Transistor behavior in LTSpice

Discussion in 'Homework Help' started by hernejj, Sep 5, 2012.

  1. hernejj

    Thread Starter New Member

    Sep 5, 2012
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    So this is not really homework, but I'm trying to understand how to work with transistors and I do not quite understand some things I am seeing in the circuit I made in LTSpice. Please refer to attached screen shot:

    Question #1:
    In the top part of the graph, you'll see that V(nb) is very close to 1V.
    Since my supply is 10V and I only have a resistor (R3), and the Q2 base-emitter junction in series with the source I would have expected V(nb) to be a lot closer to 10V than 1V. Can anyone explain the value of V(nb)?

    Question #2:
    Considering I have a voltage divider feeding 'na' I would expect V(na) to be significantly less than V(nb) as, in theory, the 10V supply should be cut down to 5V. But V(na) is somehow HIGHER than V(nB)..... Help??? :)
     
  2. bretm

    Member

    Feb 6, 2012
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    Replace your base-emitter junctions with diodes. Do the results still surprise you? What happens to the junction current once the voltage gets above about 0.7V?
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    Don't you know that the base-emitter junction of a transistor IS A DIODE?
    Don't you know the forward voltage of a diode?
    Didn't you ever see the base-emitter voltage for a transistor in its datasheet?

    R7 in your circuit has a very low current and does almost nothing.
     
  4. yakuut

    New Member

    Sep 21, 2009
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    Doesn't the R7 reduce the base voltage by acting as a resistor divider
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    Barely.
    Since the base-emitter junction is a diode then its voltage (when its current is very high at 16mA) is 0.95V. The current in R7 is only 0.95V)/500 ohms= 1.9mA. So R7 reduces the base current a small amount and reduces the base voltage only a very tiny amount.
     
  6. hernejj

    Thread Starter New Member

    Sep 5, 2012
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    Thanks for the replies. If seems as if the B-E junction is essentially a short circuit and that causes R7 to see very little load. Furthermore, I guess a voltage divider only works when the load resistance is >= the resistance of the divider?

    I think I've also figured out #1. Heres what I scratched out:

    v1 = 10 V
    r3 = 1000 Ohm

    Vbe = 0.7 (Standard diode V-drop)
    Rbe = 25 Ohm (I guess the B-E junction has some resistance on this order?)

    Iq2b = V/R
    Iq2b = (10 - 0.7) / 1000 + 25
    Iq2b = 9.3 / 1025 = 9.07 mA (this is close to the measured current at Q2 base. measured is 9.05 mA)

    V1 = Vr3 + Vq2
    Vq2 = V1 - Vr3

    v=ir
    vr3 = 9.07 mA * 1000 = 9.07V

    Vq2 = 10 - 9.07 = 930mV (this is close to the measured value although it is still a bit off. measured is 950 mV).

    Does this logic make sense? I'm assuming the differences between my calculations and the actual measured result are due to my arbitrary choice of 0.7v drop for the B-E junction and the arbitrary 25 Ohm resistance.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    A voltage divider biases the base of a transistor very well only when the transistor has a series emitter resistor for good bias stability.

    The base-emitter junction IS A DIODE, not a resistance. A HUGE difference.
    When the current through a resistance is doubled then its voltage drop doubles.
    But when the current through a diode is doubled then its voltage drop increases only a tiny amount.

    The circuits shown are wrong because their output voltage depends on the current gain of the transistor which can be high or low even for the same transistor part number.
    When an emitter resistor is added (or other ways of providing negative feedback) then different current gains do not affect the output voltage as much.
     
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  8. hernejj

    Thread Starter New Member

    Sep 5, 2012
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    Thanks everyone for the great info :)
     
  9. bretm

    Member

    Feb 6, 2012
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    Vbe of the second transistor is identically Vnb. Makes the math in your previous post much easier.
     
  10. ramancini8

    Member

    Jul 18, 2012
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    Take the Thevenin looking out of the base of Q1 and you can see that the two circuits are identical except for the resistor values.
     
  11. ramancini8

    Member

    Jul 18, 2012
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    On second thought, I advise you to concentrate on transistors first until you learn how they operate; then after you have accomplished that you can learn a computer program. I have seen many EE graduates who depended on computers and not one of them is worth much.
     
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