I have a 6V power source. I'm using a 2N3904 (NPN) transistor as a switch. My intent is to turn on a high power LED driver with a microcontroller (MC) output (~25ma) using the transistor.

I was hoping to use the MC output to control the 3904 base which would in turn allow 3v to turn on the LED driver. I'm having some troubles because the 3904 seems to have a max Ic of 200ma which seems to low for the driver. Also, how do I make sure that 3V goes to the driver ? I understand there will be some voltage drop between Collector and Emitter. How is this voltage drop determined?

I hope I included enough detail.

Thank you in advance.
Bud

 

What is a LED driver? Can you post some part numbers.

Usually, one would connect the emitter of the 2N3904 to ground.
Connect the 6V power to a current limiting resistor (220Ω in your case), to the LED, to the collector of the 2N3904, all in series.

Drive the base of the 2N3904 from the MCU through a 1kΩ resistor.

I suppose it would be easier for me to post a circuit diagram if this would help.

 

What is a LED driver? Can you post some part numbers.

Usually, one would connect the emitter of the 2N3904 to ground.
Connect the 6V power to a current limiting resistor (220Ω in your case), to the LED, to the collector of the 2N3904, all in series.

Drive the base of the 2N3904 from the MCU through a 1kΩ resistor.

I suppose it would be easier for me to post a circuit diagram if this would help.

Thanks for the reply..

Im using a Micropuck
http://www.ledsupply.com/02009-sho.php?gclid=CMb3nv2XhrUCFUQw4AodkAcAsg

A diagram would help a lot if you can.

 

The maximum voltage input to the MicroPuck is 3V. Hence you will have to reduce the supply voltage from 6V to 3V, wasting power.

You would be better to eliminate the MicroPuck and drive the LED directly from 6V through a current limiting resistor (10Ω 1W), assuming the LED can operate from the 6V supply.

Or you can try putting two LEDs in series.

Use a MOS FET logic level driver that can handle 500mA.

 

I'm using a 2N3904 (NPN) transistor as a switch.
I understand there will be some voltage drop between Collector and Emitter. How is this voltage drop determined?

When using a BJT as a switch ,we put the BJT in saturation region/mode,at saturation their is very low voltage drop between collector and emitter,you dont have to calculate it, as a typical value will be given in datasheet as \(V_{CEsat}\) ,for 2N3904 its 200mV i.e.. 0.2V.

What you are trying to do i.e.. turning a 3V load using a single transistor is not good ,in the way you are connecting the transistor it acts as a current source,it will regulate current not voltage,you can configure a voltage regulator to regulate 3V from 6V and switch that voltage regulator using the transistor,for example a linear voltage regulator LM317 ,remember linear regulator waste energy when the difference between input and output voltage is high.Check out LM317's datasheet

Good Luck