12V rectifier for LED circuit.

[John12868]

Hi Chaps,

Excuse my ignorance, but what componets would i need to rectify/regulate the 12 V ac output of a transformer so I can run a 12V DC LED as a power indicator?

Regards

John

[Herald1360]

If the LED indicator is for 12V dc it probably consists of an LED and series resistor. If it's just a straight LED, you'll need to add a resistor to limit the current to about 10mA. Something like 1Kohm 0.25W will do.

Simplest solution would be just to connect in series with a suitable diode across the 12V transformer winding. This might be a bit annoying, as it will have a 50Hz flicker. To double the flicker frequency, change to a a bridge rectifier instead of the single diode.

Chris

[pmmunro]

John,

If what you want is an LED to indicate that the supply is present, you don't need a regulator nor a separate rectifier. As stated in its name the LED is a diode and will allow only alternative half cycles to flow. The limitation of LEDs is that they have a low peak inverse voltage but this can be overcome by connecting a silicon diode in reverse parallel (cathode of LED to anode of rectifier). This way one half cycle flows in the LED and the other half cycle in the rectifier diode.

Now all you have to do is limit the current to a safe value and one which gives the required LED brightness. 20mA might be a typical value. Subtract the forward voltage of the LED from the supply voltage and divide the remainder by the LED current. The LED forward voltage will depend on the type and colour (since this is determined by the semiconductor material) but to a first approximation 1.2V is suitable for all but blue LEDS.

Finally, multiply the resistor currrent by the voltage across it to get the resistor power rating. The brightness of the LED will vary with changes in the supply voltage, but unless these variations are unusually large there will be no visibly perceptable change.

PMM.

[dominicbeesley]

Beware, not all modern LEDs have a high reverse breakdown voltage, for instance the Luxeons I've been using blow at 5V reverse so a series diode may be needed. I found this out the expensive way....

If it's 12V RMS AC then make sure your reverse breakdown voltage is over 15V

Dom

[Nickthedentist]

Originally Posted by dominicbeesley

Beware, not all modern LEDs have a high reverse breakdown voltage, for instance the Luxeons I've been using blow at 5V reverse so a series diode may be needed.

... or an inverse parallel one as PMMunro advocates?

[dominicbeesley]

Sorry!

[julie_m]

The forward voltage drop of an LED depends inversely on the wavelength of light it is emitting. Voltage (approximately) = 1240 / wavelength (in nm). The wavelength isn't constant ..... LEDs are not perfectly monochromatic, the spectrum gets broader as you feed them more power. But it's a good rule of thumb. Before blue LEDs were popular, you could generally reckon on about 2V.

Note also that "white" LEDs are actually based on blue or ultraviolet LED internals, with a phosphor coating to produce white light (and tend to change colour as the current increases, getting bluer at high current as the phosphor is overwhelmed).

Under reverse-bias, LEDs go into a "constant-voltage" state at somewhere between 5 and 6 volts. As this reverse voltage is 2 - 3 times higher than the forward voltage, it is easy to overheat an LED by supplying it in reverse. But as long as power dissipation is limited, no permanent damage will be done in this state. (The same effect is used in Zener diodes, which are used in regulated power supplies as stable voltage references; but Zener diodes are specially chosen for a particular reverse voltage. The LED manufacturer is more concerned about high luminous output in forward bias than about consistent and predictable behaviour in reverse bias.) If the voltage never reaches this figure, perhaps because something connected in parallel with the LED is limiting it, then no power will be dissipated at all.