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ViVBling
modified 9 years ago

LED-Circuit for 230VAC

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19
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04:55:23
A lossless circuit using the capacitive resistance of capacitors instead of resistors. It can be used at nearly any size of voltage as long as your diodes can handle that voltage. The simulation uses 230VAC. Close the switch to bridge the extra resistor of 470Ω.
published 9 years ago
hurz
9 years ago
A 1kOhm resistor in series to the supply will help to protect the diodes against high power on current. Have fun.
ViVBling
9 years ago
@hurz I built this circuit in real life and it works fine to me. But thanks. ☺️
BillyT
9 years ago
230vac rms = 230 * 2.828 or approximately 679 Volts peak to peak. Why not use a led as the capacitor bleed. http://everycircuit.com/circuit/5549613297172480
2ctiby
9 years ago
In this circuit shown above, there is approx 3mA total circuit usage for the half wave LED and similar usage for the half wave diode. In a conventional setup using just a resistor with the LED instead of a cap and diode, there is approx 6mA used just for the LED half wave time only. So the total average energy, ( power throughout time) appears to be similar, all other things being equal. I am not an authority, please correct me if I have got it wrong: http://everycircuit.com/circuit/5311111213088768
hurz
9 years ago
@2cent, in average 260mW vs 3.2mW! @ViVBling, yes i belive you, it was working fine. However, it will come the day and it will fail. Check this and restart always the simulation. http://everycircuit.com/circuit/4598417224892416
rich11292000
9 years ago
The main's have high frequency distortion that passes through the capacitor. The capacitor does not react the same to these frequencies as it does with 60hz. That means current spikes. If you add this distortion to the simulation you will see the issue.
hurz
9 years ago
50 or 60Hz doesnt matter a lot. But right, higher frequencies might cause high current, cuz impedance depends on frequency. And a jump from zero volt to 325 does contain in theorie all possible frequencies and thats why it cause a high current and might destroy the led. Better use a little extra resistor. The cap only is fine as long there are only 50Hz.
2ctiby
9 years ago
@hurtz My total circuit usage calculations look reasonable to me. (Compare the shown amp meters for the equal applied total circuit voltage throughout the full unit wave time, ie energy not just a selected power usage) Can you please explain your calculations....I am not saying that you are wrong, but explanations are helpful. The oscilloscope there shows my point....the current wave area shape compares similar, so the peak values shown are a reasonable thing to compare.
rich11292000
9 years ago
My calculation gets 29.3 amps
rich11292000
9 years ago
http://everycircuit.com/circuit/5119647610830848
2ctiby
9 years ago
@rich11292000 you demonstrate a good point there....but it is a different ball game to the forementioned discussion where a comparrison of energy was relating to an ac 50 Hz sine like wave, not as your few microsec pulses with long intervals between to obtain the 50 per sec. A tablespoon of salt eaten suddenly will be different to taking that amount gradually throughout a year, even though both were given the same one year period.
rich11292000
9 years ago
[BLOCKED]
hurz
9 years ago
My calculation is for what i have to pay. You checked what the total costs are, but we as end-consumer don't care. We just pay what our powermeter inhouse measures. And that is a big difference in both circuits.
2ctiby
9 years ago
Thanks Hurz, I began to see that when I was writing here about the possibility of the resistor power being much more than the capacitor usage.....and thanks for a sensible unobtrusive response.
hurz
9 years ago
Self-knowledge, is the best way to improvement!
2ctiby
9 years ago
Yes indeed, and when I see that people have developed in kindness, helpfulness and patience it is very encouraging.
zorgrian
9 years ago
2ctiby = new @bugger
mdsahinurislam567gmailcom
9 years ago
awesome
hurz
9 years ago
[BLOCKED]

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