0195136047.pdf

7.2 DIODES 347

SincevD=v 1 − 2 .6, the condition for the diode to conduct isv 1 > 2 .6V.

600 Ω v 0

(b)

Ideal diode

1.2 kΩ vD 0.6 V

D

−

+

+−+−

v 1 2 V 600 Ω

(a)

1.2 kΩ

−

+

2 V

−

+

−

+

v 1

−

+

Figure E7.2.2

EXAMPLE 7.2.3

We shall demonstrate load-line analysis to find the diode current and voltage, and then compute
the total power output of the battery source in the circuit of Figure E7.2.3(a), given the diodei–v
characteristic shown in Figure E7.2.3(b).

Solution

The Thévenin equivalent circuit as seen by the diode is shown in Figure E7.2.3(c). The load-line
equation, obtained by the KVL, is the equation of a line with slope− 1 /RThand ordinate intercept
given byVTh/RTh,

iD=−

1

RTh

vD+

1

RTh

VTh

Superposition of the load line and the diodei–vcurve is shown in Figure E7.2.3(d). From the
sketch we see that the load line intersects the diode curve at approximately 0.67 V and 27.5 mA,
given by theQpoint (quiescent or operating point). The voltage across the 10-resistor of Figure
E7.2.3(a) is then given by

V 10 = 40 IQ+VQ= 1 .77 V

The current through the 10-resistor is thus 0.177 A, and the total amount out of the source is
therefore given by 0. 177 + 0. 0275 = 0 .2045 A. The total power supplied to the circuit by the
battery source is then

12 × 0. 2045 = 2 .454 W