do they give you a value for the current source?

12. Jun 30, 2010

Staff: Mentor

Note -- it only takes a fraction of a mV to make 0.7V across the 10k Ohm R1. The rest of the current from the current source must be flowing through the transistor. If Ib were not negligible, that would change things only slightly.

13. Jun 30, 2010

kliker

no they give no value for Vcc

only for R1 and R2

in the circuit it wants actually only V0

if we assume that there is no current going through the transistor

will V0 be just I(R1+R2)?

i think im wrong again

the circuit is very confusing for me :(

14. Jun 30, 2010

Staff: Mentor

First, don't assume that there is no current going through the transistor. First, it is wrong, and second, it would keep you from being able to solve the problem.

So to solve this problem, just picture the transistor flipped so that it looks more normal. As I said, the only difference is a much lower Beta, which apparently you can just ignore in this problem.

Next, you know the voltage across R1, right? That gives you the current through R1. Does that tell you something about the current through R2 (again assuming Ib = 0)? And then what does Vo have to be?

BTW, this type of circuit (usually with the BJT in the correct orientation for the best Beta) is called a "Vbe multiplier". Quiz Question -- why is it called that?

15. Jul 1, 2010

kliker

Vbe multiplier? Hm, I don't know, actually we ve never done something like that before

I can't understand how it is possible to have current through transistor

since Ib = 0

we know that Ic = beta*Ib

so if Ib is 0 Ic should be 0 too, or no?

let's just assume that there is no current going through the transistor

we have a voltage devider

so V0 will be Vcc - I(R1+R2)? from the picture I cant understand if it will be like this or V0 = I(R1+R2)?

Last edited: Jul 1, 2010
16. Jul 1, 2010

Staff: Mentor

No, Ib and Ic are not zero. When the problem says that the Beta is very high, it just means that the Ib is negligible to support the Ic.

Google Vbe multiplier. Wikipedia covers it.