# Dropping voltage from battery supply for low current application

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:

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?

• 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? Mar 1, 2022 at 20:44
• What's the lowest supply the chip can use? Mar 1, 2022 at 20:50
• 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. Mar 1, 2022 at 20:50
• @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. Mar 1, 2022 at 20:51
• @Andyaka you probably meant voltage: 2.65 V. I'll add that critical information in the question. Mar 1, 2022 at 20:53