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Homework Help: Mode of operation for transistor

  1. Feb 10, 2012 #1
    I am trying to determine the mode of operation for the following circuit and find the voltages and currents:
    zsmgkh.png

    I am given that β = 50

    I know that VEC > VEC,sat for a transistor in active mode
    I applied KVL on the right loop and got:
    VCC = 300Ic + 1000Ic + VEC
    VEC = VCC - 1300Ic

    i'm not sure if the above is right or where to go from here. what is throwing me off is the 300 Ω resistor connected to the 1kΩ resistor. is the node between the two B or C? or are B and C equal in this case?

    thanks in advance.
     
  2. jcsd
  3. Feb 10, 2012 #2

    rude man

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    Assuming the transistor is on, what must be the base voltage without writing down any equations at all?
     
  4. Feb 10, 2012 #3
    VBE = VBE,on = 0.7 V (as defined in the text)
     
  5. Feb 10, 2012 #4

    rude man

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    That's the voltage between the base and the emitter. So what's the voltage at the base?
     
  6. Feb 10, 2012 #5

    i'm unsure. is it the voltage at that node between the two resistors?
     
  7. Feb 10, 2012 #6

    rude man

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    Why yes, the base is the node between the two resistors.
     
  8. Feb 10, 2012 #7
    what confuses me about this is the position of Vc. normally, it is the voltage across the bottom resistor, but since there are two resistors with a node in between, is Vc the voltage across the 300-ohm resistor? or does Ic occur in both the 300-ohm and 1k-ohm resistor
     
  9. Feb 11, 2012 #8

    rude man

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    Vc is the collector voltage to ground. "c" stands for "collector".

    The base voltage is Vb = Vcc - Vbe = Vcc - 0.7V.

    Ic is the current flowing out of the collector.
     
  10. Feb 11, 2012 #9
    Vcc = Vbe,on + Vb
    --> Vb = 2.5 - 0.7 = 1.8

    Vb/1000 + (Vb-Vc)/300 = 0
    1.8/1000 + 1.8-Vc/300 = 0
    --> Vc = 2.34

    Ic = 1.8/1000 = 1.8 mA

    2.5 = Vec + 300Ic + 1000Ic
    --> Vec = .16

    since Vec = 0.2 in this problem and Vec is less than this, the transistor is not in active mode so i re-solve for saturation mode?
     
  11. Feb 12, 2012 #10

    rude man

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    Yes that's exactly right, sorry for the delay in answering.

    So now what do you do? (Hint - the base current is no longer assumable to be ic/50).
     
  12. Feb 12, 2012 #11
    i solve for Ie then do the KVL for the CE loop. then solve for Ic from that equation. then get Ib from the relationship Ib = Ie - Ie. if Ib and Ic are both positive then the saturation mode assumption is correct.

    thanks for pointing me in the right direction!
     
  13. Feb 12, 2012 #12

    rude man

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    Yer' welcome! Post your results if you care to.
     
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