BJT Tutorial / Spice question

Discussion in 'General Electronics Chat' started by JStitzlein, Dec 6, 2010.

  1. JStitzlein

    Thread Starter Member

    Dec 6, 2010
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    Hi,

    I am currently working the bjt tutorial on this site with the help of pspice(orcad).

    I'm trying to emulate the following circuit (common emitter amp) in my program, but i am unsure about the current phase:

    [​IMG]

    Here is my pspice model:

    [​IMG]

    As you can see there are two current probes, one on the collector pin of the bjt and on the the left side of the 8 ohm resistor.
    ---
    I don't understand why the two probes gives me different polarity values, i thought they were at the same node?

    The probe on the left, i'm sure is ideal as i've replicated other bjt models, but i can only replicate the source graph by placing the probe on the resistor.

    Is there something i'm missing or can there be an error in the document?

    Thanks, I look forward to your help. great forum:)
     
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Not only does your amplifier produce horrible distortion because it rectifies the signal, but it has an output level that is much less than its input level.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The resistor has two terminals. If you remove resistor R6, rotate it 180°, replace it in the circuit, and re-run the simulation, the polarities will then match.

    Note that the plot in figure 4.30 was created by measuring the voltage of node 1 (high side of V13) and the current flow from the high side of V13.

    I've flipped the 8 Ohm resistor in this schematic and simulation. Note that the polarities are now the same; you can only see one trace as they overlay each other.

    [​IMG]
     
  4. JStitzlein

    Thread Starter Member

    Dec 6, 2010
    53
    0
    Rotating the resistor has no effect on the polarity for me(orcad specific?), it's only affected if i take the current reading at the collector(positive) or resistor(negative).

    The source of my confusion was that I thought I should be getting a negative plot, i'm new to electronics so i've got no electrical sense:/

    I just want to clarify, the plot was only negative because of the way you probed it, and in a real circuit it could only be a positive half wave?

    thank you kindly for the response and the simulation.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    Interesting. I have PSpice Student installed, but hardly ever use it.

    In my simulation, for voltage source V2, there is only one current path; through R2 and the collector of Q1. The current through them must be exactly equal, and in the same direction.

    You might have been measuring the voltage on the collector instead of the current from the resistor. In that case, you would get a more or less mirror image of the current plot.

    As collector current increases, the voltage on the node between the collector and the 8 Ohm resistor will decrease.

    Try probing the other side of your 8 Ohm resistor and see what you get. If it's a constant 15v, you're not measuring current.
     
  6. JStitzlein

    Thread Starter Member

    Dec 6, 2010
    53
    0
    Yes, you're right. Measuring the right side of the 8ohm gives me a constant 15V.

    Could you explain what the difference is? I don't understand how I could've generated a plot if there were no battery and resistor to give me different values.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    I take it back, you were using current probes. Sorry for the confusion.

    Try using a current probe on the right-hand side of the 8 Ohm resistor. You'll see that the current going into the right side of the resistor matches the current flow in the transistors' collector.

    On the side of the resistor towards the transistor, current is flowing out of the resistor, and into the transistor. That's why they are equal in amplitude, yet opposite in polarity.
     
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