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To control the brightness of a LED often pwm is directly used as input to the LED. Does this on/off turning by the pwm has any negative effect on the live expectation of the LED?

Would it be better for the MTTF (mean time to failure) to smoothen the current by adding a capacitor?

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  • \$\begingroup\$ So, is the idea that the capacitor would 'average' the voltage so that the LED would see that 'average' voltage, and not the PWM square wave? \$\endgroup\$ – gbulmer Dec 18 '15 at 10:38
  • \$\begingroup\$ @gbulmer: That is exactly what I expect (at least if the capacity is high enough) \$\endgroup\$ – MrSmith42 Dec 18 '15 at 10:39
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    \$\begingroup\$ Maybe if the PWM were some really low frequency you could get junction temp modulation leading to derceased reliability from thermal cycling. \$\endgroup\$ – Autistic Dec 18 '15 at 11:12
  • \$\begingroup\$ Note also that LEDs are most efficient way below I[max]. So, for efficiency, the capacitor can be beneficial. \$\endgroup\$ – JimmyB Dec 18 '15 at 12:07
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One potential problem with using a capacitor to 'smooth' a PWM voltage for the LED, is the LED has a minimum forward voltage before it turns on sufficient to be visible.

Its brightness is not controlled by voltage, it is controlled by current, and the amount of time it is switched on (i.e. the PWM duty cycle).

The capacitor might reduce the 'smoothed' PWM voltage below the minimum forward voltage, so the LED would no longer be visible, even though it would be visible using exactly the same PWM signal directly (without the capacitor).

So it would reduce the brightness range over which the LED can be controlled.

AFAIK, the bigger killer of LEDs is heat leading to a significant temperature rise, and not switching.

Typically we want to drive an LED with a constant current (or something near, e.g. a resistor), so that it is protected from too much heat leading to temperature rise and permanent damage. Edit: Depending on how the capacitor is connected, a capacitor may actually reduce the effectiveness of the constant current circuitry.

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  • \$\begingroup\$ So all the on/off switching does not really matter to the MTTF? Only the heat witch should be similar if I smooth the current or not? \$\endgroup\$ – MrSmith42 Dec 18 '15 at 10:53
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    \$\begingroup\$ To a 99% approximation, yes. Switching on and off may have a slight impact vs exactly constant current, but temperature rise is a much worse problem. \$\endgroup\$ – gbulmer Dec 18 '15 at 11:00
  • \$\begingroup\$ Many LEDs provide specs specifically for switched operation - i.e. I have a bunch of 20mA (continuous) standard LEDs and they are blessed for operation up to to 100 mA if it's 0.1ms (or less) pulses at 1/10 duty cycle (or less) datasheet.octopart.com/MC20380-SPC-datasheet-81811.pdf \$\endgroup\$ – Ecnerwal Dec 18 '15 at 14:18
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    \$\begingroup\$ Agree; there are OSRAM IR emitters which have can handle 0.1A continuous, but upto 1.5A at a low-duty-cycle. Do you think I should add this to the answer? \$\endgroup\$ – gbulmer Dec 18 '15 at 14:36
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Would it be better for the MTTF (mean time to failure) to smoothen the current by adding a capacitor?

This is usually done with an inductor: -

enter image description here

The inductor will smooth the current into the LED. Also note that average current is maintained by the sense resistor attached to the chip in the diagram. A capacitor will not smooth the current into an LED unless there is a series component like a resistor or inductor.

If you put raw PWM voltages across an LED it is likely to destroy it. Observe the LED spec regards current. Here's other examples: -

enter image description here

This one doesn't use an inductor but relies on the transistor arrrangement to regulate max current into the LED: -

enter image description here

LED forward conduction characteristics: -

enter image description here

At 2V across this "typical" LED the forward current is 20 mA. With only 1.7 volts applied there's hardly any current and the LED will be very dim. If you applied 2.5 volts, the current is off the scale and the LED is damaged. But some LEDs are designed for 1A I hear someone say and that is true but applying a few tens of millivolts more than the recommended value will kill it nonetheless. The LED brightness should always be controlled by current.

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  • \$\begingroup\$ I get it. I should not drive the led directly with the PWM signal. But does all the on/off switching not really matter to the MTTF? \$\endgroup\$ – MrSmith42 Dec 18 '15 at 10:55
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    \$\begingroup\$ Of course it matters - I'll add a picture of the characteristic of an LED. \$\endgroup\$ – Andy aka Dec 18 '15 at 10:56
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It is unnecessary to smooth the current.

If you are PWMming at a rate faster than the eye can follow, any heat cycling effects will be negligible. Smoothing the current might give a very small (but unnoticeable) increase in brightness because an LEDs emissivity vs. current is not perfectly linear, especially at higher currents. If you are adjusting the PWM to control brightness, then it's best NOT to filter it because a) effective brightness will not be linear with PWM duty cycle, and b) LED colour changes slightly with current, so you would get a colour shift also. This would be most noticeable in white LEDs.

TVs and displays that use LEDs all PWM them without filtering at 50/60 Hz or faster.

LEDs have some internal series resistance, and overall power efficiency would be higher by filtering the current, but if you are regulating current with an external resistor or current source, there is no difference (same average current consumed from supply)

If you are controlling a single LED (or single string) from a DC/DC converter, the DC/DC will smooth the current anyway. This is optimal for efficiency. TV screens use DC/DC converters to generate a sufficient voltage, but then PWM the current to maintain colour and brightness accuracy.

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    \$\begingroup\$ You might want to clarify your statement, "LEDs have some internal series resistance". To the uninitiated it sounds as though you're saying that they have internal resistors (which is false). \$\endgroup\$ – Transistor Dec 18 '15 at 15:56
  • \$\begingroup\$ OK; LEDs are not perfect diodes, and the unavoidable internal parasitic resistance in the physical structure (which is in series with the 'diode') can have a small impact on the overall power efficiency at higher currents. \$\endgroup\$ – jp314 Dec 18 '15 at 20:53

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