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Help needed with flashing brake light microcontroller Answered

Hi All

New here and hope you can help me out.

I am a biker in south africa and would like to put together an aditional brake light that flashed 4 times then go solid.

Microcontroller will most likely be the easiest and have design one as follows. (the coding has been done)

LED's to be used is 20mA at 2.2V

My concern is that for the amount of LED"S (100) i want to use 1amp is not enough and therefore i am using a LM7810 to drop the voltage from 12V to 5V with a cap to increase the 1A output to 5A

I wish to wire the LED's in parralel si if one pop's then it wont affect the rest of the LED's.

Must i therefore still wire a resistor in front of every LED? (180ohm?) or can i wire in this LED before the LED matrix to give it a 2.2-2.3V throughout the LED matrix?

Here is my design : http://i247.photobucket.com/albums/gg156/HannesNel/design_zps52f754cf.png

2nd EDIT :
How about this? reworked the design a bit?

Then i drop the 12v down to 2.5V needed for the LED's and no resistors needed?

I also opped for the LM350 due to the higher amps it can handle

http://i247.photobucket.com/albums/gg156/HannesNel/Untitled_zpsc9900901.png

3rd update:
Now using super bright LED's with a rating of 2V and therfore will use row of 6 in series to give 12V...still need resistors when connecting the 6 banks of 6 LED each in parralel?

http://i247.photobucket.com/albums/gg156/HannesNel/Untitled_zps684a2209.png

Tags:design

Discussions

Put the leds in chains of 5, and you'll only need a 47 ohm resistor, though you need 20 chains, in parallel.

You might find it doesn't get quite as bright as you expect, but you should be OK. Use a MOSFET for the drive transistor, and you should be OK.

a mosfet instead of the irf510? Or in addition?

see my second edit the OP...my concern with the chains is if one LED in the chain fails then the entire chain is dead

And you lose 5 leds out of 100 ? So what ? An LED can fail short circuit and open. You only need 400mA to run the LEDs this way.

A linear reg, like the 78XX series won't BOOST the current in anyway, if you only have 1A going in, you'll only have 1A going out