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I was wondering for a long time why in schemes which use LEDs for lighting it's very common to put a resistor to go with the LED, and finally it seems like the answer in this question explains why. (It's the easiest way to control the current through he LED to prevent LED from burning up.)

But still, isn't this a big problem? Don't those resistors waste a lot of power and isn't there really any other practical solution?

UPD: A reasonable update for the question given all the good answers I've received is to maybe provide some numbers to show just how much power is lost to heat from resistors in a typical lighting application? (Most answers say that the power loss i so small that it doesn't matter. I think it would be good if anyone could get the real numbers, to solidify that answer, then I could accept that answer and keep it on top for future interested people.)

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  • \$\begingroup\$ Here's a related question electronics.stackexchange.com/q/23974/3552 \$\endgroup\$ – sharptooth Mar 28 '12 at 5:48
  • \$\begingroup\$ As a response to your update: it's impossible to define the typical lighting application (even though the most common until now is as indicator); is instead easier to point the most critical applications, which I think to be power lighting and low low power operation (the second can be bypassed with flashing) \$\endgroup\$ – clabacchio Mar 28 '12 at 14:52
  • \$\begingroup\$ Alright, but certainly, providing a way to calculate the lost power given the configuration of a specific system and perhaps showing an example calculation for a typical system is good enough, just to give a rough estimate, how much power are we talking about... \$\endgroup\$ – Cray Mar 28 '12 at 15:01
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    \$\begingroup\$ Resistor: I^2 * R --- Dc/Dc: Pin(1-efficiency) \$\endgroup\$ – clabacchio Mar 29 '12 at 15:01

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Yes, it wastes power, but most of the time it's not enough power to matter.

In cases where effeciency matters, you use other more complicated means. For example, take a look at the schematic for my KnurdLight example project. This is battery operated and just about all the power is going into the LEDs. In this case I used a boost converter that directly regulated LED current instead of a normal power supply that regulates voltage. There is no series resistor to make a fixed voltage supply look at least partially like a current source because the power supply is a current source in the first place. R6 is in series with the LED string, but is only 30 Ω and is for sensing the current so that the boost converter can regulate it.

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    \$\begingroup\$ Your links are interesting... Google Chrome warn me: "The site ahead contains harmful programs - Attackers on www.embedinc.com might attempt to trick you into installing programs that harm your browsing experience (for example, by changing your homepage or showing extra ads on sites you visit)." \$\endgroup\$ – Blup1980 Feb 21 '16 at 10:29
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    \$\begingroup\$ @Blup: Yes, I know. Google is wrong. This is a false positive. We never install anything in your browser, never pop up anything, including ads, and all of our software comes as self-extrating EXEs that you have to explicitly chose to run. \$\endgroup\$ – Olin Lathrop Feb 21 '16 at 13:48
  • \$\begingroup\$ Whoever downvoted this, what exactly do you think is wrong? \$\endgroup\$ – Olin Lathrop Jan 13 '18 at 13:38
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Why resistors?

The reason we use resistors to set LED current is that an LED is a diode, and like most diodes, it just looks like a voltage drop when forward biased. There is very little to control current if hooked up to a voltage source; the V/I graph's slope is so steep that a 0.1 V change in diode voltage could mean a 10X change in current. Thus a direct connection to a supply without a workable current limiting mechanism will likely destroy the LED. So we put a resistor in there to make the slope shallow enough to control the current.

Typically, you figure out how much current you want in the LED based on some brightness measurement from the data sheet, or buy one and guess. For typical indicator LEDs, I start with 2 mA for normal or 0.5 mA for high-efficiency LEDs, and usually have to reduce the current further.

Once you pick a current, you take that, the voltage of your source (VS), and the forward voltage of your LED at your current (VF, try to get this from the graph in the data sheet rather than the table, which typically is characterized at 10 mA or more), and plug them into the following equation to get your resistance:

R = (VS - VF) / I

Derivation: Given that the voltage drop across the resistor is VR = I * R (Ohm's Law), that the current in the loop is constant (Kirchoff's Current Law), and that the source voltage is equal to VF + VR (Kirchoff's Voltage Law):

VS = VF + VR = VF + I * R; VS - VF = I * R; R = (VS - VF) / I

High Power LEDs

For applications where the power waste is a problem, such as in large-scale lighting applications, you don't use a resistor but instead use a current regulator to set the LED's current.

These current regulators work like switching voltage regulators, except instead of dividing down the output voltage and comparing to a reference and adjusting the output, they use a current-sensing element (current-sense transformer or low-value resistor) to generate the voltage that is compared to the reference. This can get you lots of efficiency, depending by switching element loss and switching frequency. (Higher frequencies react faster and use smaller components but are less efficient.)

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When an LED is driven with a resistor, it's necessary that the supply voltage be higher than the forward drop of the LED; the current drawn from the supply will be equal to the current through the LED. The percentage of supply power that goes to the LED will correspond to the ratio of the LED forward voltage to the supply voltage.

There are other ways of driving LEDs which will work with supply voltages below the forward drop of the LED, or which will draw less current from the supply than they put through the LED. Such techniques may e.g. reduce by half the current drawn from a 5-volt supply to feed 20mA through a 2-volt LED, but the circuitry required will almost certainly be more expensive than a resistor. In many situations, even when running from batteries, the power consumed by an LED will represent a tiny fraction of overall energy usage; even if one could reduce LED-related power consumption by 99% using only $0.05 worth of extra circuitry, the savings wouldn't be worth the cost when compared with simply using a resistor and accepting the sub-optimal efficiency.

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You wanted a calculation. Here is the basic form of the calculation.

A typical red LED has a forward voltage drop of 1.8 V, and a maximum continuous current of around 20 mA.

Now what's our voltage? Lets say we want to use a 3 V source.

So we will have a voltage drop of 3.0 V - 1.8 V = 1.2 V over our resistor. The current through the resistor will be 20 mA, so our power is 1.2 V * 20 mA = 24 mW. That is not really a lot of power, although it is a significant fraction of power consumption of the LED. The LED itself uses 1.8V * 20mA = 36 mW.

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  • \$\begingroup\$ In other words, for every 3 watts of power usefully delivered to the LED, you are wasting 2 watts of power as heat. \$\endgroup\$ – rjmunro Mar 28 '12 at 23:31
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    \$\begingroup\$ In that particular scenario yes. In other scenarios, you could have far less of the energy wasted. If your voltage is high enough, you can also put more than one LED in series, which lets them share a single resistor (but requires that they both be on or off at the same time). In that case your waste percentage can drop substantially. Also keep in mind that the lumen per watt of LEDs even when including the resistor loss is still far better than most alternatives. The choice is not resistor or no resistor, but (LED+resistor) vs (no LED) vs (other light source). \$\endgroup\$ – Kevin Cathcart Mar 28 '12 at 23:44
  • \$\begingroup\$ Nice example, upvoted you. \$\endgroup\$ – Cray Mar 29 '12 at 15:16
  • \$\begingroup\$ Please remember that the current quoted on the data sheet is typically the most the part will handle. This will generate a large amount of light, usually for illumination (i.e. flashlight) purposes, but hard to look at if you want a display or indicator. (Display currents are typically much, much lower. In this example, I would bet that 1-2 mA would be plenty.) Also, like on other semiconductor data sheets, the maximum current may be valid only with an infinite heat sink (i.e. water bath), and may generate enough heat to destroy the part without a heat sink. \$\endgroup\$ – Mike DeSimone Apr 2 '12 at 13:08
  • \$\begingroup\$ For LEDs, the max continous current normally does not assume a heatsink, since the plastic that surounds the LED is a poor heat conductor, making a heatsink rather ineffective. But you are correct that in circuit design you really should target a lower current. However, I frequenty see designs running 20mA LEDs at 15mA or more even for indicator purposes. I would be better in general to chose the current resulting in the desired brightness. \$\endgroup\$ – Kevin Cathcart Apr 3 '12 at 16:40
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Yes, it wastes power. On the other hand, in volume production a resistor will cost a fraction of a penny (US$0.01 for our international people). When the cost/benefit/difficulty analysis is done a simple resistor starts to look really nice.

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    \$\begingroup\$ What about the cost of electricity loss? \$\endgroup\$ – Qwerty Dec 28 '18 at 10:51
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The wasted power is often very small (tens of milliwatts) if you're driving the LED with 5 V or an equally small voltage.

Sure, it is a problem in systems where you have limited-capacity batteries, but then other schemes (like LED drivers using PWM) are used.

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    \$\begingroup\$ There's no such thing as "wasted current". The current through the resistor is the same current that makes the LED light up, so it's not wasted at all. What you mean is "wasted power/energy". \$\endgroup\$ – stevenvh Mar 28 '12 at 14:46
  • \$\begingroup\$ @steventh indeed. Will edit my post. \$\endgroup\$ – Renan Mar 28 '12 at 16:02
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Yes, and no. When current passes through the resistor, it generates heat and therefore wastes energy. However, if you took the resistor out (and therefore drove the LED at a higher voltage) you'd be driving more current through the circuit and thus actually burning more power than with the resister in place.

Remember that with constant voltage, current is inversely proportional to resistance. The more resistance you inline into the circuit, the less current you pass, and therefore the less power you consume. So while the resistor itself plays a part in generating heat in the circuit, its presence there actually means that less heat will be generated overall.

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    \$\begingroup\$ Well as I understand now, the problem with no resistors is that there is not good cheap way to control the voltage and that without it the LEDs will simply burn. Resistors are not used in those circuits to limit current in order to save power, but in order to keep the circuit from burning. \$\endgroup\$ – Cray Mar 28 '12 at 14:58
  • \$\begingroup\$ @Cray: It's not the voltage drop that kills LEDs, it's the heat from the current flow. Solving for the appropriate voltage drop makes the math simpler, but is's not the critical factor in the physics of it all. The internal resistance of an LED decreases as the voltage across it increases, so trying to work out the math without a traditional resistor in the circuit is difficult (and yes, of course an LED has internal resistance; it's not a superconductor). \$\endgroup\$ – tylerl Mar 28 '12 at 22:58
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I make mathematical test. I use 12V source and I connect 3 Led diodes with resistors. That one diode with Resistor have 12V, next diode with Resistor I hook up also to 12V, and last one also. At source I have 60mA. Resistor have voltage drop 9V and power consumption was in total 540mW Regardless to P = V * I I get something 720mW total in source.

But when I hook up diodes in string and I add Resistor total power consumption was only 240mW in source. I use diodes 3V 20mA.

It Sims that better is use voltage source as lower as is possible, to have power consumption only there where we want. Or use string of led diodes for higher voltage. That's why we have in computer so many outputs with different voltages from transformer.

Or another idea. We have 9V source, and I hook up one diode 3V and I need use Resistor to droop voltage. Total power will be 180mW Where diode will take only 60mW But when I hook up in string 3diodes then I have still 180mW But when I hook up 3 diodes but each one is separate connected to this same source, then I will have 540mW of used power.

Looks that better is using string rather connecting each one to source.

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There is no way to avoid the power loss with only a passive or linear active DC circuit. The reason is that the efficiency is determined by two things:

  1. The supply voltage
  2. The LED current

It doesn't matter what you put in between the LED and the supply. It could be a resistor, some diodes, a linear regulator, or a transistor-based current source. If the LED needs 10 mA for the desired brightness, and you have a 5 V supply, you're burning 50 mW total. Period.

With a fixed supply voltage and a fixed LED current, your only option for increasing efficiency is to put multiple LEDs in series. If you have a 5 V supply, and your LEDs' voltage drop is 2 V at 10 mA, you can put two in series. This comes with a limitation -- you will not be able to switch the LEDs independently.

If you have control over the power supply, there are a couple other things you can do. If your supply voltages are derived from an AC source, you can add a winding to the transformer to create a low-voltage LED supply. If you only have a DC supply, you can use a switching converter to generate a lower voltage. However, neither of these are terribly practical. If you're running off of AC (mains) power, you're probably not concerned about the efficiency of a few indicator lights. And high-efficiency switching regulators are expensive and trouble-prone.

LEDs typically consume only a small fraction of the total current of a system. It's rarely worth the trouble or expense to add a separate power supply just for them.

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  • \$\begingroup\$ " If the LED needs 10 mA for the desired brightness, and you have a 5 V supply, you're burning 50 mW total. Period.". Pretty sure you are answering another question. It is of course obvious that if the light is going to be emitted at all, it will take some power from the power source. The question is about the power wasted on heat in the resistors that are very commonly used. Also in cases where LEDs are used for lighting, they can consume not a small fraction, but most part of the total current of the system. \$\endgroup\$ – Cray Jan 7 '15 at 12:27
  • \$\begingroup\$ @Cray: If the LED has 10mA going through it and the supply is 5V, then if one is using a linear circuit the total power dissipation will be 50mW. How much power is dissipated in the resistor versus the LED will depend upon the LED's voltage drop, but the total power will be set by the supply voltage and the LED current. \$\endgroup\$ – supercat Jul 23 '15 at 13:35
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Don't think of a resistor as a power sink that diverts electricity (current). Some power is lost in the form of heat, yes, but not much (in general). Using the water analogy, just think of the resistor as making the hose through which the current flows smaller. Given the same initial force (voltage), the amount of electricity that can flow (current) is reduced. this reduces the force available at the output end of the hose (this is known as voltage drop).

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