JJBladester
Gold Member
- 281
- 2
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 IE. The book gives IE as:
I_{E}=\frac{-V_{TH}+V_{BE}}{R_E+R_{TH}/\beta _{DC}}
The circuit in question:
Homework Equations
Kirchhoff's Voltage Law --> Sum of voltage rises + drops = 0
Voltage Divider Law --> V_x=\left (\frac{R_x}{R_T} \right )E
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.