Jcarere

Ok I bought a led scanner to replace my third brake light. It uses 16 blue 5mm leds. Im replacing the blue leds to 16 red 10mm ones. The problem is that the led board regulates the voltage to 5 volts. Then uses 271ohm surface mount resistors of some size to drop it down to around 1-1.2 volts. I talk to the company that made the board and they told me I could replace the Surface mount resistors with lower value ones for more brightness. I did the calculations and found I need 100ohm 1/8 watt resistors to achieve around 3.2-3.4 volts at 20ma. The question is where do I buy surface mount resitors and what size do i need. here are some pics. The resistors say 271 on them and I tested them with a multi meter and found them to be 270 ohm resistors which is what I expected. Any input would be appriciated.

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Jcarere

I found a few assortment pack of surface mount resistors, one of them are 1206 SMT size i think. http://cgi.ebay.com/Assorted-Surface...QQcmdZViewItem

And the other is size 0603 http://cgi.ebay.com/Surface-Mount-Re...QQcmdZViewItem

Can anyone tell which size those resistors are in my previous post judging by the pictures. As you can tell im kinda new to the whole elctronics thing and need some help from someone with experience with this kinda stuff. Another thought does anyone think this will overload the controller or regulator. JCarere

mauri

The value read 271 it the same of 270 ohms
If you have to use 100 ohms, the value that you have to find and 101

The size in the photo is 0603 or 0805
0603 = 62mW
0805= 100mW

To understand the value of the resistances SMT owe:

EXAMPLE :
270 writing = 27 + 0 = 27 Ohms
271 writing = 27 + 1 zero = 270 ohms
272 writing = 27 + 2 zeros = 2700 ohms = 2K7 ohm
etc.........

I hope that you have understood me

For my prjects I use often component SMT and I buy here them:
www.distrelec.it (I alive in Italy )

or www.distrelec.com and you choose the country!

Mauri

LegoGT

They look like 0805's to me.

I did a quick search for 0805, 100 ohm, 1/8 Watt resistors on DigiKey; here are a few part numbers I found:

P100ACT-ND
311-100CCT-ND
RHM100ACT-ND
311-100ACT-ND
P100CCT-ND

There were tons more listed but they all cost about $0.08 each. You also need to order at least 10 before they'll ship which brings your grand total to a whopping 80 cents!

Just go to http://www.digikey.com/ and put one of the part numbers into the search box. You get a break on s/h costs if you break $25 (I think), so tack on a few more things if you need them. Happy ordering!

TheLlama

You can also piggyback a 158ohm (160ohm may be easier to find) ontop of the existing ones. This may be easier than trying to remove the old ones and possibly tearing up the board contacts, etc..

RPM_VR4

First, just get the 0603. They will work on any size pads.

Now some technical details:

100-ohm is not 20mA, at least not at that voltage.

If you want 3.3V in to the LEDs (which is probably too much by the way) then you would need the resistor to drop 1.7V from the 5V rail.

1.7V / .02A = 85-ohm.

What you really need to do is know the voltage drop across your 10mm Red LEDs (probably around 2.1V) and their current rating (usually 20mA). Once you have that, you can calulate the resistor you need.

(5V - 2.1V) / .020A = 145-ohm

The LED voltage drop is constant. Changing the resistor value just changes the current. The more current, the brighter the LEDs will be, but if you give them too much current they will "burn out". Therefore, you shouldn't go more that 1 or 2 mA higher than the current rating specified in the parts datasheet.

As TheLlama said, you can also stack the resistors in parallel as long as the resistance you want is less than the 270-ohms that you already got. Here's the formula for that:

R = 1 / (1/145 - 1/270) = 313.2. So you could stack a 313.2-ohm 0603 on top of each existing resistor (if you need 145-ohm total).

Say you stacked a 330-ohm, you would have 1/(1/330+1/270) = 148.5-ohm.
(5V - 2.1V) / 148.5ohm = .0195A, ...so you would be giving it 19.5mA and it would be slightly less bright than the 20mA rating.

TheLlama

BTW, if you ever need to remember the:

New = 1 / (1/Total - 1/Current)

rule, then you can just derive it from the product-over-sum rule:

Total = (New * Current)/(New + Current)

Solve for n

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