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I am experimenting with inductor so I am looking for a simple circuit that will blink the LED with oscillator circuit. Basically turn on and off LED and finally diminish. How can I do so? For power input, I am using Raspberry Pi - 5V. Here is my circuit? what am I missing here?

enter image description here

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    \$\begingroup\$ This will likely get you a blown LED. \$\endgroup\$ – Sparky256 May 21 '18 at 5:39
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    \$\begingroup\$ 270mH is a quite big inductor, there's a reason these aren't used directly for energy storage. \$\endgroup\$ – immibis May 21 '18 at 5:48
  • \$\begingroup\$ how do I calculate the required value? Is my circuit wrong? \$\endgroup\$ – Bali Vinayak May 21 '18 at 6:02
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    \$\begingroup\$ I sense XY problem! \$\endgroup\$ – winny May 21 '18 at 6:50
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    \$\begingroup\$ "Is my circuit wrong?" Probably. But we don't know what exactly you want to do apart from "finally diminish". Try to describe in several, preferable simple, sentences what you want to happen. Like: "Press button: LED on. Release button: LED off after 2.65 seconds, Run from 3.7V Battery etc.". \$\endgroup\$ – Oldfart May 21 '18 at 7:11
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I assume you try to get use of the fact that in source free RLC circuit energy will flow back and forth between capacitance (C) and inductance (L) and therefore will blink your led. Well, this is not going to work because led is diode and diode as you may know conducts current in only one direction. So what may happen at best is your diode will slowly go off. However your did not limit led current so it may already be dead.

Below circuit on the other hand will blink the leds for you with frequency determined by RC time constant:

enter image description here

How does it work? Assuming capacitor were discharged and left transistor were ON first, the left bulb will be ON while right OFF. But when left trasistor is ON it also cherges its capacitor and when it is full charged it turn right tranistor ON. When this happens the base of left transistor goes negative due to right hand capactor discharge and therefor now only left bulb lights. The whole process repeads indefinetely.

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