A particular design of a voltage regulator is below. Diodes D1 and D2 are 10-mA

Question

A particular design of a voltage regulator is below. Diodes D1 and D2 are 10-mA units; that is, each has a voltage drop of 0.7 V at a current of 10 mA. Each has n = I. What is the regulator output voltage V0 with the 150-ii load connected? Find V0 with no load. With the load connected, to what value can the 5-V supply be lowered while maintaining the loaded output voltage within 0.1 V of its nominal value? What does the loaded output voltage become when the 5-V supply is raised by the same amount as the drop found in (c)? For the range of changes explored in (C) and (d), by what percentage does the output voltage change for each percentage change of supply voltage in the worst case?

Explanation / Answer

ID=(5-0.7*2)/180=0.02 A==>V2-V1=2.3*log(I2/I1)=V2=

0.7+2.3*0.025*log(0.02/0.01)=0.71731 V==>ID=(5-0.71731*2)/180=0.01981 A

VD=0.71731 V==>ID=0.01981 A==>V0=0.71731*2=1.435 V==>IL=(0.71731*2)=

150=0.00956 A==>ID=0.01981-0.00956=0.010244 A==>V0=2*VD=

2*(0.7+2.3*0.025*log(0.010244/0.01)=

a) 1.4012 V=V0

b) V0 no load=1.435 V

c) V0=1.4012-0.1=1.3012 V==>1.3012/2=0.6506 V across the diodes

V=(1.4012-1.3012)/2=0.05 V==>ID=0.1*0.01*e(V/VT)=0.00739 A

IL=1.3012/150=0.008675 A==>I=IL+ID=0.008675+0.00739=0.0161 A

180*0.0161+1.3012=4.199 V==>5 Vo;t supply can be lowered to -4.2 V

d) new voltage:(5-4.2)+5=5.8 V==>ID=(5.8*2*0.7)/180=0.02444 A

V2=V1+2.3*log(I2/I1)=0.7+2.3*0.025*log(0.02444/0.01)=0.7223 V

ID=(5.8-2*0.7223)/180=0.0242 A==>IL=(2*0.7223)/150=0.00963 A

ID=0.0242-0.00963=0.01457 A==>VD=0.7+2.3*0.025*log(0.01457/0.01)=0.7094 V

V0=2*VD=2*0.7094=1.419 V

e) (1.419-1.3012)/(5.8-4.199)=0.073 or 7.3 %

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