Common Base: Inverted Output

Discussion in 'General Electronics Chat' started by JStitzlein, Dec 8, 2010.

  1. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    Hello all,

    I modeled the common base amplifier from the ebook, but my output is inverted and i dont know why.

    Is the spice model correct?

    This is what i'm trying to spice
    [​IMG]

    here is my orcad capture:
    [​IMG]

    i already posted today, but please bear with me - thank you for the patience and help.:confused:
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    PSpice is correct, simply plot voltage on R2. Becaues know you plot V_CB voltages
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Your transistor is saturated. Notice that Vce is only 10mV.
    Try reducing the value of R2 to 1k.
     
  4. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    I tried both of these things but it is still inverted.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Set your reference point to node 3, which is between the 15v battery + and the lower side of the 5k resistor R2. Then measure the voltage on node 4, the high side of R2 and the collector of the transistor.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What do you mean by "inverted"? What is your evidence that the output is inverted?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Ron,
    Our OP's output plot is inverted from the example plot, because their reference point is GND (node 0) instead of node 3.

    If the OP establishes their reference point as node 3, and then probes at node 4, then the plots will match.

    [eta]
    The correct nomenclature for Orcad is "voltage differential markers".
    Put the + marker on the low side of R2, and the - marker on the high side. The plot will then look like it's supposed to.

    [eta]
    See the attached.
     
    Last edited: Dec 8, 2010
  8. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    SgtWookie,
    TyTy - it worked and i learned how to use the differential V+ V- probes as well!

    Why is it that the high side of R2 shows a negative voltage? Is it because though the current flows from the emitter, the resister + voltage source on the collector "opposes & resists" this flow?

    I am also curious to know how alpha & beta are used for a common base circuit for current calculations. I have an example from a textbook that is almost identical to this except that v1 and v2 are -5v & +5V, respectively. This example finds IE using KVL and ohms law with no B, then finds IB using Ib= IE/(1+B), then Ic using Ic= (B/1+B)/Ie.

    [​IMG]

    --
    In the ebooks example, I found Ie the same way using KVL(i got 10ma, while spice says 8.9ma), but I could only solve Ic by viewing the collector side as independent and only dependent on the voltage source + resistor - am i thinking about this the right way?
    Ib was solved by Ie= Ic+Ib

    If v1 & v2 are the same can B be used for calculations?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Good!
    I'm really not very familiar with Orcad; just have Orcad PSpice Student installed, and hardly ever use it.

    The positive terminal of V1 is the most positive (+15v) node in the circuit. As Vin increases, the + terminal is connected to ground, so the emitter winds up with an increasingly negative voltage on it. The emitter resistor R1 limits the base-emitter current.

    I'll leave the rest of your questions for someone else.
     
  10. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    How does the transistor even allow emitter to collector current when the base is grounded?
     
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