Circuit analysis of a emitter follower with bootstrap

I'm trying to analyse the circuit below (from Art of Electronics, 2nd edition, p. 96). What are the small signal input and output impedance of this emitter follower?

I can analyse an emitter follower, but I'm doing something wrong with the capacitor.

Answer

Here's how I would work out the input impedance.

At the frequencies of interest, consider C2 to be a wire.

Note then, that the output voltage appears at the bottom of R3 whilst the input voltage appears at the top.

Let the voltage gain of this circuit be \$A_{CC} < 1\$.

Then the voltage across R3 is \$v_{in} - A_{CC}\ v_{in} = v_{in}(1 - A_{CC})\$

The current through R3 is then \$\dfrac{v_{in}(1 - A_{CC})}{R_3} \$

Thus, to the input voltage source, R3 "looks" bigger by factor of \$\dfrac{1}{1 - A_{CC}} \$

Looking into the base of Q1, the impedance is approximately \$(1 + \beta)\ (R_4||R_2||R_1 + r_e) \$ where \$r_e = \frac{V_T}{I_E} \$.

Then, the input impedance is:

\$r_{in} = \dfrac{R_3}{1 - A_{CC}} || \left[(1 + \beta)\ (R_4||R_2||R_1 + r_e) \right] \$

Since the left hand term in the parallel combination is typically far larger than the right hand term, the input impedance is dominated by the right hand term, i.e.,

\$r_{in} \approx (1 + \beta)\ (R_4||R_2||R_1 + r_e)\$

For example, with \$V_T = 25mV\$ and \$I_E = 1mA\$, \$r_e = 25 \Omega \$ thus, with \$R_4||R_2||R_1 = 833 \Omega \$:

\$r_{in} \approx (1 + \beta)\ 858 \Omega \$