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HOW DOES THIS WORK?? Bench light circuit has me stumped...

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JohnPen:
An approximate calculation suggests that 4 volts across C2 is correct for a 120 vac supply.  The impedance of of the 0.7 ufd at 60 hz is about 3.8K plus a small adjustment for the parallel 200K. 

Prehistoricman:
You have a ground connection on the negative side of your AC voltage source. You probably don't want that there. Even if you're calling that neutral and neutral is connected to earth, it still isn't ground in your light circuit.



--- Quote from: engrguy42 on March 27, 2020, 09:39:04 am ---But I'm still wrestling with the fact (?) that there should be a 120V full wave signal across the C2 at the output of the rectifier, right? But it's about 4v instead. So maybe the parallel RC at the bridge input on the AC side knocks down the 120VAC or something? And that damn D5 diode between the rectifier output and the battery negative just makes me insane.  |O

--- End quote ---
We keep telling you! It's a capacitive dropper. That phrase means that it drops the current/voltage using a capacitor. Bigclive has shown this circuit many times, and he likes opening up exactly this kind of product, so you should take a look at some of his vids.

D5 provides a return path for the current going into the LEDs when the unit is charging.

engrguy42:

--- Quote from: Prehistoricman on March 27, 2020, 11:24:20 am ---You have a ground connection on the negative side of your AC voltage source. You probably don't want that there. Even if you're calling that neutral and neutral is connected to earth, it still isn't ground in your light circuit.



--- Quote from: engrguy42 on March 27, 2020, 09:39:04 am ---But I'm still wrestling with the fact (?) that there should be a 120V full wave signal across the C2 at the output of the rectifier, right? But it's about 4v instead. So maybe the parallel RC at the bridge input on the AC side knocks down the 120VAC or something? And that damn D5 diode between the rectifier output and the battery negative just makes me insane.  |O

--- End quote ---
We keep telling you! It's a capacitive dropper. That phrase means that it drops the current/voltage using a capacitor. Bigclive has shown this circuit many times, and he likes opening up exactly this kind of product, so you should take a look at some of his vids.

D5 provides a return path for the current going into the LEDs when the unit is charging.

--- End quote ---

Yeah, but the capacitive dropper is dropping relative to what?? I'm thinking it's a voltage divider, but I'm not seeing what it's dividing with...

Yeah, I never watched bigclive, but noticed he's got 3,567 or more videos dealing solely with lights. Wow. When I get a couple of months free I'll scan them  :D

As far as the ground, with LTSpice I'm never quite sure what it needs for grounds. Sometimes it complains.

Anyway, apologies if I don't get this right away. I've been drawing and redrawing and trying to figure the details of this for a couple of days and I think I need to step away from it for a bit and come back at it fresh. I get to where I think I understand something, then simulate it, then grab the meter, then realize there's a tiny trace running under a resistor on the real board, etc.  |O

engrguy42:
Prehistoricman sez: "D5 provides a return path for the current going into the LEDs when the unit is charging."

BINGO !! Thanks so much. Makes perfect sense !!!!

You get 10 thumbs up for this  :-+ :-+ :-+ :-+ :-+ :-+ :-+ :-+ :-+ :-+

g0fvt:
As others have already said C1 is used as a capacitive dropper, R1 is only in circuit to bleed charge from the capacitor. All pretty normal so far, as others have said the "common" connection is misleading, I believe that D5 and D7 should be zeners and not 1N4007s. FWIW I believe that you have the collector of Q2 identified as going to the wrong place, I believe it is supposed to go to the base of Q1. This would then ensure that whilst charging that the LEDs are off due to the base of Q1 being clamped low. TBH a few aspects of the circuit may be misdrawn. Sometimes a single connection in the wwrong place can make reverse engineering very difficult.

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