I would like to power a LED off a 1,2 volt rechargeable battery,850mAh,energizer without any resistors.Since the LED has a forward voltage around 2 volts,working fine when connected to a 1,5 volt alakaline AA battery,it's obvious that the voltage isn't a problem.However,I am not so sure about the current.Would the LED be fried if I connected it to this battery?
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\$\begingroup\$ LEDs are not linear components but there are only very few where the current would actually go up if you decrease the voltage. 850mAh has nothing to do with current, it's a measure of charge (1mAh = 3.6C (Coulomb)). \$\endgroup\$– ArsenalCommented Sep 1, 2015 at 8:20
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\$\begingroup\$ Ok,I'll bear that in mind. \$\endgroup\$– Daniel TorkCommented Sep 1, 2015 at 8:25
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3 Answers
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Would the LED be fried if I connected it to this battery?
Depends on two factors. The Battery's Internal Equivalent Series Resistance, and the IV Curve of the LED. The 2V stated Forward Voltage of the LEDis likely at 20mA, the typical Forward Current for the stated expected life of the LED.
At 1.2V fully charged, the LED likely won't be damaged, but the brightness will be considerably dimmer than it would be at a voltage and current close to 2V @ 20mA.
For example, super cheap led flashlights are composed of an Led (Lets say Blue at 3.4V) and a 3V Coin Cell Battery (CR2032). The Battery's lower voltage and high ESR (about 20 Ohms) allow the LED to be nice and bright without needing a resistor.
Frankly, just sacrifice an LED in the name of scientific experimentation, and try it. If it blows, you know you need a resistor. If it doesn't, you are all good*. A quick test of a red LED and a brand new 1.5V AA Alkaline battery shows it working. How long the LED would last like would take long term experimentation to know.
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\$\begingroup\$ So you are saying that if I give the LED 700mA at 1,2 volts,it would be alright? \$\endgroup\$ Commented Sep 1, 2015 at 8:28
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\$\begingroup\$ @DanielTork no. The current that goes through the LED will be limited by the voltage across it. Sufficiently low voltage will not result in a blown led. See electronics.stackexchange.com/questions/76367/… for a full math explanation of an LED IV Curve. \$\endgroup\$– PasserbyCommented Sep 1, 2015 at 8:35
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1\$\begingroup\$ @DanielTork you cannot force 700mA at 1.2V through a LED which has a 2V forward voltage at 20mA. It will only ever draw as much current as it's resistance at 1.2V allows (and it's resistance will change with voltage if you see it like a resistor). \$\endgroup\$– ArsenalCommented Sep 1, 2015 at 8:37
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1\$\begingroup\$ Thank you everyone!It looks like it didn't turn on at all. \$\endgroup\$ Commented Sep 1, 2015 at 8:40
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\$\begingroup\$ I used 2 of those batteries with another,brighter one and this one worked just fine and it must accept just a few mA and volts past the first one. \$\endgroup\$ Commented Sep 1, 2015 at 10:50
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No, because when the supply is below the LED's forward voltage only a very small amount of current (hundreds of microamps, if that) will be able to flow; if the LED is visible at all it will be very dim.
The danger with too high a voltage is that the amount of current that is allowed to flow will be too high for the LED's power handling and the semiconductor will melt/blow up, but this is not a problem is the voltage if too low to allow much current at all.
answered Sep 1, 2015 at 8:08
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\$\begingroup\$ Doesn't current depend solely on the battery?For example,a 9 volt battery can output 550mA and say,a 5 volt battery 700mA? \$\endgroup\$ Commented Sep 1, 2015 at 8:19
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1\$\begingroup\$ @DanielTork: Each LED has its own V-I curve, which tells you how much current it will pass for a given voltage. An LED's "forward voltage" is simply the point at which the curve turns upwards sharply. \$\endgroup\$ Commented Sep 1, 2015 at 8:46
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To be sure what current the LED will draw without an explicit resistor you will have to know its characteristics, and the characteristics of the battery. It is likely that both will be so ill-determined (variable over time, temperature, recharging, etc.) that you will get a current that can vary widely. Whether this is a problem (varing brightness, maybe reduced lifetime) is up to you to decide.
answered Sep 1, 2015 at 8:07