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The waveforms of vS and vC are sketched in Fig. 2-30.
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The circuit of Fig. 2-31(a) is an inexpensive voltage regulator; all the diodes are identical and have the characteristic of Fig. 2-26(b). Find the regulation of vo when Vb increases from its nominal value of 4 V to the value 6 V. Take R 2 k.
R a b + Ib R RF1 VF1 + _
RF2 Vo
+ Vb _ Vo Vb
+ _ VF2 _ c (a) (b) c + _ _
Fig. 2-31 We determined in Problem 2.7 that each diode can be modeled as a battery, VF 0:5 V, and a resistor, RF 500 , in series. Combining the diode strings between points a and b and between points b and c gives the circuit of Fig. 2-31(b), where VF1 2VF 1 V VF2 4VF 2 V RF 1 2RF 100  RF2 4RF 200 
By KVL,
Ib
Vb VF1 VF2 R RF1 RF2 Vb VF1 VF2 RF2 R RF1 RF 2
whence
Vo VF2 Ib RF2 VF2
For Vb1 4 V and Vb2 6 V, Vo1 2 and (2.6) gives
SEMICONDUCTOR DIODES
[CHAP. 2
4 1 2 200 2:09 V 2000 100 200
Vo2 2
6 1 2 200 2:26 V 2000 100 200
Reg
Vo2 Vo1 100% 8:1% Vo1
The circuit of Fig. 2-22(a) is to be used as a dc power supply for a load RL that varies from 10  to 1 k; vS is a 10-V square wave. Find the percentage change in the average value of vL over the range of load variation, and comment on the quality of regulation exhibited by this circuit.
Let T denote the period of vS . For RL 10 , 8 10 < RL 10 5 V v vL RL RS S 10 10 : 0 (diode blocks) 5 T=2 0 T=2 VL0 and so 2:5 V T For RL 1 k, vL and so VL0 8 <
0 T=2
t < T=2 t<T
RL 1000 10 9:9 V v RL RS S 1010 : 0 (diode blocks) 9:9 T=2 0 4:95 V T
0 T=2
t < T=2 t<T
Then, by (2.6) and using RL 10  as full load, we have Reg 4:95 2:5 100% 98% 2:5
This large value of regulation is prohibitive for most applications. Either another circuit or a lter network would be necessary to make this power supply useful.
The circuit of Fig. 2-32 adds a dc level (a bias voltage) to a signal whose average value is zero. If vS is a 10-V square wave of period T, RL R1 10 , and the diode is ideal, nd the average value of vL .
R1 D +
iD RL
Fig. 2-32 For vL > 0, D is forward-biased and vL vS 10 V. vL Thus, VL0 For vL < 0, D is reverse-biased and
RL 10 10 5 V v RL R1 S 10 10 10 T=2 5 T=2 2:5 V T
CHAP. 2]
SEMICONDUCTOR DIODES
For some symmetrical input signals, this type of circuit could destroy the symmetry of the input.
Size the lter capacitor in the recti er circuit of Fig. 2-15(a) so that the ripple voltage is approximately 5 percent of the average value of the output voltage. The diode is ideal, RL 1 k, and vS 90 sin 2000t V. Calculate the average value of vL for this lter.
With Fr 0:05, (2.10) gives C% 1 1 62:83 F fRL 0:05 2000=2 1 103 0:05
Then, using the approximations that led to (2.10), we have   1 V 0:05 90 0:975 87:75 V VL0 VSm vL VSm Sm % VSm 1 2 2 2fRL C
In the positive clipping circuit of Fig. 2-17(a), the diode is ideal and vi is a 10-V triangular wave with period T. Sketch one cycle of the output voltage vo if Vb 6 V.
The diode blocks (acts as an open circuit) for vi < 6 V, giving vo vi . For vi ! 6 V, the diode is in forward conduction, clipping vi to e ect vo 6 V. The resulting output voltage waveform is sketched in Fig. 2-33.
L, V
3T/20 T/4 7T/20
Fig. 2-33
Draw a transfer characteristic relating vo to vi for the positive clipping network of Problem 2.17. Also, sketch one cycle of the output waveform if vi 10 sin !t V.
The diode blocks for vi < 6 V and conducts for vi ! 6 V. Thus, vo vi for vi < 6 V, and vo 6 V for vi ! 6 V. The transfer characteristic is displayed in Fig. 2-34(a). For the given input signal, the output is a sine wave with the positive peak clipped at 6 V, as shown in Fig. 2-34(b).
Reverse the diode in Fig. 2-17(a) to create a negative clipping network. (a) Let Vb 6 V, and draw the network transfer characteristic. (b) Sketch one cycle of the output waveform if vS 10 sin !t V.
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