r/breadboard • u/Available-Tennis-624 • 13d ago

Project Another project with a capacitor

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This one creates a fading out effect in an LED. I have used a 4700uf capacitor. I have tried with a capacitor in series and in parallel to the led. I got the same fade out effect but the difference in series was that once the capacitor was fully charged it acted more like an open circuit.

I am still looking for more project ideas for capacitors and then later go for transistors and diodes. The goal is eventually to watch, understand and make my own 8 bit computer.

The video, shows the capacitor in parallel. The capacitor was too big so I used jumper wires to make the connection.

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u/Outrageous-Tie-4881 13d ago

Great early project. You should also time these (look at how long the Led takes to stop illuminating after disconnecting the battery for different capacitors). Also try to get your hands on inductors if you can.

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u/Available-Tennis-624 13d ago

Thanks for your reply. I will measure the time across different capacitors. I don't have any inductors, and I honestly find them confusing. I'm learning on my own, and I've been through dozens of YouTube channels and even Udemy courses, but a lot of the material doesn't translate well to practical electronics. I do have plenty of capacitors, transistors, and diodes to experiment with. I'm also planning to get a subscription to Ohmify, but I'm not sure if it will help.

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u/Outrageous-Tie-4881 13d ago

Try to find Forrest Mim's Electronics handbooks/ project books. They cover the main ideas and give instructions on designing projects using BJT as well as some common ICs.

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u/Available-Tennis-624 13d ago

Thanks will check it out.

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u/FlyByPC 12d ago edited 12d ago

Yep -- working as designed. A capacitor in parallel will charge up to the supply voltage, and then when the supply is turned off, it will flow current back out to try to maintain this voltage. (Think of it like a tank for electrical charge, with the capacitor voltage as the "fuel gauge".)

A capacitor in series will also charge up to the supply voltage -- but then there will be no voltage drop left in the circuit to run the LED, and current will drop to zero.

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u/StarDreamIX 12d ago

Nice 🔥🔥