I am working through the design procedure for the AOZ2264, to buck 11.5-18V down to 9.8V 640mA. It claims:

The inductor is used to supply constant current to output when it is driven by a switching voltage. For given input and output voltage, inductance and switching frequency together decide the inductor ripple current, which is:

$$ \Delta I_L = \frac {V_O} {f L} \left(1 - \frac {V_O} {V_{IN}} \right) $$

The peak inductor current is:

$$ I_{Lpeak} = I_O + \frac {\Delta I_L} 2 $$

High inductance gives low inductor ripple current but requires a larger size inductor to avoid saturation. Low ripple current reduces inductor core losses. It also reduces RMS current through inductor and switches, which results in less conduction loss. Usually, peak to peak ripple current on inductor is designed to be 30% to 50% of output current.

Let's assume their 50% guideline to be true:

$$ \frac {I_{pkpk}} {I_O} = 0.5 $$ $$ I_{pkpk} = 320 \text{mA} $$ $$ I_{pk} = 160 \text{mA} $$ $$ \Delta I_L = 2(I_{pk} - I_O) = -960 \text{mA} $$

A negative current makes no physical sense. What gives? Did they screw up the wording and actually mean the opposite, that output current is designed to be 30% to 50% of peak to peak ripple current? Or is my interpretation of "peak-to-peak" wrong, and actually when they say "peak-to-peak" that's synonymous with \$ \Delta I_L \$?

  • 3
    \$\begingroup\$ Just never let the ripple drop the instantaneous inductor current to (and of course below) zero or else you enter discontinuous operation. That's the basis of that design guideline. So the ripple should never go above or below the output (average) current more than the average current's distance to zero current. So 1A output should not go above 2A or below 0A. That is a Ipk = 1A, and Ipk-pk = 2A. At most. \$\endgroup\$
    – DKNguyen
    May 14, 2022 at 20:25
  • \$\begingroup\$ Ipk is the peak value. The highest instantaneous current that occurs when I load is at max load. Io is the average current at peak load. When I say average, I mean averaged over one PWM switching cycle. Ipk is by definition higher than Io. So your negative number makes no sense in this case. However, negative current makes perfect physical sense in general so that statement is wrong. \$\endgroup\$
    – user57037
    May 14, 2022 at 20:41
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    \$\begingroup\$ @mkeith Generally about negative current you're obviously right, but in this case my intuition that something is deeply wrong was correct. I came to the same conclusion as Bryan did, that I was working off of an incorrect definition for peak-to-peak current. \$\endgroup\$
    – Reinderien
    May 14, 2022 at 20:47
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    \$\begingroup\$ This information is in lots of app notes and data sheets, so probably the writers just got a bit sloppy about defining everything clearly. They just figure everyone will know what they are talking about and hardly anyone will read carefully because it is just kind of a refresher course. But of course there will always be people reading it for the first time such as yourself. So it would be nice if they made it clearer. \$\endgroup\$
    – user57037
    May 14, 2022 at 20:50

1 Answer 1


Note that \$I_{pkpk}\$ is the difference between the maximum and minimum currents in the inductor. \$I_{pk}\$ however is not half of the \$I_{pkpk}\$ value, but the maximum current in the inductor, ie \$I_O + I_{pkpk}/2\$

In your final equation you should sub in a value of \$I_pk = 640+320/2\$ mA, which will give you the expected result

  • \$\begingroup\$ I figured this out from @DKNguyen's description; I think you're right. Thanks. \$\endgroup\$
    – Reinderien
    May 14, 2022 at 20:36

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