I want to get the best performance out of the Boost Regulator MCP1642B, specifically highest output current at 5v output for low voltage input (0.9v).

The datasheet says :

the inductance value can range from 2.2 μH to 6.8 μH. An inductance value of 4.7 μH is recommended to achieve a good balance between the inductor size, the converter load transient response and the minimized noise

But it does not explain any further.

It seems obvious that the lower the DCR, and higher Isat the better, (even if there are diminishing returns), and I'm wondering if it's the same for the inductance? For example: would using a 20uH inductor cause problems? And what sort of performance increase, if any might I expect?

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    \$\begingroup\$ What sort of "performance increase" are you expecting? Increasing inductance increases impedance, and therefore reduces current output capability (unless you also reduce switching frequency) \$\endgroup\$
    – user16324
    Nov 11, 2016 at 22:21
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    \$\begingroup\$ Increasing inductance reduces ripple current and maybe some iron losses, although it depends on specific inductor. Most significant difference will be space on pcb if other parameters are kept the same. \$\endgroup\$
    – user76844
    Nov 11, 2016 at 22:29
  • \$\begingroup\$ @BrianDrummond - I wanted to increase current output, since the datasheet specs say it rolls off as the input voltage decreases. What you say makes sense - I hadn't thought of it that way. Thanks \$\endgroup\$
    – Jodes
    Nov 11, 2016 at 22:29
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    \$\begingroup\$ If you think it reduces ripple current, well that's correct for a buck converter. This is a boost converter. \$\endgroup\$
    – user16324
    Nov 11, 2016 at 23:05
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    \$\begingroup\$ You need to look at overall effects. Low resistance is good. | As you increase inductance you decrease rate of current increase (dI~= V.t/L) So after time t if you double L you halve I. As Ener gy = 0.5.L.i^2 the energy is halved by doubling L (!!!) for a given charge time. At a given frequency and about the same on time % a boost converter output is inversely proportional to L (!). Agh. This is a recursive process - now go to top and start again :-). \$\endgroup\$
    – Russell McMahon
    Nov 12, 2016 at 0:03

1 Answer 1


Maximum output current is primarily determined by the current ratings of the switching FETs. Increasing inductance will only slightly reduce rms switching current due to lower ripple. So you could use a 20uH inductor, but it wouldn't significantly increase the maximum output current.

A more important factor is the physical size of the inductor. A 20uH coil with the same physical size as a 4.7uH coil will probably have around 4 times higher resistance and 4 times lower saturation current. Higher resistance causes greater loss. Lower saturation current causes higher ripple (sharply increasing as peak current approaches saturation). Both result in lower maximum output current.

So if you replace a 4.7uH inductor with 20uH inductor of the same physical size and construction the result will probably be lower maximum output current.


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