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URGENT help for an idiot

klain421

Dec 29, 2012
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Hello,

I am completely new to electronics and I'm trying to re-create a simple iron man arc reactor for new years eve.

I have bought 12 blue super bright LED's which annoyingly I found out need 5v and are 30mA

http://www.maplin.co.uk/3mm-high-bright-blue-leds-35734

I plan on powering them with a PP3 9v battery and wiring them in series and then using a transistor to make sure I don't blow the LED's.

My question is which transistor do I need to use?
 

davenn

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Sep 5, 2009
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hi there klain :)
welcome to the forums

you need a resistor to do current limiting not a transistor

have a look here for info on powering LED's

cheers
Dave
 

klain421

Dec 29, 2012
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hi there klain :)
welcome to the forums

you need a resistor to do current limiting not a transistor

have a look here for info on powering LED's

cheers
Dave

Hello,

Thanks for the quick reply :)

I had a look through that earlier and couldn't understand a word of it ha ha.

Klain
 

klain421

Dec 29, 2012
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Dec 29, 2012
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I think I have figure it out now...

R = (V1 - V2) / I

where:
V1 = power supply voltage
V2 = LED voltage
I = LED current

So for me I need x12, 133ohm resistors and then need to solder one of those to the positive anode of each led and wire in series, does that sound right?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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That will work. However you can probably put 2 LEDs in series and use a smaller resistor (and fewer of them)

In that case double V2 (since it is now for 2 LEDs) and you'll get a resistor you need to place in series with *two* LEDs.

This will mean you only need 6 resistors and also that your circuit will consume half the power.

Normally you'd be looking at VF (V2) being around 3.4V, not 5V, and typically you run these LEDs at 20mA (I can't see specs that say otherwise)
 

klain421

Dec 29, 2012
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Dec 29, 2012
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That will work. However you can probably put 2 LEDs in series and use a smaller resistor (and fewer of them)

In that case double V2 (since it is now for 2 LEDs) and you'll get a resistor you need to place in series with *two* LEDs.

This will mean you only need 6 resistors and also that your circuit will consume half the power.

Normally you'd be looking at VF (V2) being around 3.4V, not 5V, and typically you run these LEDs at 20mA (I can't see specs that say otherwise)

I used a quick calc web app to double check that I calculated it right and it gave me this image:

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-snc7/574761_10151306667412200_1903708876_n.jpg

On my original post I put a link to the store I bought the LED's from and it says 5v, 30mA I didn't think it would require that high of a voltage either to be honest, but I put it down to inexperience.
 

CDRIVE

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May 8, 2012
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That 5V spec @ Maplin is "Reverse Voltage". They spec the forward voltage as Blue Diffused= 4.5V, Blue Clear = 4V. These voltages seem high to me though. As Steve said 3.4V seems more realistic.

Chris
 
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