| ??? 12/27/05 14:09 Modified: 12/27/05 14:24 Read: times |
#106013 - Dangerous! Responding to: ???'s previous message |
This is a typical example of a circuit which should not be built in this way!
How does it work? Or better, how is it intended to work? Capacitor works as AC impedance to limit the current through the LEDs. Impedance of capacitance of 220nF is 1 / (2 x pi x f x C) = 1 / (2 x pi x 50Hz x 220nF) = 14.5kOhm. So, when 230VAC drops across this cap, current is limited to Ieff = 230Veff /14.5kOhm = 15.9mAeff. The advantage of using a cap instead of a resistor is, that at cap no heat is produced. When using a resistor of 14.5kOhm instead, on the other hand, about 3.7W would be dissipated! The big disadvantage of using a cap is, that at higher frequencies the impedance drastically lowers. So, each voltage spike on mains will result in a heavy current spike through the LEDs, which will shorten their working life. Due to the AC character of mains, only half of the shown LEDs will shine at at time, because either the current flows from the left to the right or vice versa. The metal oxid resistor of 100Ohm shall act as an additional current limiter for voltage spikes, when the capacitor looks like a short circuit, I guess. But assuming a spike voltage of only 100V dropping across this resistor, this will result in a current spike of terrible 100V / 100Ohm = 1A. And there are much higher voltage spikes on mains... What is totally ridiculous, is, that this resistor shall act as a fuse. Well, I have made some tests a while ago, to find out what overstressing a simple metal film resistor (minimelf 0204) can withstand. And I can tell you, that this little beast couldn't be damaged even by excessive overcurrents! Instead of smelting of metal film, the beast first began to glim, then to shine brighter and brighter, working finally as a lamp!! So, to make the 100Ohm/1W resistor to work as a fuse much much more than SQRT(1W / 100Ohm) = 100mA must flow! So, not the resistor will work as a fuse, but the LEDs themselves... What are the no nos of this circuit? 1. You are not allowed to connect standard capacitors to mains, which seem to show a sufficient nominal voltage. So, you must never connect a 400V capacitor to 230V mains!! Even a 630V cap isn't suited. Only special mains caps can be connected to mains, for instance 250V~X2-cap, which is made of a special metal paper. Only this type of cap will resist excessive overvoltages on mains, others can begin to burn!! 2. It's ridiculous to build this circuit without a suited varistor across the mains lines. Each voltage spike will enormously stress the LEDs. 3. Current limiting resistor is much too low to actually provide a current limiting. I would increase its value and bend a temperature fuse arround it, connected in series to resistor. 4. Why not spending a real fuse to this circuit, additionally to the "current limiting" resistor? Would be much better, than to trust on a burning resistor... Kai |
