A PNP

Exercise 4.13 A PNP

The figure below shows the symbol of a PNP transistor, identifying its emitter (E), base (B) and collector (C) nodes. Including all plus and minus signs in the element equations, the I-V relations for this device are:

\begin{array}{rcl} i_{C} = I_{C 0} \cdot (e^{- \frac{q \cdot v_{BE}}{kT}} - 1) \cdot (1 - e^{- \frac{q \cdot v_{CE}}{kT}}) \\ i_{B} = \frac{i_{C}}{βfe} \end{array}

where $I_{C 0}$ is a (negative valued) device property assuming the arrows as shown in the figure and with the listed element equations. Note that a positive conventional current flows opposite of the arrows: out of the base and out of the collector node.

a): Derive a proper small signal equivalent circuit for this pnp type BJT, including equations relating the small signal parameters to bias settings.
b): Compare the SSEC of the pnp with that of an npn. Describe the differences (if any) in value, sign, ..., equations relating the small signal properties to bias settings, ...

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