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Kirchhoff's Voltage Law for PNP Transistor Circuit

  1. Mar 27, 2013 #1

    JJBladester

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    Gold Member

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

    I seem to have no problem applying KVL to NPN transistor bias circuits, but a world of trouble getting my polarities straight on PNP transistor bias circuits. The +'s and -'s are driving me crazy. :rolleyes:

    The following circuit was presented in the "Voltage-Divider Biased PNP Transistor" section of my Electronics textbook. My task is to find IE. The book gives IE as:

    [tex]I_{E}=\frac{-V_{TH}+V_{BE}}{R_E+R_{TH}/\beta _{DC}}[/tex]

    The circuit in question:

    01.jpg

    2. Relevant equations

    Kirchhoff's Voltage Law --> Sum of voltage rises + drops = 0

    Voltage Divider Law --> [itex]V_x=\left (\frac{R_x}{R_T} \right )E[/itex]

    3. The attempt at a solution

    The first thing I did was redraw the circuit.

    02.jpg

    Then I used Thévenin's Theorem to get reduce the left-hand "window" to one voltage source and one resistance.

    [tex]V_{TH}=V_{R2}=\left (\frac{R_2}{R_1+R_2} \right )\left ( -V_{CC} \right )[/tex]

    [tex]R_{TH}=\frac{R_1R_2}{R_1+R_2}[/tex]

    The Thévenized circuit is now:

    03.jpg

    Now comes the part that I always screw up; getting the polarities correct on my KVL equation...

    [tex]V_{TH}-I_ER_E-V_{BE}-I_BR_{TH=0}[/tex]

    [itex]V_{TH}-I_ER_E-V_{BE}-\left (\frac{I_E}{\beta } \right )\left ( R_{TH} \right )=0[/itex] because [itex]I_B=\left (\frac{I_E}{\beta } \right )[/itex]

    [tex]I_E\left ( R_E+\frac{R_{TH}}{\beta } \right )=V_{TH}-V_{BE}[/tex]

    [tex]I_E=\frac{V_{TH}-V_{BE}}{R_E+R_{TH}/\beta }=\frac{\left (\frac{R_2}{R_1+R_2} \right )\left ( -V_{CC} \right )-0.7V}{R_E+R_{TH}/\beta }[/tex]

    I also drew this little diagram to help me with the PNP transistor because I tend to get confused about the polarity of the base-emitter junction:

    04.jpg

    I think my answer is the same as the book's but I'm not sure. They don't give VBE as a specific voltage level so I don't know if it's +0.7V or -0.7V.
     
  2. jcsd
  3. Mar 27, 2013 #2

    berkeman

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    Staff: Mentor

    Seeing the circuit drawn that way gives me vertigo! Flip your first diagram vertically to help with the intuition stuff. More poisitive supplies go toward the top of a circuit diagram, and more negative supplies go toward the bottom.

    If I can get rid of the vertigo, I'll try to check your equations...
     
  4. Mar 28, 2013 #3

    CWatters

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    Science Advisor
    Homework Helper

    Exactly my thoughts. The very first circuit is upside down.
     
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