Double checking my thinking on volts and amps - Page 1
Forum section: electronics / beginners

author: msknight
author_rank: Regular Contributor
timestamp: November 09, 2021, 03:42:09 pm
content: I've done some checking and testing on some bulbs.A straight replacement E10 LED for 6v requires four AA batteries to run. It measured at using 30mA. It's minimum operating voltage is 2.6v after which, it dies completely.Another LED operates on 3v and I was thinking of using two pairs of AA batteries in parallel, to light two LED's also in parallel which together measured at 30mA (15 each) ... and likewise the minimum operating voltage was 2.6v until they went dark.Now... my newbie thinking says that with the degradation curve of batteries, the four in series will last for longer before they drop below 2.6v, whereas the two parallel pairs will reach that voltage sooner and therefore I'd get more out of the 6v LED because it's sipping the same amount of power?Am I missing something here?

author: IanB
author_rank: Super Contributor
timestamp: November 09, 2021, 05:08:40 pm
content: If you look at power, the 6 V lamp is consuming 6 V x 30 mA = 180 mW.A single 3 V lamp is consuming 3 V x 15 mA = 45 mW. Therefore two 3 V lamps are 45 mW x 2 = 90 mW.From this, it appears that the 6 V lamp is consuming twice the power of two 3 V lamps.Battery life depends on the power consumption, so if you have 4 AA batteries (fixed total amount of energy available), then the pair of 3 V lamps should run approximately twice as long as the 6 V lamp, because they are consuming half the power.

author: IanB
author_rank: Super Contributor
timestamp: November 09, 2021, 05:11:17 pm
content: Of course, if you run the 6 V lamp on 3 V, it will probably consume less power (with correspondingly less light output). It all depends on the internal design of the lamp and how it regulates current. Many lamps just have a simple series resistor, in which case power will drop significantly as the voltage drops.

author: msknight
author_rank: Regular Contributor
timestamp: November 09, 2021, 05:14:50 pm
content: I see what you're saying about the mW consumption, but what caught my attention was the rate of drain from the batteries.Looking at various discharge graphs for AA batteries, they start at just over 1.5V then slide down to 1.2V for about three quarters of their life. As the LED's cease to operate at 2.6V, then when two cells in series drop below 1.3v each, then the LED will go out while there is plenty of power still in the batteries.With four AA's in series, even by the time they slip down to 1.2V, even down to 1.1V each (before they fall off the cliff) they are still providing over 4V and the LED should stay lit until near the very end of the batteries.Is my thinking right?

author: IanB
author_rank: Super Contributor
timestamp: November 09, 2021, 05:20:29 pm
content: Well, you also need to consider whether you are using alkaline batteries or NiMH rechargeable batteries.Alkaline batteries have a long gradual decline in voltage as they discharge, with no "cliff" as such. They just go down and down, to about 0.9 or 0.8 V or so.On the other hand, NiMH have a "cliff" at about 1 V. So if you put 4 NiMH in series the "cliff" will occur at about 4 V.As to 4 AA in series lasting longer, you also need to consider the current drain. If you draw 30 mA instead of 15 mA, the voltage will decline twice as fast.If you put the batteries in parallel they are sharing the load, and the current is divided between them, so the voltage will decline more slowly.

author: msknight
author_rank: Regular Contributor
timestamp: November 09, 2021, 05:30:48 pm
content: I think I'm going to run a test with each of them. Fresh sets of alkalines in each(that's what I'm most likely to get on a campsite)and see what happens. Should prove interesting :-)

author: IanB
author_rank: Super Contributor
timestamp: November 09, 2021, 05:35:36 pm
content: That's the best way to get accurate data.In my experience, battery powered LED lanterns don't tend to have any regulation, so they start out very bright on fresh batteries and then get dimmer and dimmer as the batteries drain.

author: tunk
author_rank: Super Contributor
timestamp: November 09, 2021, 05:42:49 pm
content: I guess it depend on the E10 LED. If it's a resistor in serieswith an LED, you will waste around half the power in the resistor.As for two AAs and 2.6V, you will use maybe around 40% of thebattery capacity.Edit: I don't think it's correct that you loose half the power.It will initially when it's 6V, but the loss will be reduced asthe battery is discharged.

author: Jwillis
author_rank: Super Contributor
timestamp: November 09, 2021, 05:45:11 pm
content: You also need to know that a White LED will have a forward voltage of approximately 3V . Forward voltage is the voltage required for the current to flow through the diode junction . So when your series cells deplete below the required forward voltage the LED won't light. So you might think your batteries are depleted but not really . It's just the nature of the LED . If you take a red LED and put in place of the white LED it will shine brightly because the red LED requires less than 2V forward voltage .

author: fourfathom
author_rank: Super Contributor
timestamp: November 09, 2021, 05:59:48 pm
content: Quote from: tunk on November 09, 2021, 05:42:49 pmThis is the main issue.If the LED circuit uses a switching regulator than (up to a point) there is no drawback to putting the batteries in series, and you may get better battery utilization. If the LED uses a dropping resistor then you waste power if the battery voltage is higher than necessary. You need to do the math to find the optimum battery configuration.