NNNOOOOOO 0 Report post Posted August 29, 2010 light a 2 pin bi-color LED and charge 3 capacitors (maybe 7 soon). On the LED, in one setup, it is green, when voltage is reversed, it is red. Point is, electricity can flow through it 2 ways. Due to this, everything is connected in parallel. I have a resistor connected in series to the ground pin, and a diode is connected in series to the Vcc supply. The LED can withstand the parallel ports' power and current. Here's my question. If I remove the diode, when electricity returns to the port, will it destroy the port? Share this post Link to post Share on other sites
BuffaloHelp 24 Report post Posted August 29, 2010 Yes. It's a form of short circuiting. The electricity is flowing in a closed circuit.Usually a computer can be damaged by introducing unnecessary volts. However, some computers may have been equipped with protection such as micro surge protecting circuits. If you have a spare computer that you can afford to destroy, try it. I had an old 486 computer that I submerged in the bath of mineral oil and ran for 7 days. No short circuits, even mineral oil can conduct electricity. That little experiment told me that motherboard was completely sealed. The only thing was not submerged was the AT power supply. The CPU fan created interesting swirl. Had I not done that I would have never known.I searched and obtained this information that the parallel port can produce load up to 2.6 mA at 2.4 v. While this is not enough to fry your computer, other capacitors such as ones near CPU for CPU power regulating can be damaged. Share this post Link to post Share on other sites
NNNOOOOOO 0 Report post Posted August 29, 2010 Ok. Now I have a new question. I went by radio shack and bought a PCB (one of them basic ones). After some soldering, I had 4 capacitors and 2 wires on it. Now the capacitors ain't charging and drains all energy from any circuit I put it in. Why is that? Share this post Link to post Share on other sites
