Question 4.9: Small-signal analysis of a diode circuit The diode circuit s...

V_{ T }=25.8 mV , n=1.84, V_{ S }=10 V \text {, and } R_{ L }=1 k \Omega . The iterations of the Q-point analysis in Example 4.7 gave V_{ D }=0.7148 V \text { and } I_{ D }=9.284 mA . Using Eq. (4.27), we can find the AC resistance r_{ d } from

r_{ d }=R_{ D }=\frac{1}{g_{ d }}=\frac{n V_{ T }}{i_{ D }+I_{ S }} \approx \frac{n V_{ T }}{I_{ D }} \text { since } i_{ D }=I_{ D }, \text { and } I_{ D } \gg I_{ S }                                   (4.27)

r_{ d }=\frac{n V_{ T }}{I_{ D }}=\frac{1.84 \times 25.8 \times 10^{-3}}{\left(9.284 \times 10^{-3}\right)}=5.11 \Omega

The AC equivalent circuit is shown in Fig. 4.19. From the voltage divider rule, the AC diode voltage v_{ d } is given by

v_{ d }=\frac{r_{ d }}{r_{ d }+R_{ L }} V_{ m } \sin \omega t                                 (4.33)

=\frac{5.11}{5.11+1 k \Omega} 50 \times 10^{-3} \sin \omega t=0.2542 \times 10^{-3} \sin \omega t

Therefore, the instantaneous diode voltage v_{ D } is the sum of V_{ D } and v_{ d } . That is,

\begin{aligned}v_{ D } &=v_{ D }+v_{ d } \\&=0.7158+0.2542 \times 10^{-3} \sin \omega t V\end{aligned}