A common-base and a common emitter amplifier

a)

b)

An answer:

pict

c)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{be} \cdot (R_{C} ∕ ∕ R_{in, NEXT}) \\ = + g_{m} v_{in} \cdot (R_{C} ∕ ∕ R_{in, NEXT}) \\ a_{v} & = g_{m} \cdot (R_{C} ∕ ∕ R_{in, NEXT}) \\ = g_{m} \cdot \frac{R_{C} \cdot R_{in, NEXT}}{R_{C} + R_{in, NEXT}} \end{array}

d)

An answer:
From the previously derived small signal equivalent circuit:

\begin{array}{rcl} r_{in} & = \frac{v_{in}}{i_{in}} \\ = R_{E} ∕ ∕ \frac{βfe}{g_{m}} ∕ ∕ \frac{1}{g_{m}} \approx \frac{1}{g_{m}} \end{array}

e)

An answer:
Recycling the answer to the previous question:

\begin{array}{rcl} \frac{1}{g_{m}} & = 250 Ω \to g_{m} = 0.004 [A ∕ V] \to I_{C} = 0.1 mA \end{array}

f)

An answer:
From

a_{V} = g_{m} \cdot (R_{C} ∕ ∕ R_{in, NEXT})

it can be derived that

\begin{array}{rcl} R_{C} & = \frac{a_{v} \cdot R_{in, NEXT}}{gm R_{in} - a_{v}} \\ = 40 k Ω \end{array}

g)

An answer:

pict

h)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} R_{C} v_{be} \\ v_{be} = v_{b} - v_{e} \\ v_{b} = v_{in} \\ v_{e} = (\frac{g_{m} v_{be}}{βfe} + g_{m} v_{be}) \cdot R_{E_{1}} = v_{be} \cdot g_{m} R_{E_{1}} (1 + \frac{1}{βfe}) \end{array}

Working this out yields:

\begin{array}{rcl} v_{be} = v_{in} - v_{be} \cdot g_{m} \cdot R_{E_{1}} (1 + \frac{1}{βfe}) \to \\ v_{be} (1 + g_{m} \cdot R_{E_{1}} (1 + \frac{1}{βfe})) = v_{in} \to \\ v_{be} = \frac{v_{in}}{(1 + g_{m} \cdot R_{E_{1}} (1 + \frac{1}{βfe}))} \\ a_{v} & = \frac{- g_{m} \cdot R_{C}}{1 + g_{m} \cdot R_{E_{1}} (1 + \frac{1}{βfe})} \end{array}

i)

An answer:

r_{in} = R_{B 1} ∕ ∕ R_{B 2} ∕ ∕ (\frac{βfe}{g_{m}} + (1 + βfe) \cdot R_{E 1})

j)

An answer:

\begin{array}{rcl} r_{in} & = 400 k Ω ∕ ∕ 100 k Ω ∕ ∕ (\frac{βfe}{g_{m}} + (1 + βfe) \cdot R_{E 1}) \\ \approx 80 k Ω ∕ ∕ 60 k Ω \approx 35 k Ω \end{array}

$r_{in}$ is smaller than the assumed $40 k Ω$ so the gain of the first stage will be a little lower than the numerical value derived earlier.