0195136047.pdf

346 SEMICONDUCTOR DEVICES

Solution

For determining the condition of the ideal diode, let us initially assume that it does not conduct,

and let us replace it with an open circuit, as shown in Figure E7.2.1(b). The voltage across the

10-resistor can be calculated as8Vbythevoltage-divider rule. Then, applying KVL around

the right-hand loop, we get

8 =vD+ 10 or vD=−2V

That is to say, the diode is not conducting sincevD< 0. This result is consistent with the initial

assumption, and therefore the diode does not conduct.

The student is encouraged to reverse the initial assumption by presuming that the diode is

conducting, and show the same result as obtained in the preceding.

12 V 10 Ω 10 V

(a)

Ideal diode

5 Ω vD 12 Ω

−

+

+−

12 V 10 Ω 10 V

(b)

5 Ω vD 12 Ω

−

+

−

+

8 V

−

+

+−

Figure E7.2.1

EXAMPLE 7.2.2

Use the offset diode model with a threshold voltage of 0.6 V to determine the value ofv 1 for

which the diodeDwill first conduct in the circuit of Figure E7.2.2(a).

Solution

Figure E7.2.2(b) shows the circuit with the diode replaced by its circuit model. Whenv 1 is zero

or negative, it is safe to assume that the diode is off. Assuming the diode to be initially off, no

current flows in the diode circuit. Then, applying KVL to each of the loops, we get

v 1 =vD+ 0. 6 + 2 and v 0 = 2