Electronics > Beginners
MicroSD and 18650
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Peabody:
My memory is that cards draw significantly less current in SPI mode than their maximum rated.  And older SD cards (up to 2GB) draw considerably less than SDHC.  In any case, I don't think you would ever need anywhere near 500 ma.  There's one manufacturer, SanDisk I think, that has a 100 ma maximum for all their cards.  So I would urge you do some actual measurements with the cards you will be using before assuming that 400-500 ma will be needed.

Anyway, as has been suggested, it seems the best answer may be to run everything at 3.3V.  Then you could step down from one 18650, and wouldn't need any level shifters on the SPI lines.

Some cards are supposed to go into idle mode when not being used, which draws little current, but I don't know if that's reliable, or used by all manufacturers.  What I've seen is that SDHC cards draw about 30 ma just sitting there, which is no good for your battery, so if only writing twice an hour, I think the best practice is to switch off power to the card between logging events, probably with a mosfet.

And a big capacitor for the writes.

r4r:
Thanks for replies!

I think to use TPS73630 then. 75mV Dropout Voltage i.e. to get 3.3v I need roughly 3.4 (at minimum) on 18650. Right?

beanflying, interesting module, will order couple.
Peabody:

--- Quote from: r4r on May 01, 2019, 07:32:58 am ---Thanks for replies!

I think to use TPS73630 then. 75mV Dropout Voltage i.e. to get 3.3v I need roughly 3.4 (at minimum) on 18650. Right?


--- End quote ---

Did you mean TPS73633?  Well it looks like either would work for the SD card.
r4r:

--- Quote from: Peabody on May 01, 2019, 02:29:22 pm ---Did you mean TPS73633?  Well it looks like either would work for the SD card.

--- End quote ---
Oh, sure 3.3v =)
Kasper:
LDO is good choice here. Switching regulators are often more efficient than LDO but when you have small voltage change and low current draw LDO can be more efficient partly because of lower quiescent current.  LDO is also usually smaller, easier and less noisy than switching.

Power loss in switching regulator ~= voltageOut * currentOut * ~(0.05 to 0.3)

Power loss in LDO ~= currentOut * (voltageIn - voltageOut)

Should probably use input values for these equations but I'm guessing you know your outputs better than your inputs so I used outputs. These are not exact equations, they just give an idea of the difference between switching and LDO.
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