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The drive electronics must include two basic functional elements: a power conversion circuit (essentially a transistor switch that rapidly turns on and off) and a sensing circuit, which monitors the average current through the LEDs and provides a feedback signal to regulate the proportion of time that the power-conversion switch remains turned on. In many cases, a transformer is used to change voltages and to isolate the LED from the high-voltage electrical mains.
Yokohammer wrote:Wage Slave,
You put some effort into it ... I am suitably impressed ...
But that circuit does not inspire confidence. In addition to the fact that it's off-the-shelf parts, it looks like it relies on the overall resistance of the circuit to drop the line voltage to a level that's somewhere in the ballpark for the LED, which is a pretty iffy way to go about it. There's no regulation, so if there's a bit of a spike or surge the LED (and possibly the diodes) could quite easily go pffft.
High quality LED lightbulbs are a bit more complex. There's some good information at the link:
http://spectrum.ieee.org/energy/environment/the-subtle-circuitry-behind-led-lightingThe drive electronics must include two basic functional elements: a power conversion circuit (essentially a transistor switch that rapidly turns on and off) and a sensing circuit, which monitors the average current through the LEDs and provides a feedback signal to regulate the proportion of time that the power-conversion switch remains turned on. In many cases, a transformer is used to change voltages and to isolate the LED from the high-voltage electrical mains.
Russell wrote:I agree with Wage Slave that the capacitor is doing most of the voltage dropping. Isn't the big resistor across it just to discharge the capacitor when power is switched off?
Samurai_Jerk wrote:This is officially the must boring thread in the history of FG.
Yokohammer wrote:Wage slave ... do you have an oscillator and an AC voltmeter handy?
Yokohammer wrote:Russell wrote:I agree with Wage Slave that the capacitor is doing most of the voltage dropping. Isn't the big resistor across it just to discharge the capacitor when power is switched off?
Trouble is (you know the resistance of a capacitor to AC is frequency dependent, right?) that series 0.47µF cap is going to be pretty much transparent to the AC at normal line frequencies. It depends on the impedance of the load, but I'm pretty sure it's only going to start attenuating the AC much higher frequencies.
0.47µF is a common value in audio tone control circuits. I.e. it works nicely at audio frequencies (once again, impedance dependent), but 50/60 Hz is getting pretty far out.
We'd need to either know the load impedance or actually measure the thing to be sure, but I think that's what's happening. Or at least that was the design intent.
Wage slave ... do you have an oscillator and an AC voltmeter handy?
Or ... I suppose I could see if I can find one of those bulbs at my local Daiso and do it myself (I can't believe what I'm getting myself into ...).
Yokohammer wrote:Samurai_Jerk wrote:This is officially the must boring thread in the history of FG.
SJ! I am shocked!
You mean to say you're bored by the subtleties of electronic circuit design?
(EDIT: This is why FG.com needs a geek section similar to the one I set up at the Café ... Yanpa?)
Wage Slave wrote:Yokohammer wrote:Wage slave ... do you have an oscillator and an AC voltmeter handy?
You read my mind.
I will put a voltmeter on it but an oscillator I don't have. Nor an oscilloscope - I have almost bought one several times - perhaps I will but I don't really have enough use for one.
Where do you want to measure? Voltage drop across the capacitor?
Russell wrote:Yokohammer wrote:If the capacitor would be transparent at 50/60 Hz, you could replace it by a wire, and the resistor in parallel with it would not make sense.
But it isn't transparent. Using this capacitive reactance calculator, I get an equivalent resistance of 5.6 k Ohm at 60 Hz, and 6.7 k Ohm at 50 Hz, so let's say we are talking about 6 k Ohm here. This is much less than the 470 k Ohm resistor in parallel with it, so this resistor can basically be ignored. I maintain my opinion that it is just for discharging the capacitor when power is switched off.
What the two 150 Ohm resistors are for is becoming more and more a mystery to me. The same current goes through both of them, if Wage Slave's schematic is correct (and I think it is, judging by the pics he posted), but one is likely rated for a higher power than the other, given their size differences. That does not make sense at all. If they have a function at all, they could as well be replaced by a single resistor.
Yokohammer wrote:Low tech is great if it's reliable (and compared to complex "high-tech" stuff it often is), but in this case I see a couple of problems with the circuit you traced for us.
You got full-size bulbs for 400 yen? That's an example. I don't think I've seen 'em that cheap around here.
Wage Slave wrote:The big worry remains, as you say, the lack of isolation from the mains. For an application like a led bulb that is acceptable though because there is no way that the user can touch any part of the circuit while it is in use. For a charging device that is not true of course so on further reflection this design would be a lousy idea at any price.
Coligny wrote:I have one of those:
extra limited, some only goes up to 80 volt. But good for signal watching on the whips and cheap enough to stay in the car...
Yokohammer wrote:Russell wrote:Yokohammer wrote:If the capacitor would be transparent at 50/60 Hz, you could replace it by a wire, and the resistor in parallel with it would not make sense.
But it isn't transparent. Using this capacitive reactance calculator, I get an equivalent resistance of 5.6 k Ohm at 60 Hz, and 6.7 k Ohm at 50 Hz, so let's say we are talking about 6 k Ohm here. This is much less than the 470 k Ohm resistor in parallel with it, so this resistor can basically be ignored. I maintain my opinion that it is just for discharging the capacitor when power is switched off.
What the two 150 Ohm resistors are for is becoming more and more a mystery to me. The same current goes through both of them, if Wage Slave's schematic is correct (and I think it is, judging by the pics he posted), but one is likely rated for a higher power than the other, given their size differences. That does not make sense at all. If they have a function at all, they could as well be replaced by a single resistor.
Yes, the resistor is probably just a bleeder resistor for the cap. If my theory is correct it wouldn't have much effect at bypass frequencies.
Having built numerous audio and guitar amplifiers in the past, I'm not proud to admit that I have had direct physical encounters with 100 VAC coming through 0.47µF caps (we use 'em for hum cancelling in the power switch sometimes). It hurts like hell. You don't get much current (or I'd probably be dead), but the voltage sure is there. The only thing that stops me from proclaiming that I'm definitely right in this case is the unknown load impedance. There's no way to say precisely at what frequency a cap will start attenuating AC unless you know the load impedance (which makes your reactance calculator results kinda suspicious, methinks).
Yokohammer wrote:Coligny wrote:I have one of those:
extra limited, some only goes up to 80 volt. But good for signal watching on the whips and cheap enough to stay in the car...
Nice.
Oh shit, when the geek talk starts around here all hell breaks loose.
Related but a bit off topic, I just picked up a neat little OBD II code reader for the Beamer.
I'll stop now before the thread becomes so diffuse nobody can figure out what the heck is going on, even us geeks.
Russell wrote:This is much less than the 470 k Ohm resistor in parallel with it, so this resistor can basically be ignored. I maintain my opinion that it is just for discharging the capacitor when power is switched off.
What the two 150 Ohm resistors are for is becoming more and more a mystery to me. The same current goes through both of them, if Wage Slave's schematic is correct (and I think it is, judging by the pics he posted), but one is likely rated for a higher power than the other, given their size differences. That does not make sense at all. If they have a function at all, they could as well be replaced by a single resistor.
link above wrote:The diagram shown below is a simple transformer less power supply. Here 225 K(2.2uF) 400 volts X rated capacitor is used to drop 230 volt AC. Resistor R2 is the bleeder resistor that remove the stored current from the capacitor when the circuit is unplugged. Without R2, there is chance for fatal shock if the circuit is touched. Resistor R1 protects the circuit from inrush current at power on.
Yokohammer wrote:1) Voltage across the output of the two 150Ω input resistors (before the resistor/cap).
2) Voltrage across the output end of the cap and the output of the other input resistor (i.e. across the bridge input).
... and ...
3) No 2) above with the output of the cap/resistor disconnected from the bridge.
That should give us a few clues.
Coligny wrote:I have one of those:
extra limited, some only goes up to 80 volt. But good for signal watching on the whips and cheap enough to stay in the car...
MOAR
http://dx.com/s/Oscilloscope
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