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I have some basic LEDs and recently purchased some 4MHz Quartz Oscillators and since they haven't arrived yet, I was wondering what would happen if I hooked up the crystals straight to an LED in DC?

My guess is it may explode from being continuously pulsed 4 million times a second but then hooking it straight into the power is infinitely faster as there is no delay and they don't explode.

If it does explode, is there any kind of key or calculation you can perform as to obtain how many oscillations are required to fry specific components?

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    \$\begingroup\$ The overall power dissipation is what will damage an LED, pulse width modulation where an LED is rapidly switched on and off is commonly used to drive them. Also quartz oscillators usually require some external components, the crystal is just a resonator. \$\endgroup\$ – jramsay42 Nov 2 '17 at 7:19
  • \$\begingroup\$ @jramsay42 Yes, fundamentally I'm going to hook it up to make the resonator, which is what I intend to use on the LED. I'm looking at Pulse Width Modulation now. Thankyou for the info! \$\endgroup\$ – user125207 Nov 2 '17 at 7:27
  • \$\begingroup\$ What about the datasheet of the LED? Is there any information about timing, minimal puls width and maximal frequency? LED are destroyed by too much current and too high internal temperature. If pulse frequency is too high for the LED, there will be still light from the LED, but the LED will be not fully dark between pulses. \$\endgroup\$ – Uwe Nov 2 '17 at 10:16
  • \$\begingroup\$ @Uwe Well I mean I've had them for so long off of some random eBay seller so I don't have any idea. As far as my knowledge goes, just the basic 5mm 2 pin LED's that blow at like 5v and light up red, green, blue or white. The stereotype. \$\endgroup\$ – user125207 Nov 2 '17 at 10:20
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High frequency will not explode an LED. it will simply turn it on and off very quickly. This is exactly what is happening in TV remote controls, albeit at 36kHz instead of 4MHz.

Normally, when you apply a frequency to an LED, when the signal goes high, the led turns on, and when the signal goes low, the LED turns off. Because of this, the perceived intensity corresponds to the duty cycle of the signal.

However, if one half of the period is shorter than the rise or fall time of the diode, then it will not raise to full power when the signal is high, and you can no longer compute the perceived intensity just from the duty cycle - the perceived intensity will be less than expected.

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  • \$\begingroup\$ Ah well that's not all that bad then haha. Thanks for the information. \$\endgroup\$ – user125207 Nov 2 '17 at 7:26
  • \$\begingroup\$ No problem! If this answer helped you, it would help me if you accepted the answer. \$\endgroup\$ – Billy Kalfus Nov 2 '17 at 7:27

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