I have a LED light uses three 1.5V ALKALINE batteries in series. However, the capacity of the battery is not marking. When I put a multimeter in series with these three batteries and turn on the LED, the current is 76mA. I also measured the voltage between the LEDs and it is 2.77V.

If I assume the capacity of the batteries is 2200mAh (I found it on WIKI, it said for alkaline batteries, the capacity under 50mA constant drain is 1800~2600mAh).

My question is, what is the lifetime of this LED light? Can I calculate this from 2200mAh/76mA=28.94h? Or it is also affect by LED working voltage? The other part in the LED circuit is just a PIR sensor, I think it won't draw much current.

  • \$\begingroup\$ You measured the voltage between the LEDs or across each LED? \$\endgroup\$
    – SRR
    Commented May 1, 2018 at 0:10
  • \$\begingroup\$ @S.Ramjit 6 Leds connected in parallel, I measured the voltage across them. \$\endgroup\$
    – Channing
    Commented May 1, 2018 at 0:13

2 Answers 2


An AA alkaline battery's voltage starts at 1.5 and drops to 0.8V cutoff. This makes alkaline a poor choice for powering an LED. You want a flatter discharge curve. Li-ion is a much better choice. You could easily power these LEDs with a single 18650 battery.

The lifespan is determined by the discharge rate. At 25mA expect 3000 mAH. At 100 mA expect 2500 mAH. See ENERGIZER E91 Specifications The mid point would be 2750 mAH.

Three AA alkaline in series will have a discharge curve from 4.5v to 2.6v. A better value would be 11Ω which would give you 75 mA at the 3.6V mid point between new 4.5V and discharged to 2.77V.

Each one of these LEDs must/should have a current limiting resistor. You measured 76 mA which is only valid for whatever the battery voltage was at that instant. The voltage is continually decreasing as well as current.

First you must specify the current required for the desired brightness. The use an online calculator to get the value of the optimum resistor. I like the Hobby Hour LED Series Resistor Calculator

At 75mA each LED uses 12.5 mA.

A 3.6V 18650 Li-ion will discharge to 3.2V, a much flatter curve. At 4.5V an LED with a Vf of 2.77 would use a 140Ω resistor for 12.5mA. For 12.5mA at the minimum 2.77V the optimum resistor would be 10Ω. The resistor to use would be calculated with the mid point voltage of 3.6V which is 66Ω. But the range of brightness would be noticeable. This will give you an average of 45 mW per LED or 270 mw for all 6 LEDs. With a battery capacity of ≈2750 mAH would yield a lifespan of 36.66 hours with an average 76% efficiency.

If you need a very consistent brightness then a constant current regulator would be required. something like an On-Semi NSI45060JD would work okay. For higher efficiency a switching step down regulator may do better. A TI TPS63030DSKR is very simple, low part count, small, and inexpensive. It would very work well as efficiency would vary between 90% @ 4.5V and 80% at 2.4V. It is designed especially for 2-3 AA series alkaline batteries or a single NiMH or Li-ion.

A switching regulator would yield about 10% more battery life, or about 40 hours.

  • \$\begingroup\$ The OP is not asking about battery choice or selection of limiting resistors. Nor anything about consistency of LED brightness. The OP is asking if the estimate of 2200/76 is right or not. I say it is right, and any attempt of "better" estimate will be an excess of precision for a given information, and therefore a waste of time. \$\endgroup\$ Commented May 1, 2018 at 7:41
  • \$\begingroup\$ @AliChen not a waste of time. The OP needs to understand the 76 mA measurement was only valid at the battery voltage at that instant. It also seemed the OP was connecting the batteries directly to the LEDs without any type of regulation. That too would incorrect design and battery life calculations would be inconstant. I stand by my figures of 36 to 40 hours and say your calculation is based on erroneous data and is why you are off by about 35%. LED design is a full time job for me. \$\endgroup\$ Commented May 1, 2018 at 7:53
  • \$\begingroup\$ I believe you misunderstood the question. The OP is not connecting anything, the OP has a certain off-the shelf device that already USES three batteries and a PIR sensor. And therefore very likely some switch. And maybe a constant-current driver. And maybe of DC-DC type. And maybe he is using a "heavy-duty" junk-grade batteries from no-name supplier. All your conditional considerations are off mark, unsubstantiated and based on lots of assumptions, which may or not be true. \$\endgroup\$ Commented May 1, 2018 at 8:07

It doesn't matter what is the voltage on LEDs, 2.77 V is about right for modern LEDs. However, three alkaline batteries should be delivering at least 3.3 v to 4.5 V under this modest load, so there must be some other circuit (or just a resistor) in series with this bunch of LEDs. However, nothing of this matters, the only 76 mA is important. Your estimation is right, a battery of elements with 2200 mAh capacity (I guess these are AA-size elements) will last about 28 hours at 76 mA load rate.

EDIT: You can ESTIMATE the working time as 2200/76. It will be a good rough estimate. You can't improve it much by considering LED voltage drop over gradually depleting battery, since the exact result will depend on type of LED driver (simple resistor or DC-DC converter), actual battery capacity under specified load current (and manufacturer), and overall temperature of battery and LEDs. So any "better" estimation will be dubious at most. If you want to measure, then just measure it with a stop-watch.

  • \$\begingroup\$ Wouldn't the voltage drop as time goes on? So if the voltage is dropping under discharge then there will be a point where the forward voltage of the LEDs isn't met and the batteries would still have a voltage on them (since they weren't completely drained). So wouldn't OP need to know the minimum capacity that gives a certain voltage and then decide how much capacity is actually available ? \$\endgroup\$
    – SRR
    Commented May 1, 2018 at 1:06
  • \$\begingroup\$ @S.Ramjit, you can't make any more accurate estimates since you/we don't know the driver circuit. It could be a buck converter, and as battery voltage drops, the current will INCREASE. And you don't know the actual capacity of your batteries either, nor operating temperature. So the simple estimate is the best you can do. \$\endgroup\$ Commented May 1, 2018 at 1:11
  • \$\begingroup\$ Not trying to be condescending and I'm not sure if this is "hijacking" OP's question but how will the current increase if the voltage drops ? I understand everything else you've said however. \$\endgroup\$
    – SRR
    Commented May 1, 2018 at 3:14
  • 1
    \$\begingroup\$ @S.Ramjit, if the flashlight (or whatever) uses a good constant-current LED driver (DC-DC switcher), the driver will deliver constant power to the LEDs. When battery voltage drops, the led driver will increase its input current in order to maintain the constant power output. Energy conservation at work. \$\endgroup\$ Commented May 1, 2018 at 4:09
  • \$\begingroup\$ What you say would be true if an alkaline battery maintained 1.5V over its lifespan. An alkaline has a discharge cutoff of 0.8V. That is why LEDs need 3 AA cells. Even a very good switching regulator's efficiency will vary 10% over the full discharge curve. \$\endgroup\$ Commented May 1, 2018 at 7:27

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