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Transistors, i can't understand how to solve this problem 
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#1
Jun3010, 02:07 PM

P: 106

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 thanks in advance 


#2
Jun3010, 04:23 PM

P: 106

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
Jun3010, 04:42 PM

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P: 40,717

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 upsidedown. If it really is like that, then you would need to use the reverse Beta number for the transistor in your calculation. 


#4
Jun3010, 04:44 PM

P: 106

Transistors, i can't understand how to solve this problem
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
Jun3010, 04:54 PM

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P: 40,717

As I said, the transistor is drawn upsidedown 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 emitterbase 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 (upsidedown), 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%2FMMBT3906.pdf . 


#6
Jun3010, 05:10 PM

P: 106

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
Jun3010, 05:26 PM

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P: 40,717

You're on the right track. You will get about 0.7V drop from CB, 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 (CE) 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 1020?



#8
Jun3010, 05:28 PM

P: 106




#9
Jun3010, 05:34 PM

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P: 40,717

I don't know what it means to say "βdc is too large". Especially when the transistor is being used in reverseBeta 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
Jun3010, 05:38 PM

P: 106

so the current will go from the two resistors only but im not sure if i understood this 100% 


#11
Jun3010, 05:40 PM

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P: 40,717

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
Jun3010, 05:42 PM

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P: 40,717

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
Jun3010, 05:43 PM

P: 106

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
Jun3010, 06:26 PM

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P: 40,717

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
Jul110, 02:17 AM

P: 106

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)? 


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