Hi guys I am trying to make a LED lightbar with about 10 LED's and I have seen some of the LED culculators and they just show me a diagram like this....
+----|>|----|>|----|>|---/\/\/----+ R = 56 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 56 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 56 ohms
+----|>|-----------------/\/\/----+ R = 330 ohms
Notice this has 4 resistors and what I want is all the LED's in a line with only one resistor. Is this possible? because I might not have enuff room to put all those resistors on the lightbar.
Depends on the LEDs and supply power. For white or blue LEDs, you need around 3.4 volts (varies, even from the same batch...), so 10 X 3.4 volts =34 volts just to light them, but you need a resistor to limit the current, and the forward voltage on some LEDs might be higher, so 36 volts would be good enough.
The wizard breaks down your array, to fit your supply, guessing 12 volts, as 3 blue or white is the most you can get on 12 volts.
LEDs operate on DC, not AC so you need to rectify the 120VAC.
Rectified and filtered 120VAC produces 168VDC.
Ten 3.4V LEDs need 34V to light them.
So you need to throw away 134V at about 20mA which is 6700 ohms at 2.7W.
They probably have the resistors in series to dissipate the power. You could just use 1 resistor, but it would have to be one with a higher power rating, which are more expensive and harder to find. If you make room for the extra resistors, the power that the total resistance of all the resistors is divided up evenly, so the power rating of the resistors only has to be (total power)/(number of resistors)
To reduce the resistor power dissipation and increase efficiency you can add a capacitor in series with the AC line. A 0.68uF capacitor in series with the input to a bridge rectifier gives about 45VDC out with a 20mA load and 120VAC input. Thus the resistor would only have to dissipate the power for the voltage drop between 45VDC and the LED voltage (34V in audioguru's example) instead of from 168VDC.
The capacitor should be a film or ceramic type rated at 400V or higher.
Better to use a 1N4007 forward biased to drive the LED's.
A capactors inrush current may destroy the LED's
It will reduce the power dissipation in the R's to 0.7 x RMS value.
Use two resistors in series to dissipate the heat generated.
Look under 230 Volts LED lamps in this forum. details are discussed there.
My approach has a filter capacitor at the bridge output, which I neglected to mention, to give DC to the LEDs for maximum output from the LEDs without flicker. The filter cap would absorb any inrush current.
It will reduce the power dissipation in the R's to 0.7 x RMS value.
If you have a half-wave rectifier I believe it reduces the power to 0.5 X RMS value. Of course the average current and brightness of the LED are reduced by the same factor. Also, a half-wave source causes 50-60Hz flicker which is annoying to some people.
One of the reasons for the rectifier is not so much to deliver DC vs. AC, but to protect the LEDs from a high reverse voltage. Your average red LED has a PIV rating of only around 5v, so you have to be careful. And with a series string of them all reverse-biased, one may have a higher reverse resistance than the others and will take the brunt of the reverse voltage.
You could use a current mirror and connect all your LEDs in series, so it'd maintain consistent brightness regardless of how many you hook up, to the limits of your driving voltage.
Then you could easily make them all flash in unison or vary the brightness based on what you do with the controlling driver.