Kirchhoff's Voltage Law for PNP Transistor Circuit

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JJBladester
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Homework Statement



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


Homework 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]

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.
 
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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...
 
Seeing the circuit drawn that way gives me vertigo! Flip your first diagram vertically to help with the intuition stuff.

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