An NPN amplifier

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Exercise 5.11 An NPN amplifier

a)

An answer:
pict

After replacing caps by shorts, inductors by opens, and replacing DC sources by their SS-equivalents:

pict

After replacing the non-linear device(s) by their SSEC:

pict

After cleaning up the winding ground wire we get the easiest SSEC:

pict

b)

An answer:
From the small signal derivations should be performed. Note that assuming driving the output port, all other independent sources are set to zero, here

v_{in} = 0

; the other proper way is to use

v_{in}

and calculate the unloaded output voltage and the output current if the output is (small signal wise) shorted.

\begin{array}{rcl} z_{out} & = \frac{v_{out}}{i_{out}} \\ i_{out} = i_{R 4} + gm \cdot v_{be} \\ i_{R 4} = \frac{v_{out}}{R_{4}} \\ v_{be} = 0 \\ z_{out} & = R_{4} \end{array}

c)

d)

An answer:
For input impedance, the sign of the

v_{in}

is irrelevant, according to Ohm. Because the base (in this SSEC, for this circuit) is NOT shorted to SS-ground, the small signal base voltage is NOT zero. This is different from the simplified situation that is frequently used. Tying the base to ground is wrong for this circuit. A proper derivation (one of many proper ones) is:

\begin{array}{rcl} z_{in} & = \frac{v_{in}}{i_{in}} \\ i_{in} = i_{R 3} + i_{βfe ∕ gm} - gm \cdot v_{be} \\ i_{R 3} = \frac{v_{in}}{R_{3}} \\ i_{βfe ∕ gm} = - \frac{gm}{βfe} v_{be} \\ v_{be} = - \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{in} \\ i_{in} = \frac{v_{in}}{R_{3}} + \frac{gm}{βfe} \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{in} + gm \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{in} \\ z_{in} & = R_{3} ∕ ∕ \frac{βfe ∕ gm + R_{1} ∕ ∕ R 2}{βfe ∕ gm} (\frac{1}{gm} + \frac{βfe}{gm}) \end{array}

e)

An answer:
A derivation (one of many) is:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} & = R_{4} \cdot - g_{m} \cdot v_{be} \\ v_{be} = v_{b} - v_{e} \\ v_{e} = - v_{in} \\ v_{b} = \frac{R_{1} ∕ ∕ R 2}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{e} \\ v_{be} = \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{in} \\ v_{out} & = - R_{4} \cdot g_{m} \cdot \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} v_{in} \\ a_{v} & = - R_{4} \cdot g_{m} \cdot \frac{βfe ∕ gm}{βfe ∕ gm + R_{1} ∕ ∕ R 2} \end{array}

f)

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