Tell me more ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.
up vote 12 down vote favorite
2

I was wondering a long time why in every scheme there is always a resistor to go with the LED, and finally it seems like the answer in this question explains why.

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.)

share|improve this question
Here's a related question electronics.stackexchange.com/q/23974/3552 – sharptooth Mar 28 '12 at 5:48
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) – clabacchio♦ Mar 28 '12 at 14:52
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... – Cray Mar 28 '12 at 15:01
1  
Resistor: I^2 * R --- Dc/Dc: Pin(1-efficiency) – clabacchio♦ Mar 29 '12 at 15:01

7 Answers

up vote 16 down vote accepted

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.

share|improve this answer
up vote 14 down vote

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.

share|improve this answer
up vote 11 down vote

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.)

share|improve this answer
up vote 5 down vote

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.

share|improve this answer
up vote 4 down vote

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.

share|improve this answer
2  
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". – stevenvh Mar 28 '12 at 14:46
@steventh indeed. Will edit my post. – Renan Mar 28 '12 at 16:02
up vote 4 down vote

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 3.3 V source.

So we will have a voltage drop of 3.3 V - 2.1 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.

share|improve this answer
In other words, for every 3 watts of power usefully delivered to the LED, you are wasting 2 watts of power as heat. – rjmunro Mar 28 '12 at 23:31
1  
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). – Kevin Cathcart Mar 28 '12 at 23:44
Nice example, upvoted you. – Cray Mar 29 '12 at 15:16
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. – Mike DeSimone Apr 2 '12 at 13:08
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. – Kevin Cathcart Apr 3 '12 at 16:40
up vote 3 down vote

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.

share|improve this answer
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. – Cray Mar 28 '12 at 14:58
@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). – tylerl Mar 28 '12 at 22:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.