Here's what the person was saying:

The combo led+resistor will use LESS energy compared to a led only, for the same applied voltage. (Because energy is proportional to power which is current times voltage).

Imagine a very simple circuit that just had a 3V battery and an LED that could handle up to 3V. Only two items, a battery and an LED. Now add a resistor in order to drop the voltage to the LED so that it gets 1.9V instead of 3.

Q1: All else being ignored (early burn out of the LED, etc...) will the circuits consume the same amount of power?

Q2: in order to reduce the energy of this circuit, you'd have to do something like add a buck converter or PWM, correct?

Someone is telling me that the resistor alone will decrease the energy the circuit uses. My understanding is that resistors only drop the current and dissipates current as heat and are therefore not the most efficient choice.

Q3: what exactly does a resistor do with the energy it is removing from the circuit?


10 Answers 10


Q1: All else being ignored (early burn out of the LED, etc...) will the circuits consume the same amount of power?

In both circuits, the voltage drop is the same (3V), but the currents will be different. Thus, the circuits will dissipate different amounts of power.

Below the diodes are dissipating different amounts of power. enter image description here

Below the batteries are sourcing different amounts of power. enter image description here

Q2: in order to reduce the energy of this circuit, you'd have to do something like add a buck converter or PWM, correct?

Someone is telling me that the resistor alone will decrease the energy the circuit uses. My understanding is that resistors only drop the current and dissipates current as heat and are therefore not the most efficient choice.

The resistor is limiting the current to the LED. LEDs have a non-linear I-V (current to voltage) curve, where past a certain point, small increases in voltage will cause a large change in current. The current limiting resistor helps stabilize this, it limits the maximum current through the LED. Think about what happens if the diode's forward voltage drop goes to 0V. This is the maximum current that the battery can supply. Any increase in the diode's forward voltage drop will cause the voltage drop across the resistor to decrease, decreasing the current.

Try experimenting with some simulations and see what happens if the ESR (equivalent series resistance) goes to 0Ω and there is no current limiting resistor. Can experiment in real life too if you don't mind burning out a few LEDs.

As the other answer states, driving the LED can be more efficent if instead of controlling the voltage, you control the current. Now large changes in currents cause a small change in voltage, and you don't have to have a resistor to limit the current. No resistor, no useless power dissipation.

Q3: what exactly does a resistor do with the energy it is removing from the circuit?

It dissipates as heat.


First, your questions.

Q1: All else being ignored (early burn out of the LED, etc...) will the circuits consume the same amount of power?

No. Assuming a fixed supply voltage, the one without the resistor will consume more power because the current isn't being limited by the resistor. Instead, the current is limited ultimately by the power supply and/or the LED's internal resistance (at least up until the moment it becomes a 'fry-ode' and releases the magic smoke.)

Q2: in order to reduce the energy of this circuit, you'd have to do something like add a buck converter or PWM, correct?

You can limit the LED current with a dropping resistor to limit the power, and thus, energy used.

To improve efficiency, you need to reduce the supply voltage to the lowest possible to allow lower-value dropping resistor. Or, or use a constant-current DC-DC LED driver and no dropping resistor at all.

Using PWM will also reduce the overall energy consumption, but have limited improvement in efficiency. PWM has some downsides as well, including EMI noise and strobing effects.

Fun fact: it's possible to use much higher LED current at lower duty cycle to achieve the same average luminous power output, without killing the LED. For example, if you run a 20mA LED at 100mA and 20% duty, you achieve similar light output and higher perceived brightness compared to straight DC current. But it's not a huge win. (we'll explore that below.) Nevertheless, many LED datasheets will give a max pulsed-current figure just for this reason.

A typical use case for high-current pulsed drive would be an IR remote, which drives the snot out of the IR LED at high peak current with a low-duty carrier. Because the IR output is being received by a sensor, and not a human eye, this substantially increases the reach of a remote control without increasing the power used. (Pro Tip: don't stare at an IR remote emitter with your eye - they're powerful enough to do some damage.)

Q3: what exactly does a resistor do with the energy it is removing from the circuit?

The purpose of the resistor is to set the LED current. The lower the resistor value, the higher the current in both itself and the LED, thus the more power is used, again in both. It's possible to use a resistor for current-setting because for most practical purposes, the LED forward drop (Vf) is a fixed voltage, so the LED current is simply I = (Vbat - Vf) / R.

The downside of using a resistor to set current is that the resistor dissipates power (wastes energy) that could otherwise be delivered to the LED. Where does the power go? The resistor sheds it as heat.

Now, how does the resistor value affect the amount of power wasted as a proportion of the total power used? Within a limited range of current, there will be little difference in LED efficiency vs. resistor value. This is because the IR drop of the resistor and LED don't change very much over the range of useable LED current. (We'll explore that in more detail below.)

What can you do to make an LED + resistor combo the most efficient?

  • Use the lowest possible power supply to reduce required resistor IR drop.
  • Use PWM and overdrive if the LED can handle it (and, the PWM doesn't cause other problems.)

We noted that the resistor sets LED current. How?

Think of the LED as a fixed voltage drop (which, within reason, it is.) Kirchhoff’s voltage law tells us that the current-limiting resistor will have a fixed voltage drop, too.

Let's consider two cases of a red LED with a forward voltage (Vf) of approximately 2V.

  • Case 1: Resistor = 1k, Vbat = 5V

Current will be: I = (Vbat - Vf) / R = (5V - 2V) / 1k = 3mA

Resistor power will be 3mA^2 x 1k = 9mW

LED power will be 3mA x 2V = 6mW.

So the resistor is using 60% (3/5) of the power, while LED is using 40% (2/5). Makes sense, since they're seeing voltage drops of 3V and 2V respectively, out of a total of 5V.

Now let's consider the other extreme:

  • Case 2: Resistor 1 ohm, Vbat = 5V

Current (simplistically) will be: I = (Vbat - Vf) / R = (5V - 2V) / 1ohm = 3A

Resistor power will be 3A^2 x 1ohm = 9W

LED power will be 3A x 2V = 6W.

Based on this simple fixed-Vf assumption we have the same story as the 1k ohm case: resistor power is 3/5 of the total, LED 2/5.

Conclusion? The ratio of resistor power to LED power is the same at both current extremes. Thus, both have the same relative 'efficiency'.

But we know that's not quite what actually happens. The LED forward drop increases with current. So at higher current, a larger portion of the power will be dissipated in the LED than is predicted by the simplistic assumption of fixed LED forward voltage.

Here's a chart that shows LED forward voltage vs. current, for various types of LEDs:

enter image description here

From http://lednique.com/current-voltage-relationships/iv-curves/

You'll see that the forward voltage shifts some as LED current rises. In the range shown it doesn't shift much, but it does. That's because like all semiconductor devices, LEDs in forward bias have an internal resistance that is limited by the LED substrate material itself. The bigger the LED chip, the less the Vf-vs-I shift.

Here's a quick simulation to illustrate this I-V behavior (simulate it here):

enter image description here

Falstad’s red LED model is similar to what’s shown in the chart above: Vf is about 1.8V at 20mA.

As you vary the slider between 1k down to 1 ohm, the scope plots will show you that the ratio of resistor power to LED power goes from about 2:1 down to 1:1, because the LED IR drop increases with current, much like we’d expect.

So in theory the lower resistance / higher current case is more 'efficient' since a larger portion of the total power is being shed in the LED.

As a practical matter over a reasonable constant DC current range (say, 0 up to 20mA) there it little to be gained efficiency-wise by running the LED at higher current.

You can get an efficiency improvement if you use the PWM 'overdrive' trick as I mentioned above. (I'll get into why this may be a Bad Thing in a moment.)

However, for ultimate efficiency, especially when driving a big LED, it's best to use a DC-DC constant-current driver instead of a resistor drop or using PWM. Then you avoid resistor losses entirely. While you still have some loss in the DC-DC (10-15% or so), it's still a win compared to a dropping resistor.

And now, the Bad Things about PWM.

On the electrical side, PWM switching can cause system noise issues and EMI radiation.

On the optical side, PWM strobing can be annoying, or even dangerous.

A major example where LED PWM is a cardinal sin is automotive lighting. Yet somehow many automakers still don't seem to 'get it' and continue to use PWM for exterior lighting (if you're reading this and this is your job, shame on you. Do better.)

First of all, PWM strobing becomes visible with movement. Guess what? Cars move, so other drivers see the flickering. This is at best distracting, at worst, seizure-inducing.

In a more narrow case, if you're unlucky enough to be tasked with filming a car equipped with with PWM-modulated LEDs, the strobing really messes up the footage, especially if your camera is a 'rolling shutter' type: you'll get 'banding' in your footage.

But here's where PWM becomes a hazard. LED PWM strobing can reduce the accuracy of object detection, critical for Autonomous Driving (ADAS) systems. For example, PWM strobing can influence detection of traffic light lights, reducing the effectiveness of emergency braking / collision avoidance. Imagine what PWM could do to brake light detection, an ADAS task even more time- and safety-critical than detecting a traffic light.

To combat this issue, some more recent automotive cameras have integrated anti-flicker technology to mitigate PWM artifacts. But there are plenty of ADAS cameras on the road that don't have this.


Since the LED is powered in both circuits, consider the approximation where the LED is replaced by low resistance, 1 Ω resistor in both. Let's assume the resistance of the other resistor is 99 Ω.

It's straightforward to compute total power. Electrical power is given by P = I V or P = V² / R. So the power in the first circuit is 9 W, while the power in the second is 9 cW. The power is significantly less in the second circuit.

We can also examine the power disapated by each element. The equation for electrical energy can also be stated as P = I² R. The current in the first circuit is 3 A, so we get the same power we already computed for the power across the "LED", 9 W. In the second circuit the total resistance is increased, so the current is reduced to 3 cA. Therefore, the power across the "LED" is 0.9 mW, and the power across the resistor is 89.1 mW, which together add up to the same total power as before, 9 cW.

So in simple terms, compared the the first circuit, even though some power is now wasted on the resistor (and turned into heat), the power used on the LED is much much less (and so is the light output).

P.S. Substituting a diode with a short circuit or a low resistance resistor is only useful as a rough approximation. Diodes do not have constant resistance, so the power values we computed are not accurate, but they do give a good rough picture of what's going on.


Relating to the title question of efficiency, a point the other answers haven't mentioned is that adding a series resistor will reduce the amount of current a consumed by a certain amount, and will--if available voltage is fixed--reduce power consumption by that same amount, while reducing the amount of light output by an amount that might be greater or less than that power reduction.

If adding a certain size of resistor reduces current by 50% while reducing light output by 40%, the resistor would improve the overall efficiency of the circuit since the resistor would reduce the amount of energy the LED turns into heat by an amount greater than in the LED by an amount of energy it dissipates itself. If the resistor would reduce current by 50% but reduce light output by 60%, it would reduce efficiency. Although the resistor would almost certainly reduce the amount of energy dissipated by the LED, such savings would be insufficient to make up for the energy wasted in the resistor.

If a certain LED would have optimal light-per-current efficiency at 20mA, but would be twice as bright as desired when driven with that much current, and the LED would take 11mA to produce half the brightness, then driving it with 20mA half the time would the same amount of power as driving it with 10mA, but produce as much light as driving it with 11mA.


Edit: The correct answer actually depends on the goal of the LED. Is it an Indicator LED or a lighting LED?

My answer is for the application where maximizing energy to light conversion is the goal. If the goal is minimizing energy consumption while still producing noticeable light, then the resistor does indeed help with that goal.

Does a resistor in a simple LED circuit actually make it more efficient


or does the resistor just burn off the energy?


Efficiency in this context the context of LED lighting is the amount of electrical power being reasonably converted to "visible light" instead of "heat."

The LED is the only component that converts some fraction of power going through it into light, in addition to some fraction being converted to heat.

The resistor can only convert power going through it into heat, not light.

So, given the (woefully incorrect) condition that all other things are constant and equal, having zero current limiting resistor provides the best possible efficiency because it removes all associated conversion of power into heat while not removing any conversion of power into light.

Of course, there are several reasons something this naive cannot work in real systems. Notably ...

  • LEDs are not self stable, so will break without external stabilization.
  • LEDs efficiency is absolutely not constant with respect to changing power.
  • 1
    \$\begingroup\$ Radiant power / electrical power is one way to define efficiency. However, for an indicator light for example, it often makes more sense to define efficiency as the piecewise function 1 / elecrtical power when radiant power > some threshold, 0 otherwise. \$\endgroup\$
    – Vaelus
    Commented Jan 5 at 4:39
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    \$\begingroup\$ A resistor can convert power into light, if you overdrive it hard enough. Don't expect it to last long in that configuration though. \$\endgroup\$ Commented Jan 5 at 5:29
  • 1
    \$\begingroup\$ That said, I like this answer because it's one of the few on this question which points out the important point that "brightness efficiency" does get worse with the resistor. Even if we often don't care about that kind of efficiency, sometimes we do, such as for an LED light bulb. \$\endgroup\$
    – Vaelus
    Commented Jan 6 at 16:54
  • \$\begingroup\$ Right. The problem is context sensitive to be sure. If efficiency is defined as improving the ratio between goals and resource consumption, then "goals" needs to be carefully defined. If the "goal" is "brightness efficiency" then we proceed as with my answer. However, if the goal is to minimize power consumption while still making visible light (as might be seen in an indicator LED) then the resistor does "increase efficiency" as it reduces power drain faster than it reduces light output... to a point. May add this stipulation to my answer. \$\endgroup\$
    – Charlie
    Commented Jan 6 at 17:35

The resistor paired with the LED is just a way to bias your LED at the specific point voltage;current you wish. If you use for example a backlight/LED driver you don't need the resistor inserted in the branch with the LED(s), but just the LED(s). Then you have the resistor in the bottom of tge feedback pin that is used by the driver to provide the current you set. If you use a current source (for example a MOSFET or BJT controlled by an op amp), you will need no resistor and the voltage across the LED will be function of the current flowing through it according to the characteristics of the LED. Please note that this relation voltage-current will change with temperature. For a specific LED brightness, you need to regulate the current, not the voltage. You may have a derating profile if you control the LED in extreme temperature, probably less than -10°C and more than 60°C or so, check datasheet.

So, back to your design, with the simple resistor in series with the LED, you just find the right operating point against your requirements. Example, pick an LED with a voltage Vf = 1.8 V when If = 10 mA. You can power it on 3.3 V or on 5 V. You will choose to have 10 mA across your resistor, so R = V(power - Vf@10ma)/0.01. For V(power) = 3.3 V, your resistor will be (3.3 - 1.8)/0.01 = 1.5/0.01 = 150 Ω. If V(power) = 5V, R = (5 - 1.8)/0.01 = 320. Usually, good practice leads you to choose a slightly higher resistance, 160 to 180 for the first case or 330 or more for the 2nd case. This is only to bias your LED. If you put no resistor, even with 3.3 V, you will find the operating point at 3.3 V which means a massive current and the LED wouldn't withstand it. Conversely, if you have a voltage source at 1.6 V and one of such an LED or even two LEDs in series with the 3.3 V, you shouldn't need any series resistor because if you have 10 mA at 1.8 V, it means that you will have less current at 1.6 V (the characteristics of an LED being exponential).


Efficiency and loss of energy on a resistor-led circuit depends on two factors:

  1. The difference between the supply voltage and the typical voltage of the LED. The typical voltage is the voltage at which the LED will use the desired current for a foreseeable light output. This voltage can vary slightly from one LED to another, but gives you a good idea of the voltage needed.

  2. The amount of current, therefore energy, that the LED will consume.

These two parameters will tell you how efficient it is. If you power an indicator 3V LED with only 1.2 mA, it's no big deal to connect it to 12V supply with a 10K resistor. If you want to draw 20 mA from the same LED, then you should already consider calculating the power dissipation of the resistor, yet, it's still acceptable.

Now if you want to use 150 mA, or 1A with power LED and the voltage provided is double the typical LED voltage, then you are going to waste an insane amount of energy in power dissipation through the resistor. In another word: Heat. To avoid that, LED's should be connected in series to match the supply voltage as much as possible.

A constant current supply will do that actively by adjusting the voltage to the voltage that the LED needs to draw a fixed amount of current. This is the most efficient way because no resistor are needed. But each LED should have its own power supply. This system doesn't allow LED in parallel. PWM does the same, except that it's sort of DC to DC conversion. A PWM driver after a Power Supply will also waste some energy because the circuits are redundant. It's better to use one adapted power supply than a power supply + a PWM module.

In the case of a 3V button battery, you don't have any choice but the resistor because a PWM or buck converter will use a relatively significant amount of energy for functioning.


If you define efficiency as the power consumed by the LED (and converted to light,) then the efficiency can only go down when you put a resistor in the circuit.

For this example circuit, I've simulated the efficiency using various resistor values:


simulate this circuit – Schematic created using CircuitLab

enter image description here

As you can see, the percent of power consumed by the LED starts at 100 and drops rapidly as the resistance goes up.

In that sense, the efficiency can only go down.

For comparison, the total power consumed (in the LED and the resistor together) goes down as the resistance goes up:

enter image description here

In that sense, the total efficiency of a larger circuit (where the LED is just a small part) may actually go up since the total power consumed is lower.

Looked at another way, you could get a much longer running time from a battery powered LED despite wasting more power in the resistor.

Say you have a 100mAh battery to provide an indicator light (maybe just a marker light on your big flashlight so that you can locate it in the dark when the power goes out.)

enter image description here

Without a resistor, the LED will draw about 20 milliamperes. The light will be on for 5 hours.

By putting in a 400 ohm resistor, the current drops to just 1 milliampere - the marker light will glow for 100 hours instead.

It isn't as bright, but the battery lasts longer. A modern LED at 1 milliampere can be surprisingly bright, especially in the dark.

It isn't more efficient - the resistor is wasting 30 percent of the power - but the circuit itself will do its job longer.

All of that said, an LED without a current limiter of some kind (resistor or constant current source) will soon become a DED (dark emitting diode.) It may light up (or not) or it may blow when you first turn it on.

The forward voltage of an LED is not set in stone. It varies with temperature - the forward voltage goes down as the LED gets warm, and it goes up as the LED cools.

If you use a constant voltage source, then at low temperature the LED may not light up at all. At higher temperatures, it might light up then as it warms the forward voltage will drop until it is drawing so much current that it burns out.


Hacktastical's answer is probably the best above.

In addition, the LED's corner voltage varies by color, by batch and by temperature. If you try to make the series resistor too small, you may succumb to large current variations when the diode exhibits some variation.

For example, if you're powering a white LED from 3.3V, and using a series resistor of 10ohms, the calculations say:

\begin{align} I = \frac{3.3V - 3.0V}{10ohms} = 30mA \end{align}

if the white LED drops 3.0V at room temperature. But change the temperature by 50 degrees and the corner voltage moves down by 100mV, which would increase the current to (3.3V - 2.9V) / 10ohms = 40mA! Then change the vendor of the LED, or include the variation in your power supply voltage, and that could easily change to 60mA!

To avoid this large current variation, make sure to leave a reasonable (about 1V) drop across the resistor. Or perhaps power the white LED from a 5V supply instead of 3.3V.


The two hypothetical circuits will consume the same amount of power but getting e.g. a red LED in 3V Vf is impossible from a materials standpoint.

To "reduce the energy", you'd typically use a constant current driver that switches current into an inductor to provide a mostly-constant current into the LED. Adding a resistor will reduce the power from a level that will burn out the LED to an appropriate level.

A series resistor will dissipate the excess energy as heat.

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    \$\begingroup\$ The circuits would not consume the same amount of power. The current from the battery will be different based on the series resistance. \$\endgroup\$ Commented Jan 4 at 5:54

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