Mains powered LED without a transformer.

[exxos]

I thought I would post this here as its really simple and may be useful to others.

I needed a way to power a simple LED, but I did not want to use a transformer.

So I ran a simple simulation with a simple tests circuit as shown below..

sim.JPG (22.94 KiB) Viewed 2738 times

Now my circuit simulator did not have a actual LED so I had to use simple diodes instead. But good enough for this purpose.

I have a in-line current probe to monitor the current, and all we have here is 2 antiparallel LED's alter the mains via a low value capacitor. To prevent my circuit simulator from spazing out I had to put a ground connection there, but in reality it would not be connected in a actual circuit..

When the simulation is run I get..

sim2.JPG (64.33 KiB) Viewed 2738 times

The redline is with a 220nF capacitor gives 8mA . The green line is with a 100nF capacitor and gives 4mA.

This basically works due to the capacitive impedance versus the mains frequency. I found the calculator online to calculate this anyway...

https://www.allaboutcircuits.com/tools/ ... alculator/

sim3.JPG (24.78 KiB) Viewed 2738 times

Thing to bear in mind here is that each LED will only actually see half the main cycle and will actually be half the current. So the proper calculation actually becomes..

120V / 31830R = 0.0037A or 3.7mA. This is with a 100nF capacitor.

120V / 14468 = 0.0082A or 8.2mA . This is with a 220nF capacitor.

Or if you want the actual "long-winded" calculation

(240V / 2) / 31,830R = 0.0037A

(240V / 2) / 14,468R = 0.0082A.

Of course this is just a very quick and dirty circuit, but good enough in some cases when the transformer is not really practical just one final note, the voltage of the capacitor should be rated higher than the mains voltage. In this case I would probably recommend a 400V-500V capacitor.

I hope this will be of use to someone in the future if anything, I just wasted 30 seconds of your life