Electronic – White and Pink Noise Generator Circuit – Cryptic AC Analysis

The “Electronic Devices and Circuit Theory” book (by Boylestad, Nashelsky) gives a brief description of a random noise generator circuit at the end of “BJT AC Analysis” chapter. (Btw, exactly the same circuit appeared in the IET magazine, April 1976).

DC analysis is straightforward: Q3 stage represents a 'classical' collector feedback bias. At \$V_{cc} = 21 V, \beta=100\$ we get \$A_v = 280, Z_{in} = 1.28 k\Omega\$. No surprises there. However, their description of what happens at high frequencies (where the 3 db/octave filter’s capacitors kick-in) baffles me. The exact quote from the book:

At a sufficiently high frequency all the capacitors could be replaced by short circuits, and the total resistance combination between collector and base would be reduced to about \$14.5 k\Omega\$. which would result in a very high unloaded gain of about \$731\$, more than twice that just obtained with \$R_F = 1 M\Omega\$. Because the 1/f filter is supposed to reduce the gain at high frequencies, it initially appears as though there is an error in design. However, the input impedance has dropped to about \$19.33 \Omega\$, which is a 66-fold drop from the level obtained with \$R_F = 1 M\Omega\$.

Questions:

1. Where did they get the unloaded gain of \$731\$? Shouldn't the gain actually be reduced because of the ‘heavier’ feedback?

2. How they get \$Z_{in} = 19.33 \Omega\$?

EDIT 1 (@analogsystemsrf):

Input impedance equation for collector feedback configuration is:
\$Z_i = \dfrac{r_e}{1/\beta + R_C/(R_C + R_F)} = \dfrac{20 \Omega}{1/100 + 5.6 k\Omega/(5.6 k\Omega + 14.5 k\Omega)} = 69.3 \Omega\$

(from DC analysis: \$r_e = 20 \Omega\$, \$R_F = R4 || R5 || R6 || R7 = 14.5 k\Omega\$)

\$Z_i\$ is therefore \$69\Omega\$ and not \$18\Omega\$ as stated in the book!?

EDIT 2 (@sunnyskyguy-ee75):

DC analysis:

\$I_c * R_8 + I_b * R_4 + 0.7 V = 21 V\$

\$I_c = I_b * \beta \$

\$I_c = \dfrac{21 V – 0.7 V}{R_8 + R_4 / \beta} = \dfrac{20.3 V}{5.6 k\Omega + 1 M\Omega / 100} = 1.3 mA\$

\$r_e \cong 26 mV / I_c = 20 \Omega\$

\$A_v \cong -R_C/r_e = – 5.6 k\Omega/20\Omega = -280\$

AC at some high frequency:

\$R_F = R4 || R5 || R6 || R7 = 14.5 k\Omega\$

\$A_v = -\dfrac{R_F}{R_C + R_F} * \dfrac{R_C}{r_e} = -\dfrac{14.5k\Omega}{5.6 k\Omega + 14.5k\Omega} * \dfrac{5.6 k\Omega}{20\Omega}=-202\$