Dark Alchemist said:
Back in the 80's I know I made a simple circuit with 1 led and a
resistor connected directly to the 117vac pwr main and it worked without
a hitch. Now all these years later I have forgotten what the formula
was I used and the values needed. Anyone have any ideas?
Thank you.
---------+ ----------+ ----------+ ---------+
| | | |
| | | |
.-. .-. | |
R 68k | | R 6k8 | | R 6k8 .-. R .-.
| | | | | | 6k8 - 120k | |
1/4W '-' 1.5W '-' 3W | | 1/4 - 3W | |
| | '-' '-'
| | | |
| | | |
| V LED +-----+ +----+-+ |
NE .-. - | | | A A |
( o) | si - LED V LED V | +-+
'-' | ^ - - +-)--+
| | | | | A A |
| | +-----+ +----+-+ |
| V 1N4007 | |
| - | |
| | | |
---------+ ----------+ ----------+ ----------+
fig. 1. Old fig.2 Normal LED fig.3 Normal LED fig.4 LED in
fashioned, simple, at about 20mA. The at 20mA. Si diode a bridge. Both
low(est) power 1N4007 blocks the bypasses reverse half cycles
consumption. reverse voltage. voltage of LED. used.
created by Andy´s ASCII-Circuit v1.24.140803 Beta
www.tech-chat.de
To give a direct answer to your question, One LED + one resistor does not work. There are no LEDs I am aware of that can resist over 170V reverse voltage. If your memory is correct, I guess you got a "bipolair" LED. That's two LEDs in one package connected similar to fig. 3. Neglecting the LED voltage with respect to the 127Vac mains, you can calculate the resistor using Ohms law. Knowing a general purpose LED likes to feel 20mA, the resistor becomes:
R=127/20=6k35.
Nearest common value 6k8. You also have to take the dissipated power into account:
P=20*127=2540mW
Nearest common value 3W. It's not only the resistor that has to be able to withstand this heat, it also has to get rid of the heat. So air flow or heatsink may be nessecary. Which brings me back to the most oldfashioned, simple and cheap solution: The neon lightbulb.
Common neon lightbulbs (fig. 1) ignite at 90V, glow at 60V and cease when the voltage sinks below that 60V. There are some different current ratings most of them in the range 1-2mA. Using Ohms law the series resistor for 1mA should be:
R=(127-60)/1=67k
Nearest common value 68k. Dissipated heat:
P=1*60=60mW
so a 1/4W or even an 1/8W will do. Overall power consumption is 127mW which hardly can be beaten by a LED. As a neon bulb has no polarity, both half cycles will be used.
A general purpose LED (fig. 2) does best at about 20mA. But, being a diode it will conduct current in only one direction. As its reverse voltage usually has a maximum of 10V, you need ean extra diode - the 1N4007 - the block this reverese voltage. Neglecting the LED voltage and calculating for 20mA the series resistor becomes:
R=127/20=6k35
Nearest common value 6k8. Power consumption looks like:
P=20*127=2540mW
So 3W. But, the diodes only conduct one half cycle. The other half cycle current through the resistor will be blocked. Power is dissipated only half the time so a 1.5W resistor will do. Of course, light is only emitted when the LED conducts, so it only lights half the time. At 60Hz you may not see the flicker. Even at 50Hz flicker can hardly be seen or is acceptable.
A less economical solution to prevent the LED from being blown by reverse voltage is connecting a cheap silicium diode antiparallel to the LED (fig. 2). You'll get not more light, but power consumption doubles and you have to rate the series resistor for 3W. If you can afford it (not for the money but for the heat) it's better to use two LEDs antiparallel or find a so called bipolair LED.
Another way to reduce heat is lowering the current. A lot of LEDs may give light enough for your purpose at only 10mA. Some high efficiency types seems to do well at 2mA. Of course you can rise the resistor value accordingly (and lower the power rating) but at lower currents flicker may become disturbing. A simple diode bridge with cheap (1N4148 type) diodes will reduce flicker and give almost maximum light. Almost. If you want to squeeze the last foton out of your current, use LEDs for the bridge as well.
I do not think I wrote the last word about this subject. But I'am sure I have some ideas.
petrus