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BJT Inverter w/feedback

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data
    The circuit in the figure has the form of a BJT inverter, but also has a resistor connected between the emitter lead and ground. This addition provides feedback between the output loop of the circuit and its input loop.

    (a) Find an exact expression for the Vin-Vout transfer characteristic over the region where Q1 operates in the constant-current region.

    (b) To what does your expression reduce to for large βf?

    (c) At what value of Vin does Q1 first turn on?

    (d) When Q1 saturates, what will be the value of Vout?

    (e) At what approximate value of Vin will Q1 first go into saturation?

    (f) For VCC = 12V, RC = 4.7kΩ, RE, large βF, Vsat ≈ 0.2V, and Vf ≈ 0.7V, draw the approximate transfer characteristic of the circuit.

    (g) The approximate expression obtained in part (b) will overestimate the actual gain of the circuit found in part (a). How large must βF be for the approximation to overestimate the gain by no more than 10%?

    2. Relevant equations
    KVL

    In the constant-current region:
    iC = βFiB
    Vout has the form Vout = mVin + b​

    3. The attempt at a solution
    iE = iB + iC = iB + βFiB = iB(1 + βF)

    KVL input loop:
    [1] iB = [itex]\frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}[/itex]​

    KVL output loop:

    [2] Vout = VCC - iCRC - iERE

    [3] Vout = VCC - βFiBRC - iBRE(1 + βF)

    [4] Vout = VCC - iBFRC + RE(1 + βF))​

    Part (a)

    If I combine equations [1] and [4]:

    [5] Vout = VCC - [itex]\frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}[/itex](βFRC + RE(1 + βF))​

    I want to know if I'm headed in the right direction and where to go from here. Any guidance is greatly appreciated!

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 28, 2013 #2

    rude man

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    Homework Helper
    Gold Member


    No. Vout is collector voltage with respect to ground. Vout is not Vc - Ve.​
     
    Last edited by a moderator: Mar 29, 2013
  4. Mar 28, 2013 #3
    Oh, oops. Vout = VCC - iCRC, so:

    Vout = VCC - [itex]\frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}[/itex]RCβF

    I still don't see what it would reduce to with a large beta :|.
     
    Last edited: Mar 28, 2013
  5. Mar 29, 2013 #4

    berkeman

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    Staff: Mentor

    Just put in a large beta. What does the equation simplify to?
     
  6. Mar 29, 2013 #5
    Maybe I'm blind, as I don't see what it would simplify to. Would it cause RB + RE to be ignored in the denominator? Then, the two beta values would cancel from the top/bottom? So:

    [itex]\frac{(V_{in} - V_{f})R_{C}}{R_{E}}[/itex]
     
  7. Mar 29, 2013 #6

    berkeman

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    Staff: Mentor

    Yes, Rc/Re would be the gain.
     
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