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Allowable Signal Swing in BJT?

  1. Aug 26, 2010 #1
    I'm utterly confused about this concept and the text book quite efficiently confuses me further. I'll just try to explain it with an example:

    The circuit has a V(cc) = 10 V and V(ee) = -10 V. V(c) = 2 V, V(e) = -1.7 V, V(b) = -1 V, I(c) = 1 mA, I(e) = 1 mA.

    It says that allowable signal swing or B=100 is +8 V, -3.4 V.

    How's that? I can't figure.
  2. jcsd
  3. Aug 26, 2010 #2
    If this is a homework problem, it should go there. What is the book asking for, the input peak to peak input voltage?

    The only thing I don't understand is what is the -3.4 V. B = 100 is probably referring to the beta or current gain of the transistor. The 8 V seems like a reasonable value for the output signal voltage swing. From the other data given it is possible to calculate the input peak to peak current and peak to peak voltage.

    What exactly is confusing you?
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