How to light a LED (Light Emitting Diode) from telephone line

LED is Light Emitting Diode which we can find in most of electronic devices and even used in the light indicators in luxury cars for smooth glow.  Most of us is unaware of the presence of FREE electricity that is available to us in our home. Yes, it is FREE and available 24/7 for 365 days a year. However it is not in so much amount that you can run your Juicer or laptop, but you can certainly drive a handful of LED during power cut or emergency. How does it work?

There is always a constant supply of 50V DC in your telephone line when there is no call, to keep the phone alive. However the voltage increases to a considerable level during a phone call.

Now, we are interested in the 50V DC that is available all the time. We should always remember that even when there is a power cut in the main line, the telephone line still remains active.

What we will need?We will need a few basic electronic parts which can be easily found in any store.


  • 1. Led = White of White (No mistype) 5 pc
  • 2. Resistance =33 K -2pc
  • 3. Diode = IN4007 - 1pc
  • 4. Circuit Board = A small one.
  • 5. Jack= RJ11 ( Fits in telephone outlet)
  • 6. Some wire.
  • 7. Soldering iron and wire.

How to Fix it Up?

I am attaching the circuit diagram below, please follow the same. I made the diagram in simple paint, so apologies for that.

The two 33K resistance adjusts the voltage and current, so the LED doesn't burn out. And the Diode IN 4007 prevents reverse bias in case a AC is introduced in the circuit.

The above circuit will provide enough light when used with 5 LED' s to dimly light a medium room or full light to read a book if used as a table lamp.

For those who cannot do soldering.

You can buy a Jack RJ11 and plug two or 3 LED's in series with the two wire. It will work, but not as bright as when used with the above circuit, and it will also not last for long as LED's gets burnt due to over voltage.

So, what do you think about this cool idea? Drop your comments below and also let others know if you have come across more cool stuff.


  1. Hi, The above idea is good. based on this, if you can provide some additional info it will be of great help. I need a circuit diagram which can switch on the LED lights when the power goes off (mains supply) using LDR. the lights should be off during day time and light up in case of power failure during night or when the room is dark.

  2. Assuming the voltage drop across the led is 3.5v, you would need a resistor that drops the rest of the voltage (50-3.5=46.5v). Using the recommended current (usually ~20mA) you could set up an Ohm's Law equation (V=IR) where V=46.5v I=.02A and R=resistor value. 46.5/.02=2,325?. You would also have to figure out what wattage of resistor to use (P=IV) which is .93W which is under 1 watt so you would be have to use at least a 1 watt resistor. You could add a lot more LEDs up in series if you use the correct resistor value. Say 10 of them would give a voltage drop of 35v which leaves 15v to be dropped across the resistor. So 15v/.02A=750? and 15v*.02A=.3W so I need 10 white LEDs and a 750? .5W resistor.

  3. A roommate recommended me to look at this post, brill post, fanstatic read... keep up the cool work!

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