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I'm trying to control a 7W 560MA 12-14V COB LED Strip using Arduino. When I connect the LED directly to a power supply at 14V (using 1 ohm limiting resistor) I have 0.56A / 12.3V on my LED. But if I connect it through TIP120 transistor to control it by Arduino (using 14V power supply) I have only 0.38A / 11.7V on my LED. So I have a significant drop of LED power. Did I choose the wrong transistor? If so, please help me to choose the right one.

Fritzing diagram

JYelton
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asked Feb 28, 2014 at 18:21
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  • \$\begingroup\$ Did you connect the base of the transistor directly to Arduino without a resistor? If so it's not a good idea. \$\endgroup\$ Commented Feb 28, 2014 at 19:05
  • \$\begingroup\$ I'm using 4k resistor. \$\endgroup\$ Commented Mar 1, 2014 at 5:46

2 Answers 2

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A Darlington has great current amplification but doesn't switch on with a zero volt drop: -

enter image description here

You can put a decent drive voltage on the base that will make the input transistor turn on and only drop 0.2 (ish) volts. This is recognized as the saturation limit for a BJT generally. Some are lower of course.

So, the input transistor is turned on and the voltage on its emitter is 0.2V below the voltage on the common collector point. The final transistor needs about 0.7V across its base and emitter but it can only get this voltage via the 1st transistor hence, the best volt drop across the final stage is 0.7V + 0.2V - you lose about a volt of drive.

This is the basic problem and maybe consider using a MOSFET instead.

answered Feb 28, 2014 at 18:49
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  • \$\begingroup\$ Can I use 2N2222 transistor? \$\endgroup\$ Commented Mar 1, 2014 at 5:49
  • \$\begingroup\$ Is it good idea just to increase the power supply voltage till 16V? \$\endgroup\$ Commented Mar 1, 2014 at 5:58
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The difference between 12.3V and 11.7V is 0.6V, which is exactly the voltage drop of many semiconductors such as diodes and transistors. The TIP120 transistor introduces a voltage drop of 0.6V, so you need to factor this into your calculation for a current-limiting resistor. You are still using a current-limiting resistor, aren't you?

answered Feb 28, 2014 at 18:48
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  • \$\begingroup\$ I'm using 1ohm current-limiting resistor. \$\endgroup\$ Commented Mar 1, 2014 at 5:48

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