This might sound like a stupid question, but I want a confirmation. I watched a video on Youtube about using LEDs. Those LEDs required a voltage around 2 volts at 20 mA.

In order to power one LED using a 5 volts power supply, the author used a resistor in series. He calculated he needed around 150 ohms (using U=RI, 5-2=3 volts, 3V/20mA=150ohms).

What I find disturbing is that the resistor, in order to control the voltage must be consuming energy too. P=UI, so 3x20mA=60mW, on top of the LED 2x20mA=40mW. In other words, adding +150% energy consumption to the actual need for lighting up the LED.

Am I missing something or is it typical to spend extra energy just to be able to use electronic components that require a lower voltage? And second question, is there a way to avoid doing it for this type of circuit (5V source, 2V LED)?


  1. I use the term of voltage instead of current because this is how Ben Eater in his video presented it
  2. I am mostly interested in the general issue of using extra components in order to reduce consumption locally while introducing a waste when taking the circuit as a whole. This is not just about LEDs, which serve here as an example so suggesting this is a duplicate of "How can I efficiently drive an LED?" is missing the main point (although I agree the second part of my question is related).
  • \$\begingroup\$ "Am I missing something or is it typical to spend extra energy just to be able to use electronic components that require a lower voltage?" The alternative is converting the voltage using something like an LDO, a buck converter, or a PWM module. \$\endgroup\$ – Mast Aug 14 '19 at 10:25
  • \$\begingroup\$ @Mast An LDO wouldn't make any difference though. \$\endgroup\$ – pipe Aug 14 '19 at 19:54
  • \$\begingroup\$ I would suggest 'voltage' be replaced by 'current' in the title of the question, as that is what the resistor is there to control. \$\endgroup\$ – Glen Yates Aug 14 '19 at 21:11
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    \$\begingroup\$ @GlenYates Then others with the same (common) misconception won't find the question. \$\endgroup\$ – pipe Aug 15 '19 at 1:20
  • \$\begingroup\$ @pipe And the other alternatives aren't exactly free either (just a lot more efficient). Next time I'll phrase it better. \$\endgroup\$ – Mast Aug 15 '19 at 5:32

Yes, that resistor wastes power.

If the author is using an LED for an indicator light, then they're wasting a lot more power by their choice of LED. An LED that needs 20mA to show up in a brightly-lit room is typical of 1970's technology. If you shop for higher-brightness LEDs you'll blast your eyeballs out at 20mA, and you'll find yourself stopping the thing down to 1mA or so. One such LED, with a matching resistor, would use 3.3mW at 3.3V, where a 20mA, 1.5V LED alone (never mind the resistor) would use 30mW.

The ultimate way to reduce the circuit power consumption would be to use the most efficient LEDs that you could find, and power them with a switching converter. A decent switching converter will have somewhere between 80% and 95% efficiency, so you'll use between 25 and 5% more power than just the LED. But you'd have to use one per LED (or LED string), and it's hard to justify a super-efficient switching converter for each indicator light.

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  • \$\begingroup\$ Best answer so far. What about the constant current circuit refered to by Umar? \$\endgroup\$ – Exocytosis Aug 14 '19 at 5:01
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    \$\begingroup\$ @Exocytosis: I didn't want a long complicated answer, so I left out the bit where the switched circuit would need to have a constant-current output (or at least a reasonably high impedance output). Such a constant-current source, to be efficient, would need to be switching. \$\endgroup\$ – TimWescott Aug 14 '19 at 15:41
  • \$\begingroup\$ Just out of curiosity, will those sources produce a constant flat current or an oscillating one? I am asking this because I saw oscillators in schematics and Wikipedia page description of the way charge is pulsed using capacitors let me wonder how this could end up really constant (unless a very low frequency pass filter filters the output I suppose). \$\endgroup\$ – Exocytosis Aug 15 '19 at 3:13
  • \$\begingroup\$ You can make the current ripple arbitrarily small as long as you keep adding components. Getting the ripple to less than 1% is easy-peasy. \$\endgroup\$ – TimWescott Aug 15 '19 at 4:04

You've got the right idea. Partly.

An LED used with a series resistor does waste the energy dissipated in the resistor. Depending on the voltage from the power supply, you can easily waste more energy in the resistor than you use for the LED.

So far, you are correct.

What I want to correct is the idea that the resistor is there to lower the voltage.

The resistor is there to limit the current.

LEDs are current driven devices. The forward voltage varies with current and temperature.

To get a stable brightness out of an LED, you regulate the current.

You will have noticed that the calculation for the series resistor uses the desired LED current. You take the difference of the supply voltage and the (approximate) forward voltage of the LED, and divide that by the current to find the value for the series resistor.

If you tried to regulate an LED only by regulating the voltage, then you would destroy your LED very quickly. Just below the forward voltage, the LED doesn't light up at all. Just above the forward voltage the LED becomes the next best thing to a short circuit. There is a tiny range in between where it lights up and passes only a little bit of current.

That little range is impossible to hit with just a voltage regulator - it moves with temperature and current - current makes the LED get warmer, and warmer makes the LED conduct more. You would be varying the voltage up and down wildly with some kind of feedback circuit measuring the current.

Or, just regulate the current to begin with. Provide no more current than needed to light your LED at the desired brightness, and let the voltage do as it pleases - the voltage is of no interest.

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    \$\begingroup\$ Your answer is the closest to correct. \$\endgroup\$ – GSLI Aug 13 '19 at 19:22
  • \$\begingroup\$ @GSLI: I posted my answer like nearly two hours after the others. It's still climbing. And if it stays lower voted, well, no biggie. \$\endgroup\$ – JRE Aug 13 '19 at 19:23
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    \$\begingroup\$ @Exocytosis But then again, the Stack Exchange network sites only allow a single question per, well, question... \$\endgroup\$ – Mr47 Aug 14 '19 at 13:27
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    \$\begingroup\$ @Exocytosis: This question and its answers give far more detail on efficient LED drivers. I really only posted my answer to address your assumption that you needed to regulate the voltage. That seems to have resonated with a lot of folks, although it doesn't strictly address your question. \$\endgroup\$ – JRE Aug 14 '19 at 13:52
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    \$\begingroup\$ Of course, varying the voltage with some kind of feedback circuit measuring the current is exactly what a constant-current power supply does. That kind of feedback loop is ubiquitous in the modern world with switching power supplies everywhere; we're just changing the kind of measurement that we "feed back". It's very efficient and not at all impractical. \$\endgroup\$ – hobbs Aug 14 '19 at 17:30

No, you are not missing anything. The energy consumed by the resistor is wasted but, if you were contemplating a circuit that used tens or hundreds of LEDs you might consider a buck regulator to step the LED circuit supply voltage down to maybe 3 volts and make a significant net power saving per LED drive.

You’ll still need a 50 ohm resistor but it will only be dropping around 1 volt and dissipating only 20 mW.

The good news is that many modern LEDs need only a couple of mA to obtain sufficient brightness for “standard” applications.

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One point I would like to mention is the circuits which are meant to be drive LEDs. The constant current circuitry can come in handy to save power when multiple LEDs are needed to be driven.

In general it is desired to drive the LEDs with a voltage just close to the forward vorlage drop. The voltage is only needed to bias the LED. The brightness should be controlled by the current through it. A BJT and opamp can act as a simple constant current circuit. The issue is critical when driving LEDs in series or high power LEDs.

Depending on the number of LEDs and the cost or power criteria decision should be made. Leave it as is with the resistor combination. If the power saving is a must, rethink about the whole strategy of supply voltage levels and LED options. Other answers have put light already on low power options

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  • \$\begingroup\$ That's a really valid point. Past a certain number of resistors it becomes more cost effective in terms of PCB space, component cost, and circuit complexity, to use a constant current mechanism, which only requires perhaps 2 BJTs and 2 resistors. \$\endgroup\$ – GSLI Aug 13 '19 at 19:38
  • \$\begingroup\$ @Umar: why the partial answer? I got interested with your point about using a constant current circuit whatever that is (and a short intro would have been desirable here), but you did not answer the main issue. This looks like something one would write as a comment, not an answer. Please complete your answer. \$\endgroup\$ – Exocytosis Aug 14 '19 at 7:57
  • \$\begingroup\$ But aren't those circuits just glorified resistors (still waste heat)? \$\endgroup\$ – Peter Mortensen Aug 14 '19 at 19:46

Here's what you're missing. Your power supply has a given potential: 5 V. That isn't going away. When you put 5 V across a component that can't handle it, you must do something with that extra potential or you will unbalance the voltage/current ratio in the circuit-- and the component will be damaged (in this case the LED).

The LED is not in and of itself strong enough to resist the potential applied across it - that is specifically why the datasheet is telling you it can only withstand a potential across it of 2 VDC. That's the first thing. The second thing is how much current you blow through it - in this case, the datasheet apparently says 20 mA. Understand that just because a datasheet provides a maximum current value of 20 mA, it does not mean you should actually operate the device at that level (or should you). There will usually be no appreciable difference between 8-10 mA and 20 mA, depending on color.

The resistor you use is doing a couple of things - it is taking up the slack for the voltage the LED can't handle, and it is limiting the current through the LED to what you designate you want to run through it.

The heat being dissipated by the resistor is not excess or left-over; it's actually only the heat being generated by the amount of current you are limiting the LED to. That's an important point. The resistor cannot dissipate energy that isn't flowing through it. And it certainly is not carrying the full current of your power supply.

So, Ohm's law says:

R = E / I
R = (5 V - 2 V) / 0.008 A
R = 3 V / 0.008 A
R = 375 ohms

In this case, we'll go with 360 ohms (it's the closest available).

I = E / R
I = (5 V - 2 V) / 360 Ω
I = 3 V / 360 Ω
I = 0.0083 A (or 8.3 mA).

Now that you know how much current is flowing, and what resistor to use, you can calculate how fat a resistor you need, based on what it's dissipating:

Watt's law is handy here:

P = I * E
P = 0.0083 A * 3 V
P = 0.0249 watts
P = 24.9 mW

Now that you know how much energy is dissipated, you can size the resistor. An 8th watt (1/8th of a watt) resistor will dissipate 125mW. For safety, you want a resistor that can handle twice what your power requirements are. Therefore, 2 * 24.9 mW = 49.8 mW. That small amount is far less than 125mW, so you can use an 8th-watt resistor.

I hope that helps.

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    \$\begingroup\$ It's still wasted energy because it's not being turned into useful light. \$\endgroup\$ – user253751 Aug 13 '19 at 23:24
  • \$\begingroup\$ @GSLI: Not the point. Check TimWescott’s answer, it covers my actual needs. \$\endgroup\$ – Exocytosis Aug 14 '19 at 5:00
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    \$\begingroup\$ @Exocytosis - I read his answer, which gave you a lot of information, some not accurate, but did not explain to you exactly how much energy was being dissipated, why, and where. This way you learn something so you don't have to ask again. You asked: "Will using a resistor in series with a LED to control its voltage increase the total energy expenditure?" The flat answer is 'No'. If anything the resistor reduces the energy expenditure- that's the point of the resistor. \$\endgroup\$ – GSLI Aug 14 '19 at 18:22
  • \$\begingroup\$ Are you sure it is called Watt's law (in this context)? \$\endgroup\$ – Peter Mortensen Aug 14 '19 at 21:25
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    \$\begingroup\$ Re "An 8th watt (1/8th of a watt) resistor will dissipate 250 mW": Don't you mean 125 mW? \$\endgroup\$ – Peter Mortensen Aug 14 '19 at 21:30

LEDs are designed to be driven at a particular current.

The ideal way to drive them is with a current source, such as a switching power supply operating in constant-current mode. This will be the most efficient usually, but on very small loads like this, its overhead may be worse than the resistor.

The resistor is basically a super cheap way to convert a constant-voltage source into a current source (I didn't say constant). Like any resistive regulator (think LM7805), it wastes a lot of energy as heat.

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