I have a microscope lit by a 12V 10W car light bulb, which doesn't work well because the lamp generates too much heat. So I replaced it with a four-LED array with the same spec and fitting (meant to be used in cars). This provides good lighting without heating up, but the 50Hz flicker is very annoying.

I assume a capacitor would be the solution, but it is 30 years since I soldered amplifiers and calculated circuits, and I don't know much about LEDs, so I would greatly appreciate some advice on how best to solve it. I haven't actually measured, but I assume the input voltage is about 12V AC.

 

hi,
You dont normally drive leds with 'ac', else they could die.

Use a 'dc' supply with a suitable series resistor and the flicker will go away.

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Eric " Good enough is Perfect "
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You'll need a rectifier and a capacitor. The rectifier converts the AC into a pulsating DC, then the capacitor will filter out the pulses and make it into a more steady DC. Use a full-wave bridge rectifier (4 diodes or a package device) and a sufficiently large capacitor, and then your LEDs won't flicker, and they'll last longer too.

 

How about just a wallwart of the right values? Much simpler!.

Cary

 

Try a bridge rectifier and 100u electrolytic and dropper resistor

 

Bridge rectifier as already suggested, and capacitor is best way to go.
put 4 white led's in series with series resistor.

12 Volts ac through bridge = 12 *sqrt2 = 16.92 - 1.4(U diode) = 15.5 Volts dc with 100 or 470 uF 25 Volts capacitor.

4 white led's , about 12 Volts drop.

3 Volts to disspiate in R at 20 mA.

R = 150 ohms ¼ Watt.

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The way I understood the original post he has a LED 1156 style bulb used as a replacement in a auto turn signal/brake light. They already have the resistors built in to the bulb. So adding resistors would make it dimmer.

Cary

 

A bridge rectifier and capacitor might take the 12VAC and turn it into (12*1.414) - (2*0.65). That would be 15.7VDC. An additional resistor would be necessary to bring this back down to 12V.

You might find that a simple bridge rectifier to make the 12VAC into pulsating DC is all you need. (no capacitor). The bridge would make the pulsating occur at 100 Hz instead of 50 Hz. Many people still find this objectionable, but it's equally possible that you'd find this OK.

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I don't think so because the reason to go to led's was the heat generated by the 10 Watt bulb.

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http://www.youtube.com/user/RODALCO2007 some interesting electrical stuff to watch.

 

eric, LED's will run on AC just fine as long as you stay away from their maximum reverse voltage. Replacing the power supply with a DC wallwart is the easiest solution. If the light you're using is ment for automotive applications then a 12-14 volt DC supply should work just fine.

It's not so much the frequency of the flicker that's the problem, it's the depth of it, an LED goes completely out during the negative half of the AC cycle, two strings in anti-parallel increases the frequency but it doesn't change the modulation depth of the light and they're still both completly dim during zero crossing. I ran a large array of LED's in anti-parallel directly off a wall jack just fine, the flicker was horrid though. The only reason you don't see this in incandescent bulbs is the bulb only slightly dims during the zero crossing because of the thermal energy stored in the filament, so the frequency is the same but the modulation depth is MUCH shallower. I've seen circuits that uses CDS cells to generate audio from modulated light sources and it can pick up a lightbulb flicker when fed through a high gain amplifier. Fluorescent lighting run directly off mains with inductive ballasts is somewhere inbetween LED's and incandescent. Modern electronic ballasts have no human perceivable flicker component.

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Thanks for your suggestions, all of you. As I understand it, a 12V wallwart and a 100-470uF capacitor would be the simplest solution. Or, as an alternative to the wallwart, a rectifyer after the built in transformer + a resistor to adjust the voltage. I will try one of these options.

Regarding AC or DC and LEDs: since the LED-array lamp I use is symmetrical and intended for cars, doesn't it have to contain its own rectifyer to ensure that it will work independent of which way it is put in the socket? I realize that this doesn't help me, because I need to put the capacitor on the DC side, but it would mean that I don't need to worry about burning the LEDs with my current setup.

 

Of this I am aware.

But as the original lamp is running at 12Vac, thats about 17Vpk, if the user connects LED's to this 'ac' supply via a typical current limiting resistor the LED's will MOST likely die.

Standard LED's are rated at approx 5V reverse voltage.

So I repeat: You dont normally drive leds with 'ac', else they could die.

Attached Thumbnails

 

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With the wallwart you don't need the capacitor, it's built in.

When you wire your socket put the positive wire to the center contact of the socket. In a car the contacts are positive and the main body of the socket is ground.

If you use a wallwart thats around 12VDC (a couple of volts either way won't hurt) You don't need any other resistors, they are built in to the bulb. When you use one of the bulbs in a car you don't add extra resistors.

Cary

 

I guess it really depends on the module and the LED's in question Eric. A 12V LED bulb replacement would sensibly (to me at least) be an array of at least one string of 3 LED's in series with a current limit resistor. At least as close to 12 volt drops on the diodes as possible to allow low power dissipation in the resistor. Tweak the number of LED's in series based on their forward drop. If there are 3 LED's in series, then that's a maximum reverse voltage of 15 volts, which is only going to allow a very small fraction of the 17VAC cycle to be over voltage and the current limit resistor will still limit the reverse current to some extent so destruction is not likely in that case as the RMS power dissipation is very low.

If it is a single LED and resistor for each element then yes destruction is just about assured.

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A point that you are overlooking is that LED's and diodes in series strings do no always share the reverse voltage equally, this is a common mistake by hobbyist project builders.

If you really wanted to be sure that you would not exceed the reverse rating of individual LED's in a string, its common practice to add a 1megR across each LED.

The same method , ie 1megR across each capacitor or diode when the components are not individually rated for the overall reverse voltage.

I would advise the OP to go buy a 12Vdc 'wall wart' psu and power the LED's via suitable series resistors.

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