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| HOW DOES THIS WORK?? Bench light circuit has me stumped... |
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| g0fvt:
Just FYI the capacitive dropper works simply, the capacitor will have a given impedance at 50/60Hz. For a given applied AC voltage you will get X amount of current flow (with phase shift). The reason capacitive droppers are used in AC circuits is that they are "wattless". IE a resistor with perhaps 100v rms applied and a current flow of 100mA would need to dissipate 10w, a substantial amount of heat to deal with in a tiny circuit. A suitable sized capacitor could have the 100mA current flow with no heating. |
| engrguy42:
Ahhh...okay...I saw a website discussing capacitive droppers and I think what you guys are saying is that the capacitance is the major player, providing an almost constant current or current limited source to the LED's. So the LED's are almost like what the capacitor is "tuning" to. Since you're down at such low values of amps (in this case like 120VAC divided by the 3k or so impedance of the input capacitor), the voltage on the secondary of the rectifier is a bit irrelevant, since what's being delivered is low amps (in this case like 360mA) to the LED load. Okay, even if I'm wrong, that's good enough for me :-DD |
| engrguy42:
--- Quote from: g0fvt on March 27, 2020, 12:14:29 pm ---Just FYI the capacitive dropper works simply, the capacitor will have a given impedance at 50/60Hz. For a given applied AC voltage you will get X amount of current flow (with phase shift). The reason capacitive droppers are used in AC circuits is that they are "wattless". IE a resistor with perhaps 100v rms applied and a current flow of 100mA would need to dissipate 10w, a substantial amount of heat to deal with in a tiny circuit. A suitable sized capacitor could have the 100mA current flow with no heating. --- End quote --- Thanks. Yeah, the transistors have no markings whatsoever, so I'm kinda guessing on what they're doing. And also with regards to Q2, I'm not showing a resistor that gets inserted between + and the base of the resistor when the switch goes into LOW mode, like it is with Q1 (the circuit is shown in HI position). I assumed that's just a biasing resistor to allow you to turn on Q2, and when it's switched out then Q2 can't turn on. And when it does turn on in LO mode it pulls more current thru the LED's? Makes no sense, but it's all I could figure. |
| engrguy42:
Okay, I greatly simplified my circuit to include only the AC source and rectifier, feeding a filter cap and a resistive LED load. And with the right choice of LED ohms (in this case 300), I get a very ripply 4 volts across the LED. Beauty. So I guess it's just a matter of selecting the right C for the load. Wow, what a piece of crap. :D And to think I was originally gonna not even include the RC input since I thought it was just a filter for the AC. :D |
| Prehistoricman:
These kinds of products are built down to such a price that they need every sneaky transistor/diode trick they can use. And that can make them surprisingly hard to analyse for being simple. You say it doesn't charge the batteries if it's plugged in and the lights turned on, right? That would be an interesting thing to investigate. Why did you use a resistor in the schematic instead of the LED array? You can use a single diode and edit its properties to make it behave like the whole array. |
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