Question 4.7: Finding the Q-point of a diode circuit The diode circuit sho...

If the initial guess of v_{ D } is not specified, we can use the default value of v_{ D }=0.70 V . But it should affect the final results as long as the initial guess value is reasonable around 0.7 V.

(a) V_{ D }=v_{ D }=0.61 V . From Eq. (4.16), i_{ D }=\left(V_{ S }-v_{ D }\right) / R_{ L }=(10-0.61) / 1 k \Omega=9.39 mA .

i_{ D }=\frac{V_{ S }-v_{ D }}{R_{ L }}                           (4.16)

(b) R_{ L }=1 k \Omega, n=1.84, V_{ T }=25.8 mV \text {, and } v_{ D }=0.61 V \text { at } i_{ D }=1 mA .

Iteration 1: Assume v_{ D }=0.61 V . From Eq. (4.16),

i_{ D }=\frac{V_{ S }-v_{ D }}{R_{ L }}=\frac{(10-0.61) V }{1 k \Omega}=9.39 mA

From Eq. (4.5), the new value of v_{ D } is

v_{ D }=n V_{ T } \ln \left(\frac{i_{ D }}{I_{ S }}\right)                                     (4.5)

\begin{aligned}v_{ D ( new )} &=n V_{ T } \ln \left(\frac{i_{ D }}{I_{ S }}\right) \\&=1.84 \times 0.0258 \ln \left(\frac{9.39 mA }{2.682 \times 10^{-9}}\right)=0.7153 V\end{aligned}

Iteration 2: Assume the values of v_{ D } from the previous iteration. That is, set v_{ D }=v_{ D (\text { new })}=0.7153 V . From Eq. (4.16),

i_{ D ( new )}=\frac{V_{ S }-v_{ D }}{R_{ L }}=\frac{(10-0.7153) V }{1 k \Omega}=9.2847 mA

From Eq. (4.5), the new value of v_{ D } is

\begin{aligned}v_{ D ( new )} &=n V_{ T } \ln \left(\frac{i_{ D ( new )}}{I_{ S }}\right) \\&=1.84 \times 0.0258 \ln \left(\frac{9.2847}{2.682 \times 10^{-9}}\right)=0.7148 V\end{aligned}

Iteration 3: Assume the values of v_{ D } from the previous iteration. That is, set v_{ D }=v_{ D (\text { new })}=0.7148 V . From Eq. (4.16),

i_{ D ( new )}=\frac{V_{ S }-v_{ D }}{R_{ L }}=\frac{(10-0.7148) V }{1 k \Omega}=9.285 mA

From Eq. (4.5), the new value of v_{ D } is

\begin{aligned}v_{ D ( new )} &=n V_{ T } \ln \left(\frac{i_{ D ( new )}}{I_{ S }}\right) \\&=1.84 \times 0.0258 \ln \left(\frac{9.285}{2.682 \times 10^{-9}}\right)=0.7158 V\end{aligned}

Therefore, after three iterations, V_{ D }=v_{ D ( new )}=0.7158 V \text { and } I_{ D }=i_{ D (\text { new })}=9.285 mA . Note that the results of iteration 3 do not differ significantly from those of iteration 2. In fact, there was no need for iteration 3.

(c) Substituting for the given values in Eq. (4.18), \left(V_{ S }-v_{ D }\right) / R_{ L } \simeq I_{ S } e^{v_{ D } / \eta V_{ T }} , we get \left(10-v_{ D }\right) / k \Omega= 2.682 \times 10^{-9} \times e^{v_{ D } /\left(1.84 \times 25.8 \times 10^{-3}\right)} which, after solving by MATHCAD software function, gives V_{ D }= 0.7148 V and

i_{ D }=\frac{\left(V_{ S }-v_{ D } / R_{ L }\right)}{R_{ L }}=I_{ S }\left(e^{v_{ D } / \eta V_{ T }}-1\right) \simeq I_{ S } e^{v_{ D } / \eta V_{ T }}                                (4.18)

I_{ D }=I_{ S } e^{-V_{ D } / \eta V_{ T }}=\left(2.682 \times 10^{-9}\right) e^{-0.7148 /\left(1.84 \times 25.8 \times 10^{-3}\right)}=9.2845 mA

NOTE: Four-digit answers were used to control computational errors and the number of iterations needed to reach the solution. In reality, resistors will have tolerances, and such accuracy may not be necessary.