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I was designing the PCB for a 10 W LED driver with the AL1676 controller. From the datasheet, the application circuit shows that the driver takes its VCC supply through R1 and R2. The maximum value for VCC is 18 V out of the approximately 381 V mains after rectification.

Can someone help me figure out the formulas/calculations for these resistor droppers and for the value of capacitor C3?

The evaluation module shows the values of the droppers as R2=R3=180k. The IC uses startup current: 100 µA, operation current: 170 µA (5 kHz switching) which means with 180k*2 and 170 µA, they should be dropping only 61.2 V out of the 381 V supplied to it, so 319 V is available at the VCC pin when it can handle only 18 V.

How does that work?

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    \$\begingroup\$ The ic has an internal zener clamp diode. So that node won’t exceed the 15 or so Volts. At 360k, the current is around 1mA depending on the actual mains voltage. You can size the capacitor to achieve an acceptable ripple at a load of 170uA \$\endgroup\$
    – Kartman
    Commented Mar 14, 2022 at 12:38
  • \$\begingroup\$ Thanks! That makes sense. \$\endgroup\$
    – John
    Commented Mar 14, 2022 at 19:36
  • \$\begingroup\$ If it were not given on the evaluation module, how would we know the value of these resistors? Is it just a guessing game such that the current exceeds 170uA? \$\endgroup\$
    – John
    Commented Mar 14, 2022 at 19:42

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