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If I connect a LED to a 1.5V battery without a resistor and it blows up, is that because of the voltage or the amperage? I think amperage, but I'm a beginner at this.

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    \$\begingroup\$ If you fire a gun and it damages something, was it because of the bullet or because of pressure on the trigger? \$\endgroup\$ Commented Oct 9, 2016 at 8:57
  • \$\begingroup\$ LED voltage and battery voltage must match closely.. A good example is 3.0V Lithium and 3V White LED. A bad example is 9V battery on a 2V Red LED. When they do not match a current limiting R in series is needed following Ohm's Law. \$\endgroup\$ Commented Oct 9, 2016 at 15:17
  • \$\begingroup\$ The IV curve gets quite steep in the end and the current delivered by a fresh battery is impressive. Add resistance if you have no other current limiting. \$\endgroup\$
    – KalleMP
    Commented Oct 9, 2016 at 20:11

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It's because of the power.

  • Power can be calculated by \$ P = VI \$ (volts x amps) and the answer is in watts.
  • The power will be dissipated as heat.
  • Heat will increase temperature until equilibrium is reached (heat lost to surroundings = power input) or the device fails.

A typical small LED can take about 50 mA for a while and a red LED will have a forward voltage drop of about 1.8 V. We can calculate the power as \$ P = VI = 1.8 \times 50m = 90~\mathrm mW \$.

LEDs normally use a series resistor to limit the current to a safe value of 5 - 20 mA for this reason.


An LED is unlikely to "blow up" (whatever that means) at 1.5 V. I suspect that the actual voltage was higher which it could be for a fresh alkaline battery.

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    \$\begingroup\$ Still, I'm a little confused about 1.5V. I can wire up a RED LED to my bench supply and set it to 1.5V and I don't think the LED will blow up. And I think the reverse voltage zenering is higher than 1.5V, as well. (And yes, I'm still on the late shift!) \$\endgroup\$
    – jonk
    Commented Oct 9, 2016 at 9:18
  • \$\begingroup\$ Yes, I spotted that but failed to address it. See the update. \$\endgroup\$
    – Transistor
    Commented Oct 9, 2016 at 9:23
  • \$\begingroup\$ hehe. okay. Well, I've measured the brand new AA alkalines often and I almost never see more than 1.59V or so. Maybe European ones are better? ;) \$\endgroup\$
    – jonk
    Commented Oct 9, 2016 at 9:29
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    \$\begingroup\$ Infra-red LEDs can be lower than 1.5V. \$\endgroup\$
    – user16324
    Commented Oct 9, 2016 at 9:41
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The relationship between voltage and current is not always clear at first. I like using water analogies for basic examples. Let's say you have a open 55 gallon drum of water sitting in your driveway with a valve on the bottom and you are really thirsty and want a drink. The volume of water represents the current and the pressure of the water represents the voltage. If you open the valve slightly the water will trickle out because even though there is a lot of water above it (current) the pressure (voltage) is low and the valve being barely open restricts the flow of water (resistor). This flow rate is slow enough where you can easily wrap your lips around the valve and drink comfortably. If you were to now cover this barrel with a lid and connect it to your air compressor you will see the water start to spray out of the valve at a higher rate which would make it difficult for you to drink. Say the lid is now stuck and you can't change the pressure (voltage), so your other option is to reduce the flow to a comfortable level by closing the valve (increase resistance). If you were to place a pressure gauge at the bottom of the tank and another on the output of the valve you would see a large difference when the valve is closed and a small difference when the valve is open, this is the same as a large voltage difference when resistor is infinite and zero voltage when the resistor is 0ohms. (Note: An open resistor is like a valve connected to a closed pipe, you can open the valve all the way and water will not flow just like you need a something connected to the other end of a resistor or current will not flow.)

Taking the above analogy back to an electric circuit, think of the barrel of water as a 12V car battery you want to use to run your LED. A typical car battery can provide 500A of current but our LED only needs 20mA so how do you connect a tiny LED to a current source which has 25,000 times more current than we need without vaporizing it? If you use a resistor the LED will be supplied with the proper amount of current and run indefinitely. If you were to decrease the resistance (open the valve) or increase the voltage (attach the air compressor) the current would increase to a point where the LED would burn out or explode.

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  • \$\begingroup\$ The volume of the water is analogous to the current? How does that analogy work? In electronic circuits, current is voltage divided by resistance; it seems like in order for your analogy to be sound, the water volume would have to be proportional to the pressure divided by the resistance of the valve. \$\endgroup\$ Commented Dec 27, 2019 at 2:27

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