# Kirchhoff's Voltage Law for PNP Transistor Circuit

1. Mar 27, 2013

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.

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:

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

The circuit in question:

2. Relevant equations

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

Voltage Divider Law --> $V_x=\left (\frac{R_x}{R_T} \right )E$

3. 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.

$$V_{TH}=V_{R2}=\left (\frac{R_2}{R_1+R_2} \right )\left ( -V_{CC} \right )$$

$$R_{TH}=\frac{R_1R_2}{R_1+R_2}$$

The Thévenized circuit is now:

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

$$V_{TH}-I_ER_E-V_{BE}-I_BR_{TH=0}$$

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

$$I_E\left ( R_E+\frac{R_{TH}}{\beta } \right )=V_{TH}-V_{BE}$$

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

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

2. Mar 27, 2013

### 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...

3. Mar 28, 2013

### CWatters

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