user30282: I have an Arduino Uno controlling an array of 24 LEDs (datasheet).  I understand these LEDs are ultraviolet and have a forward current of 0.5A.  My power source (12V 2A) is wired to a switch which is then wired to the LED array and Arduino.  I should probably use a 12V 4A power source but was worried if too much current would go to the Arduino.  As per my attached diagram, the Arduino and LED array are in parallel.  I would like to get more current through my LEDs to increase their brightness.  What would be the best way to do this?  
Updated: I tried adding a 100 Ohm resistor (where R11 is located) but the Arduino wasn't drawing enough power to turn on my LCD.  And then I tried a 1 Ohm resistor instead but the resistor burned out.  Unfortunately, I threw out the resistor and don't have a picture of it.  I believe the correct value should be ~47 Ohm so I have added that resistance as R11.  However, I haven't tested it yet as I want to make sure it's correct.  (12V - 6V) = 130mA * R >>> R = 46.15 Ohm.  I used a voltage drop of 6V so that at least 5V is still going into the Arduino (12V - 6V = 6V which is greater than 5V).  I measured the current through a multimeter to be 130mA.


user20245: The current a bipolar transistor configured as a common emitter amplifier will allow from its collector to its emitter is equal to its beta time the base to emitter current. 
A beta of 100 is common for many transistors. Assuming this information is close enough to your transistor, we can choose a base series resistance suitable for the collector to emitter current desired.
For example if you were using a 5 volt microprocessor and the voltage drop across the base to emitter was the usual 0.7 volts for silicon semiconductors and you had a base series resistor of 10Kohms:
V = I * R
V = 5 - 0.7 = 4.3 volts
R = 10000 ohms
I = 4.3/10000 = 0.43mA
I(ce) = Beta * 0.43mA
I(ce) = 100 * 0.43mA = 43mA

You could pass up to 43mA from the collector to the emitter.  If that is not enough,you can lowered the base resistor to 1Kohms:
R = 1000 ohms
I = 4.3/1000 = 4.3mA
I(ce) = Beta * 4.3mA
I(ce) = 100 * 4.3mA = 430mA

And increase the current to 430mA.  You still need to verify that the processor can supply the current to control the transistor (4.3mA as calculated above) as well as pick a transistor of sufficient capacity to handle the load.
If a beta of about 100 is not enough, consider using a Darlington transistor arrangement. This type of "2 transistor" device has a gain of the product of each individual transistor's beta.  So, it is common for such a device to have a beta of 10000!  Using this device, you can use a much larger base resistor and still be able to switch more current.

30282: <p>I have an Arduino Uno controlling an array of 24 LEDs (<a href="http://www.mouser.com/ds/2/427/vlmu3500xxx120-968628.pdf" rel="nofollow noreferrer">datasheet</a>).  I understand these LEDs are ultraviolet and have a forward current of 0.5A.  My power source (12V 2A) is wired to a switch which is then wired to the LED array and Arduino.  I should probably use a 12V 4A power source but was worried if too much current would go to the Arduino.  As per my attached diagram, the Arduino and LED array are in parallel.  I would like to get more current through my LEDs to increase their brightness.  What would be the best way to do this?  </p>

<p>Updated: I tried adding a 100 Ohm resistor (where R11 is located) but the Arduino wasn't drawing enough power to turn on my LCD.  And then I tried a 1 Ohm resistor instead but the resistor burned out.  Unfortunately, I threw out the resistor and don't have a picture of it.  I believe the correct value should be ~47 Ohm so I have added that resistance as R11.  However, I haven't tested it yet as I want to make sure it's correct.  (12V - 6V) = 130mA * R >>> R = 46.15 Ohm.  I used a voltage drop of 6V so that at least 5V is still going into the Arduino (12V - 6V = 6V which is greater than 5V).  I measured the current through a multimeter to be 130mA.</p>

<p><a href="https://i.stack.imgur.com/Rkc4V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rkc4V.png" alt="Updated schematic with resistor values"></a></p>
 20245: <p>The current a bipolar transistor configured as a <a href="https://en.wikipedia.org/wiki/Common_emitter" rel="nofollow noreferrer">common emitter amplifier</a> will allow from its collector to its emitter is equal to its <a href="https://en.wikipedia.org/wiki/Bipolar_junction_transistor#Transistor_parameters:_alpha_.28.CE.B1.29_and_beta_.28.CE.B2.29" rel="nofollow noreferrer">beta</a> time the base to emitter current. </p>

<p>A beta of 100 is common for many transistors. Assuming this information is close enough to your transistor, we can choose a base series resistance suitable for the collector to emitter current desired.</p>

<p>For example if you were using a 5 volt microprocessor and the voltage drop across the base to emitter was the usual 0.7 volts for silicon semiconductors and you had a base series resistor of 10Kohms:</p>

<pre><code>V = I * R
V = 5 - 0.7 = 4.3 volts
R = 10000 ohms
I = 4.3/10000 = 0.43mA
I(ce) = Beta * 0.43mA
I(ce) = 100 * 0.43mA = 43mA
</code></pre>

<p>You could pass up to 43mA from the collector to the emitter.  If that is not enough,you can lowered the base resistor to 1Kohms:</p>

<pre><code>R = 1000 ohms
I = 4.3/1000 = 4.3mA
I(ce) = Beta * 4.3mA
I(ce) = 100 * 4.3mA = 430mA
</code></pre>

<p>And increase the current to 430mA.  You still need to verify that the processor can supply the current to control the transistor (4.3mA as calculated above) as well as pick a transistor of sufficient capacity to handle the load.</p>

<p>If a beta of about 100 is not enough, consider using a <a href="https://en.wikipedia.org/wiki/Darlington_transistor" rel="nofollow noreferrer">Darlington transistor</a> arrangement. This type of "2 transistor" device has a gain of the product of each individual transistor's beta.  So, it is common for such a device to have a beta of 10000!  Using this device, you can use a much larger base resistor and still be able to switch more current.</p>
