The datasheet shows the formula for calculating Peak inductor current is: enter image description here

(Input voltage) Vi=3.2; efficiency=85%; OuputPower=12w

I(peak) = (OuputPower/(Vi x efficiency)) => 12w/2.72v => 4.4118A .... (i)

How I thought I would calculate is as follows;

Required Input power = Output Power + Loss power => 12w x 1.15 => 13.8 watts

I(Peak) = (Required Input power)/Vi => 13.8w/3.2v => 4.3125A ....... (ii)

what's wrong with the equation (ii)? What's being add up in (ii) to make (i)? I am a little confused here. Ignore my ignorance.


1 Answer 1


Firstly, the equation IS NOT the formula for calculating peak inductor current, it's the equation for calculating average DC input current; just look at its structure - it doesn't contain any mention of duty cycle and duty cycle is needed to be known to calculate inductor peak current.

So, if your output power is 12 watts and your efficiency is 0.85 then input power is 14.12 watts.

If input voltage is 3.2 volts then average input current is 14.12/3.2 = 4.4118 amps.

What you appear to be proving is that 12 x 1.15 does not equal 12/0.85. Or put more simply, 1.15 does not equal 1/0.85.

  • \$\begingroup\$ Thanks for the reply. And you are right, it's not peak current but an average dc current. But how can input power = 14.12watts (as you just said)? 12 x 1.15 => 13.8 watts or in other words 12 + (12x0.15) => 13.8watts... isn't ? \$\endgroup\$ Feb 9, 2017 at 8:55
  • \$\begingroup\$ Because 1/0.85 DOES NOT equal 1.15 it equals 1.1765 \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 8:57
  • \$\begingroup\$ InputPower = OutputPower + ExtraNeededPower; ExtraNeededPower = 15% of (12w) => 1.8watts; InputPower = 12watts + 1.8watts => 13.8watts \$\endgroup\$ Feb 9, 2017 at 9:06
  • 1
    \$\begingroup\$ My point is that you are using it both ways. At one point you use 1.15 and at another point to use 1/0.85. Output power = input power * 0.85. Not the same as output power * 1.15 = input power. \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 9:12
  • 1
    \$\begingroup\$ Your question edit has partially invalidated my answer - I am trying to help you here and I don't need to have to put in extra work to counter moving goalposts. Please roll-back your question. If you can't do it, I'll do it. \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 9:15

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