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I want to power an LED that has a 3.3V forward voltage using two 3V CR2032 coin cell batteries to create a 6V supply. I set the resistor for a 20 mA current. For my application, the LED will be turned on for about five minutes a day.

The CR2032 has a capacity of 235 mAh (Energizer datasheet: http://data.energizer.com/pdfs/cr2032.pdf) The datasheet only has a "continuous discharge characteristics" curve for a 0.19 mA drain.

How do I know what the capacity is at the current draw level of 20 mA? How can I estimate the battery life, besides dividing the mAh capacity by the current draw?

It's confusing because on some sites I've read that coin cell capacity seriously degrades if you continuously draw more than a few mA... but I've seen many keychain LED flashlights that even use the smaller capacity CR2016 battery.

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  • \$\begingroup\$ Note that because of how different your voltages are (~3.3V and 4-6V), your series resistor will waste a considerable amount of power. The internal resistance of your battery ranges from 10 to 140 ohms and they're rated for pulses of 1/3 the current you want to run, so they probably won't do well with 3 times that current continuously. \$\endgroup\$
    – K H
    Commented Aug 9, 2018 at 4:36
  • \$\begingroup\$ With a Red or Yellow you only get about 150mAh out of possible 225 mAh or about 1/2 day bright, 1/2 less bright , 2 days very dim. With White a bit less. This directly across CR2032 Current is limited by internal ESR then by voltage drop below Vf of White LED of 3V down to 2.8V But white is not using 20mA., perhaps 5mA depending on part. Because brightness, LED voltage and battery voltage are very dynamic, and lifespan in seconds is hard to model without a well defined test procedure and cutoff. \$\endgroup\$
    – D.A.S.
    Commented Aug 9, 2018 at 5:02
  • \$\begingroup\$ See the application manual data.energizer.com/pdfs/lithiumcoin_appman.pdf for 3mA curve. \$\endgroup\$
    – Passerby
    Commented Aug 9, 2018 at 5:18
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    \$\begingroup\$ Tony's comment should be an answer. For once. \$\endgroup\$
    – Passerby
    Commented Aug 9, 2018 at 5:22
  • \$\begingroup\$ Ok thanks for the advice, I'll have to see how bright the LED is with a smaller current like 5 mA...or go for the bigger CR2477 coin cell! It seems like there's no obvious answer here... \$\endgroup\$
    – donut
    Commented Aug 9, 2018 at 5:39

1 Answer 1

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You can only rely on the (Amp Hour) the battery has. Or for small batteries the mAh (milli amp hour). Anyway all these figures are estimates anyway. The lifetimes depends e.g. on temperature.

I don't know what you need the LED for but you can get LEDs which are a lot more efficient: 1 mA would give you 20x as long.

As K H said: you are wasting energy in a series resistor: Try to find a switching current source. A simple PWM circuit with an inductor comes to mind.

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  • \$\begingroup\$ I need the LED to illuminate a box. This isn't just an indicator LED, so I don't think 1 mA would be sufficient. \$\endgroup\$
    – donut
    Commented Aug 9, 2018 at 4:59
  • \$\begingroup\$ Also, are you referring to PWM for the purpose of dimming the LED? \$\endgroup\$
    – donut
    Commented Aug 9, 2018 at 5:14
  • \$\begingroup\$ @donut you'd be surprised how bright a led is at 1mA especially at night/dark. Depends on the size of the box tho. Try 5mA. \$\endgroup\$
    – Passerby
    Commented Aug 9, 2018 at 5:23
  • \$\begingroup\$ With PWM I am thinking of some sort of unregulated switching current supply. Combined with an L you can save a lot of energy. (Note that LEDs are controlled by current, not voltage). \$\endgroup\$
    – Oldfart
    Commented Aug 9, 2018 at 6:17
  • \$\begingroup\$ LEDs also are more efficient at lower instantaneous voltages, meaning that if you Give an LED a 10V square wave with 32% duty cycle, it will have an effective voltage of 3.2V and the LED probably won't burn out, but if you increase the frequency of the square wave and run it through an inductor as @oldfart is describing, you can have the efficiency of switched mode and constant voltage output similar to a linear supply. \$\endgroup\$
    – K H
    Commented Aug 9, 2018 at 7:36

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