Comment on Scales that refuse to measure if the battery isn't brand new

• magikmw@lemm.ee ⁨2⁩ ⁨months⁩ ago

  Except the cheap part. But likely not by much.

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  • chonglibloodsport@lemmy.world ⁨2⁩ ⁨months⁩ ago

    You can order 3000 3.3V low drop out (LDO) voltage regulators on LCSC for $25.50. That’s less than a penny each.

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    • Ghoelian@lemmy.dbzer0.com ⁨2⁩ ⁨months⁩ ago

      Yeah, but it’s more than 0 pennies each.

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    • B0rax@feddit.org ⁨2⁩ ⁨months⁩ ago

      Doesn’t help when you need 3.3V and the batteries are now down to 2.5V they are not putting a boost converter in there.

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      • chonglibloodsport@lemmy.world ⁨2⁩ ⁨months⁩ ago

        Right. If your design requires 3.3V minimum then putting in a 3.3V battery and no boost converter is just dumb (or extremely user-hostile).

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    • itsmect@monero.town ⁨2⁩ ⁨months⁩ ago

      Batteries have one advantage over over supplies: extremely low noise. Even an good LDO will bump up the noise floor, and a cheap lcsc part will do so too. Plus you’s want a reasonably low dropout and quiescent current, which also increases price. Maybe 10ct in volume is reasonable for such a part - and yes, that will absolutely eat the margin

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      • chonglibloodsport@lemmy.world ⁨2⁩ ⁨months⁩ ago

        That’s definitely true. But I would definitely pay more for a scale with ultra long battery life.

        I made the mistake of buying an off brand digital calliper and now like an idiot I find myself removing the battery when it’s not in use just to avoid damn thing running flat in one month thanks to its atrocious standby current which enables the display to turn on instantly when I move the slide (rendering the on/off entirely moot).

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• Buffalox@lemmy.world ⁨2⁩ ⁨months⁩ ago

  Absolutely, My kitchen scale remains accurate and can use the same battery for years.

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• XeroxCool@lemmy.world ⁨2⁩ ⁨months⁩ ago

  Eventually, the battery drops lower than the device needs. So no, none of which is fixed by a voltage regulator.

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• iamdefinitelyoverthirteen@lemmy.world ⁨2⁩ ⁨months⁩ ago

  No, not voltage, current.

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  • B0rax@feddit.org ⁨2⁩ ⁨months⁩ ago

    Which in this case means: you need a constant supply voltage

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