0195136047.pdf

356 SEMICONDUCTOR DEVICES

Forv→−∞,i=0 sinceD 1 will be off andD 2 on. WithD 1 at its breakpoint, the circuit

is drawn in Figure E7.2.7(b). It follows thati=0 andv 1 =v+12; but one does not know the

value ofv 1 and the state ofD 2. If one assumesv 1 >10 V, thenD 2 will be off and there is no

source for the currenti 1 =v 1 /4. Hence one concludes thatD 2 must be on andv 1 =10 V. Then

the correspondingi–vbreakpoint is ati=0 andv=−2V.

v 1 4 Ω

(a)

2 Ω

10 V

−

+

v

12 V

i D 1

Ideal diode

D 2

− Ideal diode

+

−

+

−

+

v 1 = v + 12 4 Ω

(b)

2 Ω

10 V

−

+

v

12 V

i D 1

0A

0 V

i 1 D 2

− −

+

−

+ +

0A

−0 V+

−

+

4 Ω v 1

(c)

i, A

v, V

D 1 breakpoint

D 2 breakpoint

i =+ 2

1

6

v

6

D 1 off, D 2 on

D 1 on,

D 2 off

i = 0

2.5

2.0

03

(d)

− 2

2 Ω

10 V

−

+

v

12 V

4

i D 1 v 1

−

+

−

+

−

+

Figure E7.2.7