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Let's assume I have 3 Leds which have forward voltage of 3.3v and current rating is 300mA. All the leds are in series and I have a power supply of 12v. when I series a resistor to limit current it draws much power.

Calculations:

Voltage through leds =\$3.3v*3=9.9v\$

voltage drop through resistor=\$12v-9.9v=2.1v\$

Resistor value for \$300mA =2.1v/0.3A=7\Omega\$

Power loss through resistor \$= 7*0.3=2.1W\$

I need to reduce this power loss. Is there any way with MOSFET or BJT transistors? I read this http://www.instructables.com/id/Power-LED-s---simplest-light-with-constant-current/ . But i don't understand how to calculate values. Here is the circuit diagram in above site.Current limiter circuit

I don't need exactly this. I need a proper way to do that.

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  • \$\begingroup\$ This answer might be helpful, especially the last part about high power LEDs. If it isn't, let me ask you why you need to reduce the power loss. The above schematic will not reduce the power loss but distribute it over different parts (the mosfet mainly) \$\endgroup\$ – Arsenal Jan 5 '18 at 13:01
  • \$\begingroup\$ … and you can achieve the same distributing of losses by having e.g. four 7/4Ω resistors in series. The advantage of the above circuit is that it's actually a current source (or sink, depending on how you look at it) and less dependent on the actual supply voltage. \$\endgroup\$ – Marcus Müller Jan 5 '18 at 13:03
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    \$\begingroup\$ As far as I know the ONLY way to reduce power loss (compared to using a resistor or a current source like shown here) is to use a switched regulator. That's much more complex and you will need to use some chip for that. \$\endgroup\$ – Bimpelrekkie Jan 5 '18 at 13:06
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    \$\begingroup\$ Your expectations of efficiency and simplicity are unfortunately very naive and impossible to meet. I propose that the highest efficacy and energy efficiency can only be done when the supply voltage matches the load and for highest efficacy operate well below rated currents. It would be far more efficient to operate at 3V per LED than 3.3V then use more LEDs to achieve lumens required \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 '18 at 13:45
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    \$\begingroup\$ BTW . re Power loss through resistor =7∗0.3=2.1W WRONG that's voltage dropped across resistor 2.1V, Power low is 2.1 * .3 = 0.63W. i.e 82% efficiency. \$\endgroup\$ – Trevor_G Jan 5 '18 at 16:42
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The circuit you have indicated does not lower the power loss. It merely replaces your fixed resistor with a current limiter that ensures you have closer to the 300mA or whatever you need, under a wide range of LED forward voltages and supply voltages.

As such the current limiter is really just a smart resistor and will still dissipate the same sort of heat a simple limiting resister would.

To be low power you need a circuit that uses some sort of switch mode regulation to generate the required current in the LEDs at a high efficiency conversion factor.

UPDATE

By the way your math is wrong.

You stated..

Power loss through resistor \$= 7*0.3=2.1W\$

That is incorrect, that is the formula for voltage drop across resistor = \$2.1V\$

Power loss through resistor \$= 2.1 * 0.3 =0.63W\$

You are already running at 82% efficiency.

With a switch mode current regulator you might be able to boost that up a few percentile, but it may not be worth it. It is still wise to use a current limiter rather than relying on a resistor though.

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  • \$\begingroup\$ what is the basis of the regulator like this? can i make such regulator without IC? \$\endgroup\$ – Lakindu Akash Jan 5 '18 at 13:20
  • \$\begingroup\$ @ultimatex you would need some sort of driver IC, there are various drivers on-line. \$\endgroup\$ – Trevor_G Jan 5 '18 at 16:26
  • \$\begingroup\$ @ultimatex even then you may only get 85% efficiency, compared to your current 82%. \$\endgroup\$ – Trevor_G Jan 5 '18 at 16:31
  • \$\begingroup\$ @ultimate see my updated answer. \$\endgroup\$ – Trevor_G Jan 5 '18 at 16:41
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    \$\begingroup\$ I'm sorry. I understood. Thanks for showing me. And I lernt about buck converters and they are much efficient as you mentioned. \$\endgroup\$ – Lakindu Akash Jan 7 '18 at 6:31
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You could add one more LED and thus reducing the voltage drop across the resistor. Or choose another led, with a bigger forward voltage ex: 3.5V - 3.9V.

Because the power loss is directly proportional to the voltage drop across de resistor.

P = U x I

In your case

P = 2.1V x 0.3A

P = 0.63W

with a bigger forward voltage (Led with forward voltage of 3.9v) In series, 3 x 3.9 = 11.7V then the remaining voltage is 12 - 11.7 = 0.3 V across resistor.

Then:

P = 0.3V x 0.3A

P = 0.09W

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  • \$\begingroup\$ Good point in pointing out the mistake in the resistor power dissipation calculation. The suggestion of adding a fourth LED makes sense. However, it makes the current too sensitive to changes in the supply voltage. Any small change in the 12V supply will result in a significant change in LED current. Simply using a resistor that can handle 0.63W, or using multiples resistors in series or parallel to distribute the power sounds like a better solution to me. \$\endgroup\$ – joribama Jun 27 '19 at 2:25

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