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I have a device (IC) with a built in DC-DC regulator that is rated for 1.8v - 3.8v with an absolute max of 4.1v. I would like to use this device with a lipo cell but, my understanding is that these are at 4.2v when fully charged. Is this asking for trouble? If so, is there a simple way to "trim" that top 0.2v or so off the top without having to add another regulator? Thanks.

Here is a link to the device datasheet

I'll add a little more detail on my system requirements. It's an MCU/RF transceiver that has a built-in DCDC converter. It will consume 3-4mA 90% of the time and <30mA while transmitting. The converter as stated above is rated for 1.8-3.8v. I also need 5v for a small LCD display at <1mA. I would like to power all of this from a small lipo cell which provides 3-4.2v. I recently designed a similar system without the built-in converter. I used s step-up converter to 5v and a step-down to 1.8v. The system worked great, except that these converters are almost $5 each. I'm looking to reduce my BOM cost if possible. I would like to utilize the built-in converter and find the cheapest and most efficient way to get this done. I'm wondering if a Zener would be a good way to trim the voltage down to 3.8v. And then I guess I would just bite the bullet and pay for the single step-up to 5v.Thanks for any ideas.

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    \$\begingroup\$ Yes, you could probably add a series diode. Please link the datasheet of the device you are powering. \$\endgroup\$ – uint128_t Feb 11 '16 at 5:36
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    \$\begingroup\$ If it's rated for 3.8V, absolute max is 4.1V, and you knowingly expose it to 4.2V, then you are politely asking for trouble. \$\endgroup\$ – Nick Alexeev Feb 11 '16 at 7:00
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If you use a smart charger such as the imax 1010b+ you can set it to charge your lipo battery to 4.1 volts or less instead of the usual 4.2 volts.

Your lipo battery would have a longer lifespan if charged to only 4.0 volts (i.e. more charge/discharge cycles) and you won't lose a lot of capacity from the lower starting voltage.

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