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Equivalent Circuit

  1. Aug 8, 2016 #1

    1. The problem statement, all variables and given/known data

    Why are the two circuits below equivalent?
    c7521dabf48d8801a085b8aa5407156f.png

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Background:
    I am studying BJTs, and in my textbook the equivalent resistance of resistor 1 and 2 is noted to be in parallel. The text provides no explanation why. In my search for an explanation I stumbled upon a video from David Williams (https://goo.gl/VCCuCy) where he redraws the circuit on the left. Once it is in the form on the right I can identify why the resistors are in parallel, but I do not understand how Mr. Williams redrew the circuit.
    I would greatly appreciate if someone could walk me through their thought process. For now I can simply remember to do that transformation, but I want to understand the fundamentals so I can apply that technique to different circuits.
    If my usage of the terms transformation and equivalent are incorrect for this case please let me know.
     
  2. jcsd
  3. Aug 8, 2016 #2

    Tom.G

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  4. Aug 8, 2016 #3
    Tom I have some exposure to thevenin equivalent circuits, but in this case I do not see how that applies. In the link you provided for me there are no examples that closely resemble the sample problem I provided.
    My confusion stems from the role that Vcc in the circuit. Why is it acceptable to redraw the circuit with Vcc on both sides? How does redrawing it take into account the direction the current flows in?

    Ultimately I did not think this involved thevenin's theorem because I am not reducing anything. My understanding is TT involves reducing multiple resistors to a single resistor that is in series with an equivalent voltage. As you can see in the picture above the number of resistors does not change.
     
  5. Aug 8, 2016 #4

    LvW

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    In both circuits, the resistors R1 and Rc are connected to a voltage source Vcc.
    Assuming (a) that the value of Vcc is the same and (b) that it is an IDEAL source (indpependent on the drawn current), there is no reason that both circuits should NOT behave identical.
     
  6. Aug 8, 2016 #5

    Tom.G

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    The significant current in the Base circuit is the the current going into the Base, regardless of where it is coming from. If the total current from the original Vcc source was needed, then the R1 current would be needed. but in this case no one is asking for the current from the original Vcc source.

    The added source and its Thevenin resistance have the same current loop thru the Base, B-E junction, Emitter and Re as the original circuit does; therefore the same effect on the total circuit.
     
  7. Aug 8, 2016 #6
    Thank you both for your replies. Upon looking at it again, it made more sense to me when I paid attention to the fact that R2 and Re are connected to ground.
    Applying KVL also made me more comfortable in recognizing nothing changed.

    One last follow up question:
    By examining the figure on the left, would you have immediately recognized that the resistors are in parallel? I am very novice, and have always relied on parallel resistors sharing 2 branches. This caught me by surprise, and without transforming it like Mr. Williams did I would not have recognized the resistors were in parallel.
     
  8. Aug 8, 2016 #7

    Tom.G

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    As far as the Voltage source is concerned, the resistors are physically in series, no doubt about that. The trick is if you look at the source and series resistors from the R1-R2 junction, the equivalent resistance from that point is that the resistors are in parallel. Keep in mind that an ideal voltage source has zero resistance and can be replaced with a wire when considering current.

    To help convince yourself,
    1) Find the open circuit voltage at the R1-R2 junction when disconnected from the transistor
    2) Find the current from the R1-R2 junction if you short that point to ground (equivalently short R2 and find the current.)
    3) Divide the voltage from step 1) by the current from step 2)
    This will give you the equivalent, or Thevenin, resistance at the R1-R2 junction.

    I agree, it's not an obvious transformation. It is however quite useful in some circumstances, especially once you get used to looking for it.
     
  9. Aug 8, 2016 #8

    NascentOxygen

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    Staff: Mentor

    It depends on ones point of view. As far as the Vcc supply is concerned, R1 and R2 certainly are in series.

    But place yourself at the base---what does the base see its current coming from? The base doesn't see a voltage source of the full Vcc voltage, the base sees its current apparently coming from a source of reduced voltage. This voltage is determined by the R1, R2 potential divider, and this apparent source seems as far as the base is concerned to have an impedance given by the Thévenin equivalent of that potential divider arrangement, viz., R1║R2.

    If you were to replace the base biasing circuit (comprising Vcc and R1 and R2) with a voltage source of the Thévenin-equivalent volts in series with the Thévenin-equivalent resistance, the base would be unable to tell the difference. That's why it's termed "equivalent", because it really is equivalent.
     
  10. Aug 8, 2016 #9

    CWatters

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    In the AC small signal model R1 and R2 would be in parallel because the impedance of Vcc is considered to be zero.
     
  11. Aug 8, 2016 #10

    CWatters

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    You can split VCC in two like that because of the "rule" that says you can connect two nodes with a wire (without affecting a circuit) if they are at exactly the same voltage.
     
  12. Aug 13, 2016 #11
    Thank you all for your help. Sorry for the late response.

    I feel more comfortable. This thread can be locked/marked as resolved.
     
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