I was trying to make an LED lamp with 16 series 3528 SMD LED chips. I am planning to use a transformerless capacitive power supply to step down 230V. As per my calculation 16 led will require the voltage of 56V and each LED can take 20mA. So how to calculate series capacitor values for stepping down 230V to 56V.

  • \$\begingroup\$ What kind of circuit schematics do you have in mind? \$\endgroup\$ Apr 16, 2018 at 6:58

1 Answer 1


Calculation of a capacitive power supply

Schematic uses single pulse rectification

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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 = II*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 = 1/2PifXC = 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


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