- #1

JJBladester

Gold Member

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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.

The following circuit was presented in the "Voltage-Divider Biased PNP Transistor" section of my Electronics textbook.

**My task is to find I**The book gives I

_{E}._{E}as:

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

The circuit in question:

## 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.

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:

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:

**I think my answer is the same as the book's but I'm not sure.**They don't give V

_{BE}as a specific voltage level so I don't know if it's +0.7V or -0.7V.