Low power and low cost battery cut-off IC? - Page 1
Forum section: electronics / projects

author: Artlav
author_rank: Frequent Contributor
timestamp: October 02, 2015, 08:27:10 pm
content: I tend to wander around at night in the dark, and i want to see things without turning up the lights.So, i made these little LED nightlights:They run on a single Li-ion coin cell, and last for about a month on one charge.However, when they do discharge, they drain the cell down to 2V or so, which is not exactly healthy.So, i was wondering - is there some sort of a undervoltage cut-off IC or a solution, that would leak less than a few microamps and fit on a coin cell-sized light?I looked through ICs like LTC1998, and they alone cost more than all the lights together, and don't even come with a switch.Are there any other solutions?Does not have to be terribly precise, but must be low power.

author: KL27x
author_rank: Super Contributor
timestamp: October 02, 2015, 09:14:51 pm
content: What I would use is1. a voltage detector. They make these preset at li ion voltages, I would assume specifically for purposes like this. Usually in conjunction with a microcontroller. You can buy them with a preset voltage of, say, 2.7-3.0V, for instance. Extremely low current draw, avail in both open drain or push/pull rail-to-rail out.2. a logic level trench MOSFETThe voltage detector has only 3 pins, and come in SOT 23 packages. Vin, ground, and signal out. When Vin crosses the preset voltage the output switches your FET. The only issue is you might get some switching on/off at the cutoff point, sans hysteresis. But it will protect the battery.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 02, 2015, 09:44:39 pm
content: A voltage detector with push/pull output could run the LED directlySearch Digikey for "Voltage detector" - approx a zillion options, power draw down to a couple of uA.An issue is that they may start flashing as when the load is turned off, the battery voltage may recover to above the threshold.

author: Maxlor
author_rank: Frequent Contributor
timestamp: October 02, 2015, 09:45:38 pm
content: Rohm BD48xxx/BD49xxx series. It's a voltage detector available from 2.3V upwards in 0.1V steps. It uses less than a uA.

author: mariush
author_rank: Super Contributor
timestamp: October 02, 2015, 09:50:16 pm
content: Could always go for a cheap tiny microcontroller, like PIC12LF1552 for example:http://uk.farnell.com/microchip/pic12lf1552-i-ms/mcu-8bit-pic12lf-32mhz-msop-8/dp/2382712Works with as little as 1.8v, can output up to 25mA through its pins so enough for those leds, and has ADC which would allow you to measure the input voltage every few minutes or so and just stop turning on the leds when voltage drops too much (or make one or all of them blink to tell you to change/recharge the battery).At 500kHz - 1 Mhz or even lower, the mcu would use less than 1mA, nothing compared to how much the LEDs would consume.Of course, there's the issue of buying programmer for it (about $50-60), but could be a good starter project to get into microcontrollers.

author: Artlav
author_rank: Frequent Contributor
timestamp: October 02, 2015, 10:04:21 pm
content: BD49K30 looks great, and can power the LEDs directly, so basically a one part solution.The blinking of the battery recovery can even be a feature - easier to notice what to replace.However, i was thinking whether i'm asking the right question.The battery is not drained to zero.Why...?Because the LEDs have a voltage drop, a minimum voltage they can work at. Which happened to be 2.2V.So, what if i add more diodes?Specifically, a diode in series, to increase this voltage?I tried it, and even the crappiest 4007s drop only 0.4V at these currents, but two of them in series drop 0.8V at full load, and 0.6V at near zero.current.It seem to work perfectly - at nominal 3.6V we still have 2.8V available, so 0.6V to drop across the resistor, but it can never drop to less than 3V.And no transcontinental part delivery necessary.Am i missing something?This sounds a little too easy to be true.

author: KL27x
author_rank: Super Contributor
timestamp: October 02, 2015, 11:20:01 pm
content: No that works fine, except as the Vbat approaches FVD, your current and LED brightness will start to drop off, some. And you will obviously want to examine your series resistors for the new voltage curve. You may be able to improve the performance with a tweak.

author: Artlav
author_rank: Frequent Contributor
timestamp: November 14, 2015, 11:56:19 pm
content: Followup.I ended up using BD49K30, powering the LEDs directly off it's output, and it seem to work great.There is virtually no voltage drop across the IC at that current (30uA or so), even though the datasheet say 0.5v will always be dropped.I noticed a few of the lights being off a few days ago, and there is no flashing.The voltage recovered to 3.1v over a day or two, which is still below the hysteresis. It might flash up in a few days, but nothing fast or noticeable. I guess at this sort of a low load the battery's "recovery capacity" gets used up along the way. Or something.Anyway, it works nice and thanks for the advice.