Q. It si needed to make a shunt regulator to give a regulated voltage of approximately 10V. The available 10V, 1Ω zener of kind !N4740 is specified to own a 10V drop at a test current of 25mA. At this current its rz is found 7 W. The raw supply has a nominal value of 20V but can change as much as ±25%. The regulator is needed to provide a load current of 0 to 20mA. Design a minimum zener current of 5mA.
a). Calculate Vzo.
b). Determine the required value of R.
c). Find out line regulation. What is the difference in Vo expressed in %age, matching to the ±25% change in Vs.
d). Determine the load regulation. By what percentage does Vo alter from no load to full load situation.
e). What is maximum current that zener in your design must be able to produce? What is the zener power dissipation in this situation?
Solution:
The given data is
Vz=10V, Iz=25mA, rz=7Ω, Vs=20V±25%, Izk=5mA, ΔIL=20mA.
a). Vzo=Vz-Izrz=10-25m(7)=9.82V
b). Minimum current Izk= 5mA will take place when Load current IL is maximum, that is, 20mA and Vs is at minimum, so that Vs = 20-5=15V.
Thus
c). Line Regulation=ΔVo/ ΔVs=rz/(R+rz)
7/212=33mV/V.
±25% change in Vs= ±5V
Thus Vo alters by ±5x33m
= ±165mV
Percentage= ±0.165(100)/10= ±1.65%.
d). Load Regulation=-rz||R=-7||205
=-6.77W=-6.77V/A
ΔVo=-6.67(20m)=-135.4mV
Which in percentage =-135.4(100)/10=-1.35%
e). Maximum zener current takes place at no load and Vs=20(1+0.25)=25V
Current=(25-9.825)/(205+7)=71.6mA
Zener Power Dissipation =71.6m(10)=716mW
Or more accurately
Vz= 9.825+71.6m(7)=10.326V
Pz=71.6m (10.326)=739.4mW.