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Initially I used a 9 V 170 mAh battery for powering a 3 W LED which was rated 9 V and 333 mA. Calculating the power consumption of the battery based on how long the LED light will last came to around:

(9*.170) Watt hours/(9*.333) Watts = 0.51 hours.

LED Specs,

Battery specs : http://www.amazon.in/Duracell-Rechargeable-170-mAh-Batteries-Pack/dp/B0002FQXJK

However the LED lasted for more than one hour, infact it was glowing for more than 6 to 7 hours before it became appreciably dim.

I foresee this to be somewhat similar to this diagram below where the left hand side estimates the theoretical value at full discharge and the right hand side is the observed value of the run time of the LED. How do I find this observed run-time through calculations or datasheets?

enter image description here

How do I calculate this value before it becomes appreciably dim?

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  • \$\begingroup\$ Please provide links (in the question and not in the comments) to the datasheets for the LED and the battery. Also your title is backwards. It should be "The actual battery discharge time through LED". Are you able to measure the current with your multimeter? \$\endgroup\$
    – Transistor
    Jul 28, 2016 at 6:56
  • \$\begingroup\$ Impossible to answer, because "appreciably dim" is not defined, nor is it easy to define. You could change it to a specific set current, but we need the datasheet for a given LED. \$\endgroup\$
    – pipe
    Jul 28, 2016 at 7:03
  • \$\begingroup\$ I don't know what value you consider dim, but based on what you stated, it would have been a third of full brightness within minutes of connecting. \$\endgroup\$
    – Passerby
    Jul 28, 2016 at 7:26
  • \$\begingroup\$ The maximum output brightness is 300 lumen. When I meant dim, it was coming around 50 lumen or so. \$\endgroup\$
    – shadesofpurplegreen
    Jul 28, 2016 at 7:29
  • \$\begingroup\$ @pipe I have inputted the specs of the LED and the battery \$\endgroup\$
    – shadesofpurplegreen
    Jul 28, 2016 at 7:29

2 Answers 2

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In response to, expansion of, and disagreement with Passerby's answer, you need to start by looking more closely at the data on the link you provided.

The LED is actually a series of 5 LEDs, with no series limiting resistor. The combined LEDs have a Vf (forward voltage) listed as 9.6 to 10.5 volts, and applying 3 watts to 10.5 volts gives a nominal operating maximum current of 283 mA, which differs from the listed value of 260 mA. Why this is so is a mystery, but I'd guess it's an indicator that you might not really want to buy from these folks. Oh well, too late now.

Since you have applied your 9-volt battery directly to the LED, with no series resistor, you are relying on both the internal resistance of the battery and the fact that the battery voltage is actually operating well below 9 volts. While commodity batteries don't have easily-found data sheets, this one would seem to apply, and notice that the open-circuit voltage is 8.4 volts.

So the reason the LED lasts so long at 9 volts/333 mA is simple - it's at neither. I'd guess you're running at about 7-8 volts and somewhere in the vicinity of 30 mA.

If you want more light (and you should be able to get it) you'll need a bigger battery or a DC power supply - something on the order of 12 volts or more. If you do this, you MUST provide a series limiting resistor on the order of 4 ohms per volt over 9 volts. If you do not provide this limiter, and you connect to a source which will provide more current, you will almost instantly kill your LED. So don't do it.

If you're going to mess around with this stuff, you absolutely need to get a cheap DMM and actually measure what you're doing. You can't see voltages or currents, and you'll have no idea of what you're seeing without a meter.

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  • \$\begingroup\$ 6 diodes, not 5. And likely not in series unless each diode only has a VF of 1.5V. More likely 2 parallel strings of 3 diodes in series. \$\endgroup\$
    – Passerby
    Jul 28, 2016 at 21:23
  • \$\begingroup\$ Plus, what do you disagree with of my answer? \$\endgroup\$
    – Passerby
    Jul 28, 2016 at 21:24
  • \$\begingroup\$ @WhatRoughBeast '4 ohms per volt over 9 volts'. What does that mean? Based on these ratings, what do you think is the series limiting resistor? \$\endgroup\$
    – shadesofpurplegreen
    Jul 29, 2016 at 2:28
  • \$\begingroup\$ @shadesofpurplegreen - I meant 4 ohms for every volt over 9 volts. So a 12 volt supply would start with a 12 ohm resistor. This is a very conservative value, but it guarantees that you won't kill the LED. More precisely, use 4 ohms for every volt above the operating Vf, and this will limit the current to 250 mA. At 10.6 volts Vf, a 12 volt supply and 12 ohms will only pull 1.3 watts ((10.6) x (12 - 10.6)/12 , but you can always cut the resistance once you know what to look for. \$\endgroup\$
    – WhatRoughBeast
    Jul 29, 2016 at 3:15
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How do I find this observed run-time through calculations or datasheets?

You wouldn't. An led datasheet wouldn't graph out a battery life, and a 9V battery datasheet wouldn't graph out twice it's actual capacity as a load. 9V batteries are designed for low current, long life applications, not high current short life. The 170mAh listed is at some current n, where n is undoubtedly nowhere near 333 mA. At 333 mA, the life would be considerably shorter than 170mAh.

To calculate the runtime, a voltmeter and ammeter would be used. Graph the current and voltage across the led and battery, to find the actual runtime specifics.

That said, it's likely your led is a fake, and not pulling the stated current.

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