Transistor problem

  • Thread starter bottlesheep
  • Start date
  • #1
http://home.netvigator.com/~cyberforce/img/phy.jpg [Broken]
I have learn transistor just few lessons..

For a)
what i learn is that R1 and R2 would share Vin
that's mean there is p.d. when current flows via R1, R2
thus, Vin = VR1 + VR2
and i am confused if the current flows via them is the same
that's mean I1 and I2 are the same...
but if they are the same...
how come the question ask me both of the seperately??
..Should they be different??

and for IB, I am confused too.
Because when i learned Transistor,
there is a Base bias resistor,say RB, in the circuit
thus, VR2 = (IB)(RB) + VBE
but now there is no RB, so VR2 = VBE..
then I think that there is no potential difference, so no IB
but it seems that I am wrong...
coz part b) assume the current gain is 100..
that means (IC)/(IB) =100, which implies there is IB...

so I don't know how to do the question at all..
I am not good at English and presentation...
I don't know if you understand what I don't understand...

my physics is poor...
please help...
are my concepts totally wrong..? > <..
thank you for your help...
 
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Answers and Replies

  • #2
Guybrush Threepwood
520
1
When you studied the transistor did you learn something about VBE?
 
  • #3
Originally posted by Guybrush Threepwood
When you studied the transistor did you learn something about VBE?

yes, I know..
then..??
 
  • #4
Guybrush Threepwood
520
1
Originally posted by bottlesheep
yes, I know..

what exactly do you know?
 
  • #5
Originally posted by Guybrush Threepwood
what exactly do you know?

current IB starts to flow when VBE is greater than 0.5V...
in my graph...VBE is the potential difference between any points on the wire which between resistors R1and R2 and any points on the earthing wire...
 
  • #6
Guybrush Threepwood
520
1
ok, when I was doing that kind of problems in the college VBE was always considered to be 0.6V. That's a reasonable approximation for the real world.
I don't know if this is acceptable for your problem (you have to check similar problems you've made before in class).
 
  • #7
Originally posted by Guybrush Threepwood
ok, when I was doing that kind of problems in the college VBE was always considered to be 0.6V. That's a reasonable approximation for the real world.
I don't know if this is acceptable for your problem (you have to check similar problems you've made before in class).

ok..
but this circuit is quite different from my previous exercises..
as I mention, there is no Base bias resistor in this circuit which always appears in the exercises I have done before...

and also, I really want to know if I1 and I2 are the same or not?? can you tell?
 
  • #8
Guybrush Threepwood
520
1
Originally posted by bottlesheep
and also, I really want to know if I1 and I2 are the same or not?? can you tell?

no they're not equal, I1 = I2 + IB
start by writing how VBE depends on I2
 
  • #9
Originally posted by Guybrush Threepwood
no they're not equal, I1 = I2 + IB
start by writing how VBE depends on I2

OH..IC..
that means here, I2=IB..
since VR2=VBE...
Is it correct?
 
  • #10
Guybrush Threepwood
520
1
nope
VBE = VR2 = R2*I2 and you get I2. I2 would be equal to IB if you'd had a series circuit, but it's not the case...

then you find out VR1, I1 and after that IB
 

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