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I wanted to design a custom LED light bulb, so I bought a couple and tore them down in the hope to reverse engineer them.

Most of them implement a circuit with a chip shown as below, with the circuit the green area implemented on the PCB, with 1 additional 20 Ohm resistor in heathrink, shown in red, making the connection between the bulb base and the PCB.

Is this a current limiting resistor? How come there is no fuse and no varistor? It's a bit surprising because even bulbs from reputable brands (like OSRAM) don't implement fuses or varistors. Shall I add one on my custom circuit? Would it be before or after the resitor?

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Thanks everyone!

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The 20 ohm resistor (which would pass amps of current in the event of a short) is probably the fuse. Hence they put it in the base where it will blow in the event of a short and completely disconnect the bulb from mains.

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The resistor is usually a fusible resistor and it might be marked with an extra color band that it is a fusible resistor. It also doubles as the inrush current limiting resistor. So therefore, no fuse is necessary.

The question about the varistor on the other hand is more interesting one. Small overvoltage spikes are limited by the series fusible resistor, and since the bridge rectifier output is connected directly to a capacitor, it will limit the surge voltage by simply charging up a bit. So that may be enough of a filter. Besides the SM2082EAS chip is a constant current driver that can work with up to 500V of voltage, so as a whole, the lamp may just handle any typical overvoltage spikes so a TVS is not necessary.

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