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I’m making a voltage booster to boost li-ion voltages 3v-4.2v to 5v with high current and a small footprint. I’m using TI’s Webench tool to help. It suggest a 1uH inductor for a 12A load, but if I increased the load to 15A, it recommends an inductor of at least 1.3uH. Out of curiosity,If I tried to run a 15A load through a circuit designed for 12A, what would happen? Would the inductor overheat trying to deliver the current or would it limit the current?

Below is a reference to what I’m making. A few components vary somewhat in value compared my own design, but remains vastly the same. enter image description here

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    \$\begingroup\$ Don't forget that you want to avoid saturating the inductor, too. \$\endgroup\$ – Hearth Oct 4 '19 at 2:06
  • \$\begingroup\$ @Hearth Thanks Hearth! I’ve found an inductor with an ISAT of 55A so I don’t think this will be a problem :) \$\endgroup\$ – Ryan Oct 4 '19 at 6:18
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For a 12 A load at 5 V you will have an input current of more than 12 A * 5 V = 60 W => 60 W / 3 V = 20 A average current.

As the current isn't continuous the peak currents will be much higher. I didn't take any losses (switch resistance, wire resistance) into account which will be significant due to the high current and low voltage (even a small voltage drop can be significant). This means you'll need an inductor with a very high saturation current as Hearth mentions in a comment. My guess is that you'll need an inductor with a saturation current higher than 30 A.

When the inductor saturates, it behaves as a short and current will shoot up and efficiency drops. The DCDC converter is then unable to deliver more power to the output so the output voltage will drop. Depending on the DCDC converter implementation the chip might detect the high current though the inductor and shut everything off or just "do what it can" meaning you will not get 5 V at the output but some lower voltage. In a converter without overcurrent protection it is possible that the switch will be damaged. Overheating of the inductor is indeed also a possibility.

For a proper design, you simply want to avoid the inductor from saturating.

For your design goal with such a low input voltage but a high output current (meaning very a high input current) it will be challenging to make a good design as it will be very difficult to keep all the series resistances small enough. Combined with a "small footprint" it might be impossible (depends on what you call small).

Try to find a commercial product with a similar specification, if you cannot find any then you now know why.

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  • \$\begingroup\$ thanks for the insight. Luckily I found a tiny 1uH inductor that has a max current (IRMS) of 43A and the saturation current (ISAT) is 55A! octopart.com/xal1010-102meb-coilcraft-18814479 \$\endgroup\$ – Ryan Oct 4 '19 at 6:14
  • \$\begingroup\$ With the max current rating so high, do you think it will have any problems handling the extra current? Webench estimates the efficiency in the upper 80%’s. Most of the losses comes from the 510mv drop across the diode. The resistance of the inductor itself is only ~1mohm \$\endgroup\$ – Ryan Oct 4 '19 at 6:30
  • \$\begingroup\$ No, I do not expect issues resulting from such an inductor. A challenge will be to make sure that you don't add several milli ohms series resistance elsewhere in the circuit. For example, what is the series resistance of the battery and the connections between battery and the boost converter? \$\endgroup\$ – Bimpelrekkie Oct 4 '19 at 6:48

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