To calculate the value of RS
V − Vout
RS = in
, where IS = IZ+IL, using an average value of IL.
IS
• Effect of input voltage variation
• Effect of load variation
Example 1:
A 5.0V stabilised power supply is required to be produced from a 12V DC power supply
input source. The maximum power rating PZ of the zener diode is 2W. Using the zener
regulator circuit above calculate:
a) The maximum current flowing through the zener diode.
b) The minimum value of the series resistor, RS
c) The load current IL if a load resistor of 1kΩ is connected across the Zener diode.
d) The zener current IZ at full load.
Voltage Regulation:
a) Line Regulation
In this type of regulation, series resistance and load resistance are fixed, only input
voltage is changing. Output voltage remains the same as long as the input voltage is
maintained above a minimum value.
Percentage of line regulation can be calculated by =
∆Vout
× 100
∆Vin
where Vout is the output voltage and Vin is the input voltage and ∆Vout is the change in
output voltage for a particular change in input voltage ∆Vin
b) Load Regulation
In this type of regulation, input voltage is fixed and the load resistance is varying. Output
volt remains same, as long as the load resistance is maintained above a minimum value.
V − VFL 
Percentage of load regulation =  NL
 × 100
 VNL 
where VNL is the mini. load resistor voltage and VFL is the full load resistor voltage.
Example 2:
Say Vin =20V, Vout required is 15V over a load current range from 10mA to 20mA.
Design the regulator circuit.
Solution:
Average load current =(10+20)/2 = 15mA. Let the diode has IZT = 6mA. Therefore, zener
must be able to handle maximum IL(average)+6mA = 15+6=2mA.
Therefore, PZ = 15Vx21mA =315mWatt. You need a Zener diode having power rating
more than 315mW.
RS = (Vin – Vout)/Iin(max) = (20-15)/21 = 238 ohms
Resistor power rating = V2/RS = 52/238 = 0.1W
** Calculate load regulation (home work)