LED Help

[nath98]

Hi there,
Im putting together some LEDS, but when I run them in series off the same power source they are dimmed, how can I have them at their full brightness off the one power source.. There is a way right? Is it a matter of how I wire them together?
Thanks,
Nathan

 

[I²R]

Depends on the LEDs and the power supply. You need to know the forward voltage drop of the LEDs; if you have the data sheet, it will tell you the Min, Max and Typical value. A normal gallium arsenide red LED will have a typical forward voltage drop of around 1.9 volts. If your power supply is 12.0 volts, you can run 6 of them in series. More than that, and your total of the voltage drops add up to more than the supply voltage, and so each LED will get less than its full turnon voltage. If you want to power more LEDs than your supply will allow in series, you need to arrange them in a series-parallel configuration. If you have a series which totals much less than the supply voltage, you'll need a current limiting resistor to keep the current below the max the LEDs can handle--typically around 20 mA or so for your average LED.

 

[nath98]

Im using a 12v AC/AC Adaptor. It says Output: 12V~850mA 10.2VA
I dont know much about this stuff, I am new to it all so you might have to explain it in laymans term for me
Any ideas on what I can do?
Thanks for your help,
Nathan

 

[I²R]

How many LEDs are you trying to run in series? And what color are they?

Your power supply's specs mean it outputs 12 volts AC at a maxiumum current of 850 milliamps (.85 amps) RMS. Total power output is 10.2 volt-amps. A volt-amp is a unit of apparent power, which is (as the unit name suggests) volts multiplied by amps (12 x .850 = 10.2). Real power is measure in watts and is apparent power times power factor.

 

[nath98]

I want to run 5 (5mm) Leds which are blue and red, will this cause any problems? If so Ill just use the blue ones (4) and I also have 2 leds which shine blue green and red at the same time (3mm). I have resistors that came with them, which are meant for 12v but appart from that I don't know much about them.
Thanks,
Nathan

 

[I²R]

Blue and red are going to have different voltage drops. Reds are typically around 1.8-2.0 V, while blues are anywhere from about 3 V to 5 V, depending on chemistry. If you're not sure, use a variable resistor of about 1 kOhms in series with ~ 6 VDC to power the LED. Start with the VR at max and slowly turn it down until the LED's stops getting noticeably brighter, then measure the voltage across the leads of the LED. For best accuracy, you should also measure the current through the LED, to ensure it doesn't exceed its maximum forward current--this can often be anywhere from 15 to 50 mA, with values around 20 mA being typical.

The LEDs intended for 12 V already have a current-limiting series resistor; these should be run in parallel rather than in series.

 

[Hero999]

No wonder they're dim, they're only using half of the wave and there isn't enough voltage.

The most blue LEDs you can have in series is four, giving a total voltage drop of 14V, less than the peak voltage of 15.6V, after the bridge rectifier.

Another thing is, calculating the resistor to use isn't as simple as Ohm's law.

For a start, the duty cycle isn't 100% as the LEDs only conduct when their peak voltage is exceeded and the current varies throughout the sinewave too.

Here's how I would do it:

First we need the duty cycle but to get that we need the angle at which the LEDs start conducting.

[latex] \theta = arcsin \frac{Vdrop}{Vpeak}[/latex]
Vdrop is the total voltage drop of the LEDs plus the rectifier loss, in this case we'll assume there are fou LEDs giving a total drop of 14V, plus the rectifier losses which gives 15.4V

Vpeak is the peak power supply voltage which is equal to 12*1.414 = 17V.

Putting the aforementioned into the equation gives:
[latex] \theta = arcsin \frac{15.4}{17}=64.9^o[/latex]

Now let's calculate the duty cycle.

[latex]Duty = \frac{90-\theta}{90} = \frac{90-64.9}{90} = 27.89%[/latex]

Here's where my maths leaves me a bit, we need to calculate the RMS current but the problem is it varies during the part of the cycle when the LED is conducting. Oh well I won't bother, since the peak current is a close enough approximation.

Now suppose the LED is being powered from 17V continiously; first we calculate the resistor value for that:

[latex]R = \frac{V-Vrdop}{Iforward}[/latex]
Where If is the Iforward is the LED forwad current; let's set this at 10mA. As the LED is only conducting for 27.89% of the time we can safely increase this to 1/27.89%.

0.01/0.2789 = 35.9mA.

[latex]R=\frac{17-15.4}{0.0359} = 45\Omega[/latex]