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Common-Emitter Amplifier (Collector Voltage)

  1. Jun 1, 2013 #1

    My question is regarding an example of a common-emitter transistor amplifier shown in my electronics book. Here is the diagram of this particular amp operating from a DC power supply:


    My book says: "the transistor has β = 99, with no signal applied the voltage at the base of the transistor is measured to be 6 volts, so the dc voltages at the emitter and collector of the transistor are Ve=5.3 V, Vc=7 V respectively."

    So I don't understand how you can have 7 volts at the collector in this particular circuit. :confused:

    My book always makes the assumption that b-c voltage drop is 0.7. So we have

    ##I_e = \frac{V_e=6-0.7}{5.3 \ k\Omega} = 1 \ mA##​

    And ##I_e \approx I_c##, therefore ##V_c=I_cR_c = 5 \ V##! So is the book wrong, or is there something wrong with my calculation?

    We know that ##V_{cb} = 12-6 =6 V##. Hence Vc can't be greater than 6. I'm very confused and I think it might be a typo. Any help would be greatly appreciated.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 1, 2013 #2
    If IC = 1mA then the voltage drop across RC is 5V. Your book is correct.
  4. Jun 1, 2013 #3

    The Electrician

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    Gold Member

    Why do you think that Vc can't be greater than the base voltage?
  5. Jun 1, 2013 #4
    I agree with The Electrician, you are wrong to state that Vc-b =12-6=6V
  6. Jun 1, 2013 #5

    jim hardy

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    What polarity?

    double check that statement. i'd have expected them to say that about voltage b-e not b-c.
  7. Jun 2, 2013 #6
    My mistake, I meant to say b-e voltage drop is 0.7 volts. That is why we had ##V_e =6-0.7= 5.3 \ V##.
    Last edited: Jun 2, 2013
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