Assume that the silicon diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7 V) of its I–V characteristics. Also assume that the voltage across the diodes is independent of the current above the knee point.
(a)    If V_B = 5 V, what should be the maximum value of R so that the voltage is above the knee point?
(b)    If V_B = 5 V, what should be the value of R to establish a current of 5mA in the circuit ?
(c)    What is the power dissipated in the resistance R and in the diode, when a current of 5 mA flows in the circuit at V_B = 6 V.
(d)    If R = 1 kΩ, what is the minimum voltage V_B required to keep the diode above the knee point?

(a) Here,
Minimum current required by the silicon diode = 1mA= 10^–3 A. 

Minimum voltage across the diode is 0.7 V (so that it is above the knee point of the characteristic curve). 

So, voltage drop across the resistance R = (5 – 0.7) V = 4.3 V.

The minimum current i = 1 mA .

$\therefore$ Maximum value of resistance, $\mathrm{R} =\frac{\mathrm{v}}{\mathrm{i}} = \frac{4.3}{{10}^{-3}}\mathrm{\Omega } = 4.3 \mathrm{k\Omega }$ 

(b) Current through resistance R = 5 mA = 5 x 10^–3 A.

Voltage drop across R = (5 – 0.7) V = 4.3 V

$$\begin{gathered} \therefore \mathrm{R} = \frac{4.3}{5\times {10}^{-3}} \\ = 0.86 \times {10}^{-3}\mathrm{\Omega } = 860 \mathrm{\Omega } \end{gathered}$$ 

(c) Here, V_B = 6 V. 

Voltage V' across the resistance = 6 – 0.7 = 5.3 V 

Current, i = 5 mA = 5 x 10^–3 A 

Power dissipated through the resistance R, P = i V’  
                                                                   = 5 x 10^–3 x 5.3 W
                                                                   = 26.5 x 10^–3 W
                                                                   = 26.5 mW 

Power dissipated in the diode is P’' = i x 0.7 W
                                                    = 5 x 10^–3 x 7 x 10^–1 W
                                                    = 3.5 x 10^–3 W
                                                    = 3.5 mW

(d) For keeping diode above the knee point, the minimum current required is 1 mA.

Voltage drop across R (= 1 kΩ), V_R = 1 x 10³ x 10³
                                                     = 1 V 

Minimum voltage drop across the diode = 0.7 V

Therefore,
Minimum voltage required, V_B = (1 + 0.7) V = 1.7 V.