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My battery might supply 3.7 V, while absolute maximum for an IC it is supplying is 3.55 V and lowest operating voltage is 2.65 V. The IC is at sleep most of the time, consuming 10 nA. Every now and then <1 % of it's lifetime it wakes up to consume 1 mA. Voltage could be dropped with a series (Schottky) diode. Here are curves for I-V curves for forward currents at different temperatures from Toshiba CUS10S30 Schottky diode datasheet:

enter image description here

But only if it was always 25 C and current draw would be stable. I do need to make a sacrifice to one way or the other, but adding an LDO regulator type of a device would drain the battery fast. How do I achieve not damaging the IC and only sacrificing ~tens of % of battery life?

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    \$\begingroup\$ What is your battery that has max 3.7V? Can you drop more voltage with a standad diode or LED? What is the chip with 3.55 maximum voltage, what is the nominal maximum, can you provide datasheet? \$\endgroup\$
    – Justme
    Mar 1, 2022 at 20:44
  • \$\begingroup\$ What's the lowest supply the chip can use? \$\endgroup\$
    – Andy aka
    Mar 1, 2022 at 20:50
  • \$\begingroup\$ Battery: Li-SOCl2 chemistry with elevated temperature. Regardless of manufacturer to meet other spec I run into the problem. The rest of the circuit I simplified by saying it's one IC, it's actually a more complicated module with one pin being problematic. I do get frustrated too when there is no actual circuit in the posting, but it would have over-complicated the matter by a lot, so I tried to explain the situation well. And it is one pin in a module and the module MFG refuses to give any extra info than what I put in the question and the main datasheet is no help. \$\endgroup\$
    – Ralph
    Mar 1, 2022 at 20:50
  • \$\begingroup\$ @Andyaka lowest state the chip goes to is 10 uA (by the little spec I was able to get). When it operates it uses 1 mA. \$\endgroup\$
    – Ralph
    Mar 1, 2022 at 20:51
  • \$\begingroup\$ @Andyaka you probably meant voltage: 2.65 V. I'll add that critical information in the question. \$\endgroup\$
    – Ralph
    Mar 1, 2022 at 20:53

1 Answer 1

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What's wrong with using a really low quiescent current LDO regulator like this one: -

enter image description here

It sounds like the 2.8 volt version might suit.

Here's a few more.

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  • \$\begingroup\$ I seemed to have messed up the units in the question for sleep more, it was actually 10 nA, not 10 uA, sorry. This is a great answer nevertheless. I might be able to use the rest of my circuit to take use of the 1 nA Shutdown Supply Current too. \$\endgroup\$
    – Ralph
    Mar 1, 2022 at 21:41

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