This Is a capacitor power supply. This is a very small currents of energy-effective, because it has very little loss, much smaller than the miniature transformer. Capacitor impedance is inversely proportional to frequency. This is a great disadvantage when the grid appear in the short interference pulses caused by switching on large appliances with inductive load character. For such a limiting pulse capacitor (C1 here) is practically a short circuit and load through a short period of much higher current. LEDs are the glitches very sensitive and easily damaged or destroyed directly.
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Current pulses damaged LED is rated at significantly less power and small streams are starting to shine from a stream. Oshin especially white LEDs will be damaged very easily. Therefore, the circuit is completed by the Zener diode and resistors R2 and R3. Resistor R2 limited to a maximum pulse current of 300 mA. Zener diode for reducing interference pulse voltage of 12 V and resistor R3 ensures that the LED current will not flow more than 40 mA, LED after a short period of time without lasting damage.
When connected correctly, the circuit components should work on first connection. If you're not sure you're getting involved in correctly, try the following parts of the circuit at low voltage. First connect the regulated power to the Zener diode in accordance with the polarity of the electrolytic capacitor. Slowly zoom voltage. When voltage is about 6 V would be to light up the LEDs. When voltage is about 11 V should be a current consumption of about 30mA. When voltage is about 12 to 13 in the current consumption should increase sharply, because the current will pass through the zener diode. Tension just zoom into the current 100-200 mA. Do not worry about big long zener diode current, could be destroyed. Then connect the source to the rectifier to the terminals marked with a tilde. The device should behave the same, only all voltages will be about 1 V greater. Connect the power rectifier with the opposite polarity. The...
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