How to operate a chip very close to its absolute maximum voltage?
I want to operate a BLE chip at very close to its absolute maximum voltage.
The chip is nRF52810
This is intended for a battery powered device and I want to use a 3.6V non-rechargable battery directly wired to the chip.
However the datasheet in page 409 mentions:
The battery has an open-circuit voltage 3.6V - 3.65V
What should I do in that case?
I was thinking of a series resistor 1Ω - 3Ω between the battery positive line and the VCC of the chip.
After the resistor a large decoupling capacitor 10uF - 47uF will form a low pass RC filter.
Is that going to work?
PS: I tried connecting the power directly to the chip and it works but I'm worried about the status of the device in long-term.
I want to avoid using an LDO due to power loss, the device must consume as low energy as possible. Currently the device consumes around 2.5uA while in sleep mode.
I have seen similar practice: 3.6V battery directly to a microcontroller that also had an absolute maximum value of 3.6V
I think it was a TI sub-GHz but I found it strange to be used like that.
The diagram of what I suggest:
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| User | Comment | Date |
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| DeadMouse | (no comment) | Jul 14, 2024 at 06:00 |
3.6 V is not the absolute maximum voltage. The table you show explicitly says "Recommended Operating Conditions".
Therefore, running the chip up to 3.6 V is perfectly fine.
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| DeadMouse | (no comment) | Jul 14, 2024 at 06:14 |
The term absolute maximum ratings has a well-defined meaning in datasheets, meaning stress values that the part will endure for a short period of time without breaking. And the opposite: if you exceed asbolute maximum ratings, it may break.
The absolute maximum ratings of this particular part could be found in chapter 9, "Absolute maximum ratings":
The section you quote is not that. Parts of a datasheet called electrical characteristics, operating conditions or similar refer to normal use and is what you should design according to.
So using 3.6V would be fine. However a battery rarely holds an exact voltage and if it is rated for 3.6V then that's just a nominal voltage. A fully charged battery may very well exceed that voltage, meaning that the design would be questionable and perhaps the chip would start to misbehave or break over time. To ensure proper voltage, you would have to place a zener on the battery voltage which burns away all energy exceeding 3.6V. That's not energy-efficient at all.
A low dropout linear regulator might not be such a bad idea - they are only inefficient if working with large voltage spans. One which gives you a clean 3.0V out of some ~3.3 to 3.8V wouldn't likely be that inefficient. The advantage of the LDO is that they are ideal for powering RF parts.
Otherwise for pure low current consumption purposes, a switching buck regulator is obviously the way to go. They have worse EMI characteristics but are the most power efficient solution.
So it's a design decision between low noise and current consumption.
I was thinking of a series resistor 1Ω - 3Ω between the battery positive line and the VCC of the chip. After the resistor a large decoupling capacitor 10uF - 47uF will form a low pass RC filter.
Not sure what you are going with that. What are you trying to filter? 1/(2π * 2 * 22uF) would give a cut-off frequency of 3.6kHz. What's that good for?
It would seem that filtering away the 2.4GHZ Bluetooth would make far more sense, in which case we are talking about radiated emissions caught by your PCB. To deal with that, sprinkling small decoupling caps here and there would be a better idea. For example placing a 47pF cap next to the usual 100nF decoupling cap on your IC supplies.
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