0195136047.pdf

7.2 DIODES 355

vout=

RS

RS+RZ

(

VZ+

RZ

RS

VS−RZiout

)

ForRZ<< RS,RZ|VS/RS−iout|<< VZ, in which case

vout∼=VZ

Thus, the zener diode regulatesvoutby holding it at the fixed zener voltageVZ, in spite

of the possible variations ofVSoriout.

(b) Foriout=0,

vout=

100

100 + 4

[

20 +

4

100

( 25 )

]

=

100

104

× 21 = 20 .19 V

we have 20. 19 − 20 = 4 iZ,oriZ= 0. 19 / 4 = 47 .5 mA.

Foriout=50 mA,

vout=

100

104

[

20 +

4

100

( 25 )−( 4 × 0. 05 )

]

=

100

104

× 20. 8 =20 V

we have 20− 20 = 4 iZ,oriZ=0.

Breakpoint Analysis

When a circuit consists of two or more ideal diodes, it will have several distinct operating
conditions resulting from the off and on states of the diodes. A systematic way of finding those
operating conditions is the method ofbreakpoint analysis. For a two-terminal network containing
resistors, sources, andNideal diodes, and driven by a source voltagev, thei–vcharacteristic
will in general consist ofN+1 straight-line segments withNbreakpoints. Thei–vcurve can be
constructed by following these steps:

1. Forv→∞, determine the states of all diodes, and writeiin terms ofv; do the same for
   v→−∞.


2. With one diode to be at its breakpoint (i.e., having zero voltage drop and zero current),
   find the resulting values ofiandvat the terminals; do the same for each of the other
   diodes.


3. Plot thei–vbreakpoints obtained from step 2; connect them with straight lines and add
   the end lines found in step 1.
   Note that in step 2, if two or more diodes are simultaneously at breakpoint conditions, the
   numbers of breakpoints and line segments of thei–vcurve are correspondingly reduced.

EXAMPLE 7.2.7

Determine thei–vcharacteristic of the network shown in Figure E7.2.7(a) by the use of breakpoint
analysis.

Solution

Forv→∞,D 1 will be forward-biased whileD 2 will be reverse-biased (becausev 1 >10 V).
Hence, withD 1 on andD 2 off,v= 2 i− 12 + 4 i,ori=v/ 6 +2.