well.. i think it will slow down de charging cos de bulb got resistance...
I believe it is because of the LED and resistor connection.Originally posted by GoldenHawk@Jan 31 2005, 09:28
Reposting (in summarized form :P )...
1. Created circuit as shown below.
2. Compensate 1 Farad capacitor with capacitor bank (total = 27+++µF)
3. Tested on 12v power supply (indoor adapter) - worked fine
4. Tested on car battery - worked fine
5. Started car - worked fine for about 2 minutes
6. Large capacitors (2200µF@16V@85°C) blew up in sequence, one by one
7. All that was left were my 1000µF@16V@105°C capacitors & Mica caps.
My question... WHY!!! :(
Please help... ThanX!
branded VS from Pivot, u need at least RM200...Originally posted by n305er@Jan 31 2005, 08:36
Wah... Wouldn't that costs more then buying a new and branded Voltage Stabalizer??
Dude... where did you find the LED in the circuit? :huh:Originally posted by Vincew@Jan 31 2005, 09:49
I believe it is because of the LED and resistor connection.
Normal LED circuit: You need to connect the positive of the battery to the positive of the LED THEN from the negative of the LED to the Resistor and the Resistor to the negative of the battery. Take note that this is serial connection.
(your circuit will sure to blow because your LED will burn as it is directly connected to the power source and not buffered by the resistor "it is not connected in serial")
However I do not recommend to put led to the VS but if you really want it. You can attached the positive of the LED to the positive of the VS , Then the negative of the LED to the resistor. Then the resistor to the Positive of the Capacitor.
The LED will light up as the capacitor charge up. It will turn off if the capacitor charge is full (Stop charging).
Again "Do it at your own risk". "DIY rule" :) Good Luck.