Question 12.11: Consider the circuit of Figure 12.15(a). A diode is in serie......
Consider the circuit of Figure 12.15(a). A diode is in series with a resistor with resistance R. The voltage across the diode is denoted by v and the current through the diode is denoted by i. The i-v relationship for the diode is non-linear and is given by
i=I_{\mathrm{s}}\left(\mathrm{e}^{40 v}-1\right)where Is is the reverse saturation current of the diode. Given that the supply voltage, V_s , is 2 \mathrm{~V}, I_{\mathrm{s}} \text { is } 10^{-14} A and R is 22 kΩ, calculate the steady-state values of i and v.
There are several ways to solve this problem. A difficulty exists because the diode i-v relationship is non-linear. One possibility is to draw a load line for the resistor superimposed on the diode i-v characteristic, as shown in Figure 12.15(b). The load line is an equation for the resistor characteristic written in terms of the voltage across the diode, v, and the current through the diode, i. It is given by
\begin{gathered} V_{\mathrm{s}}-v=i R \\ i=-\frac{1}{R} v+\frac{V_{\mathrm{s}}}{R} \end{gathered}This is a straight line with slope -\frac{1}{R} \text { and vertical intercept } \frac{V_{\mathrm{s}}}{R} \text {. When } v=0, i=\frac{V_{\mathrm{s}}}{R} \text {. }
This corresponds to all of the supply voltage being dropped across the resistor. When v = V_s, i = 0. This corresponds to all of the supply voltage being dropped across the diode. Therefore, these two limits correspond to the points within which the circuit must operate. The solution to the circuit can be obtained by determining the intercept of the diode characteristic and the load line. This is possible because both the resistor characteristic and diode characteristic are formulated in terms of v and i, and so any solution must have the same values of i and v for both components. If an accurate graph is used, it is possible to obtain a reasonably good solution. An alternative approach is to use the Newton-Raphson technique. Combining the two component equations gives
Now R=2.2 \times 10^4, I_{\mathrm{s}}=10^{-14}, V_{\mathrm{s}}=2 and so
-v+2=2.2 \times 10^4 \times 10^{-14}\left(\mathrm{e}^{40 v}-1\right)Now, define f (v) by
f(v)=2.2 \times 10^{-10}\left(\mathrm{e}^{40 v}-1\right)+v-2We wish to solve f (v) = 0. We have
f^{\prime}(v)=2.2 \times 10^{-10} \times 40 \mathrm{e}^{40 v}+1=8.8 \times 10^{-9} \mathrm{e}^{40 v}+1Choose an initial guess of v_1 = 0.5:
v_2=v_1-\frac{f\left(v_1\right)}{f^{\prime}\left(v_1\right)}=0.5-\frac{2.2 \times 10^{-10}\left(\mathrm{e}^{20}-1\right)+0.5-2}{8.8 \times 10^{-9} \mathrm{e}^{20}+1}=0.7644With an equation of this complexity, it is better to use a computer or a programmable calculator. Doing so gives
v_5=0.6895, \ldots, v_{10}=0.5770, \ldots, v_{14}=0.5650which is accurate to four decimal places. It is useful to check the solution by independently calculating the current through the diode using the two different expressions. So,
\begin{aligned} & i=10^{-14}\left(\mathrm{e}^{40 \times 0.5650}-1\right)=6.53 \times 10^{-5} \mathrm{~A} \\ & i=-\frac{0.5650}{2.2 \times 10^4}+\frac{2}{2.2 \times 10^4}=6.53 \times 10^{-5} \mathrm{~A} \end{aligned}and therefore the solution is correct.