Use 2v Signal to Control 12v LED strip

[ClassyTurkey]

I'm sorry if this title is confusing, but let me try to explain. If this is in the wrong section please let me know.

I am currently working on a DIY project where I currently have a board that runs on 3v that is a receiver for motorcycle signals. So whenever my blinkers activate, the device will turn on the 2v LEDs for that signal. Ex Left, Right, or Brakes.

I want to modify this and attach it to 10 LED strips that are 12v with 60ma per strip. The LEDs have their own separate power supply, but I am needing to control/complete the circuit with the 2v LED signal. I am willing to actually remove the LEDs on the board with transistor or MOSFET, but I am confused on what is the correct way to do this. Also if I am even going about it the correct way. Any help would be amazing.

 

[Bluejets]

Mosfet would be good if you can find one with a low enough turn on...don't know for sure what the requirement is for the L series. I've often used them for logic circuits (5V)
Could make up a voltage doubler(or trippler) and drive the gate with that I suppose.
Other than that, a transistor should do ok if you are willing to live with a small amount of voltage drop.
If you need a circuit, Google along those lines.

 

[ClassyTurkey]

Mosfet would be good if you can find one with a low enough turn on...don't know for sure what the requirement is for the L series. I've often used them for logic circuits (5V)
Could make up a voltage doubler(or trippler) and drive the gate with that I suppose.
Other than that, a transistor should do ok if you are willing to live with a small amount of voltage drop.
If you need a circuit, Google along those lines.

I researched into the MOSFET, but am coming up short on knowing which one would work.

 

[Bluejets]

Look at requirements for logic level mosfets.

 

[12vdc]

http://www.jameco.com/1/3/relay-3-volt

 

[Harald Kapp]

Nah, don't use an electromechanical relay. It will work for some time but will probably wear out rather quickly, especially when used on a motorbike with the accompanying shake, rattle and roll
An electronic solution using a transistor is way better. At only 2V of signal level, a MOSFEt that turns reliably on is not that easy to find. You may consider using a bipolar transistor which requires only ~0.7V (at the cost of some base current). We have a ressource on how to use this kind of transistor to switch a load here.

 

[ClassyTurkey]

Nah, don't use an electromechanical relay. It will work for some time but will probably wear out rather quickly, especially when used on a motorbike with the accompanying shake, rattle and roll
An electronic solution using a transistor is way better. At only 2V of signal level, a MOSFEt that turns reliably on is not that easy to find. You may consider using a bipolar transistor which requires only ~0.7V (at the cost of some base current). We have a ressource on how to use this kind of transistor to switch a load here.

Thats what I was worried about for using a relay. Also that the relays would be slower than a transistor or MOSFET.

So I started looking into Bipolar Transistors and came across http://www.farnell.com/datasheets/1700327.pdf
Would that work for what I am trying to do with my current Voltage/Current going through it? Or am I looking in the wrong place?

 

[Harald Kapp]

This is a PNP transistor. You'll probably want to use an NPN transistor, depending on the polarity of your 2V drive signal. You'll also need a current limiting resistor in series with teh base, see the ressource I linked.
The relevant parameters of the transistor are:
Vceo >= 20V (as you are going to drive 12V LEDs and you'll want some headroom)
Ic >= 100mA (you want to drive 60mA, also some headroom is desirable)
For Ic=60mA and assuming a current gain of >=100, your base current is 60mA/100=600µA. The base-emitter voltage drop is approx. 0.7V. That leaves 2V-0.7V=1.3V across the base series resistor. 1.3V/600µA translates into Rbase=2.2kΩ.