0195136047.pdf

348 SEMICONDUCTOR DEVICES

12 V D

iD, mA

60

40

20

0 0.5 1.0 1.5 2.0 2.5

(b)

vD,V

iD, mA

60

40

20

41.38

27.5

01.01.50.5

0.668

Qpoint

Load line

2.0 2.5

(d)

vD,V

10 Ω

(a)

50 Ω 20 Ω

20 Ω

−

+

−

+

vD

iD

VTh==10 (12) 2 V D

60

(c)

RTh=++ 20 20 (10||50)=48.33Ω

Figure E7.2.3

EXAMPLE 7.2.4

Consider the circuit of Figure E7.2.4(a) withvS(t)=10 cosωt. Use the piecewise-linear model

of the diode with a threshold voltage of 0.6 V and a forward resistance of 0.5to determine the

rectified load voltagevL.

Solution

Figure E7.2.4(b) shows the circuit with the diode replaced by its piecewise-linear model. Applying

KVL,

vS=v 1 +v 2 +vD+ 0. 6 +vL or vD=vS−v 1 −v 2 − 0. 6 −vL

The diode is off corresponding to the negative half-cycle of the source voltage. Thus no current

flows in the series circuit; the voltagesv 1 ,v 2 , andvLare all zero. So when the diode is not

conducting, the following KVL holds:

vD=vS− 0. 6