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

Calculations:

Voltage through LEDs =\$3.3v*3=9.9 V\$

Voltage drop through resistor=\$12v-9.9v=2.1 V\$

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

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

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, 2018 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, 2018 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, 2018 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 EE75
    Jan 5, 2018 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, 2018 at 16:42

3 Answers 3

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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, 2018 at 13:20
  • \$\begingroup\$ @ultimatex you would need some sort of driver IC, there are various drivers on-line. \$\endgroup\$
    – Trevor_G
    Jan 5, 2018 at 16:26
  • \$\begingroup\$ @ultimatex even then you may only get 85% efficiency, compared to your current 82%. \$\endgroup\$
    – Trevor_G
    Jan 5, 2018 at 16:31
  • \$\begingroup\$ @ultimate see my updated answer. \$\endgroup\$
    – Trevor_G
    Jan 5, 2018 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, 2018 at 6:31
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From what I understand of this circuit is that when the voltage of R3 exceeds the voltage that is required to turn on Q1, the gate of Q2 will be pulled to GND and the Q2 will be turned off.

Hence, there will never be a voltage higher than that across R3. Let's assume the voltage to turn on Q1 is approximately 0.7 V.

Now, if we want to limit the current through the resistor to e.g. 0.3 A, you can calculate what R3 should be. R = U / I = 0.7/0.3 = 2.33 ohm.

The power rating should be such that it can dissipate P = I^2*R = (0.3 A)^2 * 2.33 ohm = 0.21 watt.

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  • \$\begingroup\$ Good answer! Some uggestions: R3 and its power dissipation seems correct for the current limiting feature of the schematic. Before the overall dissipation, the MOSFET will control the current, distributing the voltage drop - it would be worth mentioning it. And possibly using a cite_’”’ feature, to highlight the Original poster (OP) question. About the OP’s first calculation - is about dissipation of a simple resistor “R”, but it could be good to mention that has an error calculating R_power and the overall value is the same using a simple Resistor or the shown current limit circuit. \$\endgroup\$
    – EJE
    May 21, 2022 at 13:57
  • \$\begingroup\$ @EJE Thanks for your comment, I think it's a valuable addition. There will be a voltage drop over the Mosfet indeed and it will subsequently get warm and needs to be properly rated as well (and cooled). Feel free to improve the answer. \$\endgroup\$
    – Anne van Rossum
    May 23, 2022 at 14:27
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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, 2019 at 2:25

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