BJT problem

Discussion in 'Homework Help' started by ADCapacitor, Feb 3, 2016.

  1. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    hi
    I have some problems about solving the BJT circuit
    Here is my problem
    For the circuit, IA = 3 mA and VCE = 5.4 V
    VBE(on) = 0.7 V and  btea= 50,
    (a). Find RA, RC, and IC3
    (b). Find the quiescent power dissipated in transistor Q3 (DC bias)

    part A

    VBE(on) = 0.7 V and beta = 50,
    IA=0-(VBE-VEE)/RA
    RA=3100 ohm

    IA=Iref+IB1+IB2
    Since
    Iref=IC1=IC2,
    IB1=IB2,

    IA=IC2+IC2(1/beta)+IC2(1/beta)
    IC2=2.88mA

    I am confused about whether Io is equal to Ic3 or not.(Io=Ic3)
    And I assume Io=Ic3
    VCC-VEE=IC3*RC+VCE
    20=2.88m*RC+5.4
    RC=5069 ohm

    part B
    P = IB3*VBE + IC3*VCE
    P=0*0.7+5.4*2.88m
    P=0.016W

    Is my calculation correct?
     
    • BJT.JPG
      BJT.JPG
      File size:
      26.2 KB
      Views:
      13
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    What about Q3 beta ?? Is Ie current equal to Ic current ?
    Also are you sure about Rc part ? What is the voltage drop across Rc resistor ?
     
  3. Russmax

    Member

    Sep 3, 2015
    81
    12
    Jony130 is asking the right questions. You were doing well until you got to Q3, and forgot about its beta. Then you botched the equation for calculating Rc. You're making it too hard, ADCap. V is given. Find Ic3. Rc = (5.3V / Ic3). For part A, you don't care about the supply voltages or the voltages across other devices.

    You're using the right equation for part B, but you neglected beta of Q3, just like you did in part A.
     
  4. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    Thank you
    May be I use this one to find Ic3?

    Io=Ic2
    then
    Io=Ib3+Ic3
    Io=Ic3*(1/beta)+Ic3
    Ic3=Io/(1/beta+1)
    and
    Ic3/beta=Ib3
    So I can find the Ic3 and Ib3 ?
    then calculate the Rc and the power of q3?

    But the ac signal is considered as short circuit in dc biasing ?
    Then, the base of Q3 connect to the ground and Ib is 0A?
    Therefore, I don't understand why Q3 beta still need to be considered and think (Ic3=Io) previously.
     
  5. Russmax

    Member

    Sep 3, 2015
    81
    12
    You are correct until the second to last line.
    Yes, the AC signal is a short in DC biasing. Since when does a short block DC current? The AC signal has a DC voltage of zero, but the current will be whatever it needs to be. If the AC source blocks current from flowing from ground, how can you call that a short?

    If you are correct that Ib3=0, then Ic3 must also equal zero, since Ic3=Ib3(beta). A transistor does not suddenly have infinite beta because its base is tied to ground. No, the beta is still 50. The answer is that Ib3 is NOT zero, and current is certainly able to flow from ground (0V) to -0.7V across the Vbe3 diode.

    Perhaps you have the idea that current can flow into, but not out of, ground. No--ground is not an ideal diode. Current can flow both ways. Perhaps you are confusing ground with the virtual ground created by an ideal op amp, when the non-inverting input is tied to ground and the inverting input voltage and current are both zero. This is not virtual ground, but real ground, like the ground you get when you stand on a wet floor and insert your finger in a light socket. Ground just means that the voltage is zero, but if you short ground to a supply, you approach infinite current.
     
    ADCapacitor likes this.
  6. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    Thank you for the detailed explanation, Russmax
    I get it.
     
Loading...