Calculation of a capacitive power supply
Schematic uses single pulse rectification
For the function of a Zener diode: During the positive half-wave, D1 operates as a voltage-limiting component. The required output voltage can be acheived by adjusting the zerner diode value, in your case its 57V, Since zener is before D2 we need to consider voltage drop across D2 (0.7V), You should choose 57+0.7V zener diode D1.
During negative half cycle of input large amount of current flow through D1, it should be limited which can be done by R1.
R1 = Peak input voltage / max current through D1
Peak input voltage = 1.414 * 230V
Max current through D1 can be tacken from datasheet say 1A
R1 = 325.22 ohm.
Choose some nearest value say 330 Ohm
Since load current pass through R1,we should consider power dissipation.
Form factor of single pulse rectifier 2.2,
Actual load current 20mA *2.2 = 44mA
P = I*I*R
p = 44*44*330 = 0.6388W
This component will get heated up need to consider derating wrt temperature, lifetimne etc consider 2x
Power ratingof R1 = 1W
Voltage drop across resistor at full load
Vr1 = 0.6388/44mA = 14.5V
Now its time to calculate capacitor value
Capacitive reactance = 230V-23-14.5-57.7/44mA = 3063.63 Ohm
Capacitance C1 = 1/2Pi*f*XC = 1/(2*3.14*50*3063.63) = 1.03uF
C1 voltage rating should be higher than ac input, select x rated cap usually called as box cap or film cap.
D2 can be any diode 1N4001 commonly used.
C1 smothering of rectified AC.
Time period of 50Hz 20ms
During negative half cycle output should be tack care by C1, Hals cycle time period 10 mS
Ripple voltage can be considered based on application requirement and available size, economical factor, say 2% of output voltage ~ 1V
Load resistance = Voltage/ load current = 57/20mA = 2850 Ohm
C1 = -10ms / (2850 * ln(56/57)) ~ 200uF, nearest 220uF, 100V
Add some 0.1uF or lesser for noise elimination parallel to C1
Add fuse for protection
Note: This circuit doesn't provide any galvanic isolation
Schematic uses single pulse rectification
For the function of a Zener diode: During the positive half-wave, D1 operates as a voltage-limiting component. The required output voltage can be acheived by adjusting the zerner diode value, in your case its 57V, Since zener is before D2 we need to consider voltage drop across D2 (0.7V), You should choose 57+0.7V zener diode D1.
During negative half cycle of input large amount of current flow through D1, it should be limited which can be done by R1.
R1 = Peak input voltage / max current through D1
Peak input voltage = 1.414 * 230V
Max current through D1 can be tacken from datasheet say 1A
R1 = 325.22 ohm.
Choose some nearest value say 330 Ohm
Since load current pass through R1,we should consider power dissipation.
Form factor of single pulse rectifier 2.2,
Actual load current 20mA *2.2 = 44mA
P = I*I*R
p = 44*44*330 = 0.6388W
This component will get heated up need to consider derating wrt temperature, lifetimne etc consider 2x
Power ratingof R1 = 1W
Voltage drop across resistor at full load
Vr1 = 0.6388/44mA = 14.5V
Now its time to calculate capacitor value
Capacitive reactance = 230V-23-14.5-57.7/44mA = 3063.63 Ohm
Capacitance C1 = 1/2Pi*f*XC = 1/(2*3.14*50*3063.63) = 1.03uF
C1 voltage rating should be higher than ac input, select x rated cap usually called as box cap or film cap.
D2 can be any diode 1N4001 commonly used.
C1 smothering of rectified AC.
Time period of 50Hz 20ms
During negative half cycle output should be tack care by C1, Hals cycle time period 10 mS
Ripple voltage can be considered based on application requirement and available size, economical factor, say 2% of output voltage ~ 1V
Load resistance = Voltage/ load current = 57/20mA = 2850 Ohm
C1 = -10ms / (2850 * ln(56/57)) ~ 200uF, nearest 220uF, 100V
Add some 0.1uF or lesser for noise elimination parallel to C1
Add fuse for protection
Note: This circuit doesn't provide any galvanic isolation
