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For blue LEDs with a forward voltage of 3.3 V and supply voltage of 3.3 V, is a series resistor still needed to limit current?

Ohm's Law in this case says 0 Ω, but is this correct in practice?

Perhaps just a small value like 1 or 10 Ω just to be safe?

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8 Answers 8

up vote 25 down vote accepted

No, it's not correct, if only because neither the LED nor the power supply are 3.3V. The power supply may be 3.28V, and the LED voltage 3.32V, and then the simple calculation for the series resistor doesn't hold anymore.

The model of a LED is not just a constant voltage drop, but rather a constant voltage in series with a resistor, the internal resistance. Since I don't have the data for your LED let's look at this characteristic for another LED, the Kingbright KP-2012EC LED:

LED characteristic

For currents higher than 10mA the curve is straight, and the slope is the inverse of the internal resistance. At 20mA the forward voltage is 2V, at 10mA this is 1.95V. Then the internal resistance is

\$R_{INT} = \dfrac{V_1 - V_2}{I_1 - I_2} = \dfrac{2V - 1.95V}{20mA - 10mA} = 5\Omega\$.

The intrinsic voltage is

\$V_{INT} = V_1 - I_1 \times R_{INT} = 2V - 20mA \times 5\Omega = 1.9V.\$

Suppose we have a power supply of 2V, then the problem looks a bit like the original, where we had 3.3V for both supply and LED. If we would connect the LED through a 0\$\Omega\$ resistor (both voltages are equal after all!) we get a LED current of 20mA. If the power supply voltage would change to 2.05V, just a 50mV rise, then the LED current would be

\$ I_{LED} = \dfrac{2.05V - 1.9V}{5\Omega} = 30mA.\$

So a small change in voltage will result in a large change in current. This shows in the steepness of the graph, and the low internal resistance. That's why you need an external resistance which is much higher, so that we have the current better under control. Of course, a voltage drop of 10mV over, say, 100\$\Omega\$ gives only 100\$\mu\$A, which will be hardly visible. Therefore also a higher voltage difference is required.

You always need a sufficiently large voltage drop over the resistor to have a more or less constant LED current.

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"Always"? There isn't a more efficient way to drive them? –  endolith Oct 26 '11 at 13:38
@endolith - there are ways of controlling the current with a smaller current sense resistor and a transistor (BJT or MOSFET), but there the transistor replaces the usual resistor, and also needs the voltage drop. –  stevenvh Oct 26 '11 at 14:38
There are also ways of controlling the current with a switch-mode current regulator, which will be more efficient, but that's probably not necessary for readers of this question. –  Kevin Vermeer Apr 17 '12 at 12:24

You always need a current limiting device. When using a voltage source, you should always have a resistor, think about what happens when the voltage changes by a small amount. With no resistor, the LED current would shoot up (until you hit a thermal based limit due to the LED materials). If you had a current source, then you would not need a series resistor because the LED would run at the current source level.

It is also unlikely that the forward voltage of the LED is always exactly the same as the supply. There will be a range mentioned in the datasheet. So even if your supply exactly matched the typical forward voltage, different LEDs would run at vastly different currents, and hence brightnesses.

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Also think about what happens if the LED fails and shorts. Limiting resistor is not just a good idea, its the law! (well ok maybe not) –  freespace Apr 11 '11 at 13:55
Surely a fuse would be better for higher powered applications? Otherwise you'd need pretty hefty resistors able to dissipate more than the rated power of the led! –  avl_sweden Jul 16 at 18:46

The I-V relationship in a diode is exponential, so applying a voltage difference of 3.3 V +/- 5% to an LED with a nominal 3.3V drop is not going to result in a 5% variation in intensity.

If the voltage is too low, the LED may be dim; if the voltage is too high, the LED may be damaged. As Hans says, a 3.3V supply is probably not enough for a 3.3V LED.

When driving an LED, it is better to set the current, not the voltage, since the current has a more linear correlation with the light intensity. Using a series resistor is a good approximation of setting the current through the LED.

If you can't use a supply with enough headroom to allow a current-setting resistor, you might be able to use a current mirror. That still requires some voltage drop, but possibly not as much as you'd need for a resistor.

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A current mirror also has a required voltage drop over the transistor. It may be able to follow the current quite well (so you set a 10mA reference and get very close to that on the other side), but it will need a little bit of voltage to work. –  Hans Apr 11 '11 at 17:14
@Hans good point, I updated to make that clearer –  Andy Apr 11 '11 at 17:25

You need a voltage drop over the current limiting resistor for it to work. And that voltage drop should be substantial to avoid high currents when your 3.3V is a bit off (maybe 3.45V for a while). If you would drive a LED with 1V voltage drop across a resistor and the supply is 1V higher, you would have approx. the double current.

A LED needs a constant current to shine. However, a constant current source probably needs more than 3.3V for a blue LED, unless you're using a buck-boost version.

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If the power source was exactly 3.3V and the voltage drop across the LED was 3.3V then you would not need a current limiting resistor. However, the world is not perfect and there are imperfections in everything!

You only can calculate an appropriate safety resistor value after accounting for the power supply configuration and variation in the LED forward voltage. If you think \$V_{SOURCE}-V_{LED}\$ may have an error/variation up to \$0.5 \mbox{ }V\$, then you would calculate the resistor value for that. As an example the case of \$ \pm 0.5\mbox{ }V\$, which isn't unreasonable for a 10% 5V source:

$$\frac{V}{I} = \frac{0.5\mbox{ }V}{20 \mbox{ mA}} = 25\mbox{ }\Omega $$

Just note that this is probably not a good idea in practice, but it is possible.

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If you don't use a current limiting resistor on a LED, it will typically splice itself in two (personal experience). –  Hans Oct 25 '11 at 18:41

Even if the voltages were the same you would still need to add a resistor. The only time you don't add a resistor is when the current output from source is less then or equal to the amount needed, for example connecting a white LED to a CR2023. No resistor needed, since the battery's internal resistance limits the current to an acceptable level.

Don't worry about adding a resistor because it is the cheapest thing you can add to protect your LED, unless you're dealing with high current LED.

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If the forward voltage and supply voltage are nearly equal, using a resistor will yield results which are very sensitive to variations in supply voltage or LED characteristics. If the resistor is sized to avoid damaging the LED if it turns out the supply voltage is at maximum and the LED intrinsic voltage is minimum, the LED will only light with a fraction of its possible brightness if the supply voltage is at its minimum and the LED intrunsic voltage is at its maximum.

Using some type of current-regulating circuit will yield much better results, although most simple current-regulating circuits have a certain amount of compliance voltage. Probably the easiest thing to do in many cases is use an LED driver chip with a built-in booster circuit. Some of those can do a good job regulating LED brightness independent of supply voltage.

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LEDS can handle much more PEAK current than steady state. Study the LED datasheet and then PWM the LED within its PEAK duty cycle limits and you then will not need a resistor

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This LED has 160mA **Absolute Maximum Ratings**(!) at 10% duty cycle. AMR means that it should not work continuously under the given conditions. How are you going to limit the current to less than 160mA without series resistor? –  stevenvh Apr 17 '12 at 12:58

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