V_{ S }=15 V \text { and } R_{ L }=250 \Omega .
(a) From Eq. (4.16), the load line is described by
i_{ D }=\frac{V_{ S }-v_{ D }}{R_{ L }} (4.16)
i_{ D }=\frac{\left(V_{ S }-v_{ D }\right)}{R_{ L }}
The diode characteristic is defined by a table of data. The Q-point can be determined from the load line and the data table by an iterative method, as discussed in Sec. 4.5.3.
Iteration 1: Assume v_{ D }=0.7 V . From Eq. (4.16),
i_{ D }=\frac{\left(V_{ S }-v_{ D }\right)}{R_{ L }}=\frac{(15-0.7)}{250}=57.2 mA
which lies between 50 mA and 60 mA in the table. Thus, we can see from the table of data that the new value of diode drop v_{ D ( new )} lies between 1.173 V and 1.212 V. Let us assume that the diode voltage v_{ D }(k) corresponds to the diode current i_{ D }(k) and the diode voltage v_{ D }(k+1) corresponds to the diode current i_{ D }(k+1) . This is shown in Fig. 4.20. If i_{ D } lies between i_{ D }(k) \text { and } i_{ D }(k+1) , then the corresponding value of v_{ D } will lie between v_{ D }(k) \text { and } v_{ D }(k+1) . Thus, v_{D(n e w)} can be found approximately by linear interpolation from
v_{ D ( new )}=v_{ D }(k)+\frac{v_{ D }(k+1)-v_{ D }(k)}{i_{ D }(k+1)-i_{ D }(k)}\left[i_{ D }-i_{ D }(k)\right] (4.34)
=1.173+\frac{1.212-1.173}{60 mA -50 mA }(57.2 mA -50 mA )=1.201 V
Iteration 2: Use the value of v_{ D } from the previous iteration; that is, set v_{ D }=v_{ D ( new )}=1.201 V . From Eq. (4.16),
i_{ D }=i_{ D ( new )}=\frac{(15-1.201)}{250}=55.2 mA
From Eq. (4.34), the new value of v_{ D } is
v_{ D ( new )}=1.173+\frac{1.212-1.173}{60 m -50 m }(55.2 m -50 m )=1.193 V
This process is repeated until a stable Q-point is found. After two iterations, we have V_{ D }=v_{ D ( new )}=1.193 V and I_{ D }=i_{ D ( new )}=55.2 mA .
(b) Since R_{ D } is the slope of the tangent at the Q-point, we get
\begin{aligned}R_{ D }=\left.\frac{\Delta v_{ D }}{\Delta i_{ D }}\right|_{ at Q- \text { point }} &=\frac{v_{ D }-v_{ D }(k)}{i_{ D }-i_{ D }(k)}=\frac{v_{ D }(k+1)-v_{ D }}{i_{ D }(k+1)-i_{ D }} \\&=\frac{1.193-1.173}{(55.2-50) \times 10^{-3}}=3.9 \Omega\end{aligned} (4.35)
The diode threshold voltage is
V_{ TD }=V_{ D }-R_{ D } I_{ D }=1.193-3.9 \times 55.2 mV =0.98 V
(c) From Eq. (4.27), the small-signal AC resistance r_{ d } is
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 }>>I_{ S } (4.27)
r_{ d }=\frac{n V_{ T }}{I_{ D }}=\frac{1 \times 25.8 mV }{\left(55.2 \times 10^{-3}\right)}=0.5 \Omega
NOTE: The difference between rd and RD is due to the fact that a diode follows the Shockley diode equation, whereas the values in the data table are quoted without regard to any relationship. The correlation will depend on how closely the tabular data match with the Shockley equation [Eq. (4.1)].
i_{ D }=I_{ S }\left(e^{v_{ D } / n V_{ T }}-1\right) (4.1)
