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I've calculated everything up and came up with the following.

Vin,max – 4.2v , Vfb = 0.1v , 1.8 <= Vf <= 2.4V, Vout – Vf + Vfb

fsw = 1.5MHz, 30mA <= I_led <= 250mA, I_led = Vfb / Rs

I set I_led to be 160mA and I_ripple is 40% of I_led ( 64mA). So, I got 10uH inductor. I guess I could get smaller inducter if I set I_led to be higher like 200mA or 250mA. Then, what would be the consideration. power dissipation? enter image description hereenter image description here

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  • \$\begingroup\$ You did not show a cap spec with ripple current rating >> design ripple nor Rs value in question. This affects results and choice of L. All power dissipation values have temp rise and MTBF consequences. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 13 at 4:06
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I would do a simple P=I*V calculation to find the power dissipation of the LED. The voltage will be fairly constant with some ripple from the switcher.

If the average LED current is 160mA, the voltage will be ~2.25V. The power would then be 0.160A*2.25V=360mW

For 200mA and 2.35V it would be 470mW

enter image description here
Source: https://media.osram.info/media/resource/hires/osram-dam-4134009/asset_pdf_3587997.pdf

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