I broke apart a cheap flashlight to look at how it works. It is a rechargeable flashlight you can just stick into a wall supply, but you're supposed to do this only when the flashlight is in "OFF" state. By mistake it was being charged in "ON" state and seemed to stop working, hence the breaking.
I have drawn the circuit of the flashlight when it's charging:

Can you please explain how it is working? There doesn't seem to be stepping down, rectification, so how does it convert the AC to DC and charge the battery?

The battery is pretty big in size. Also, R1 in the circuit is quarter watt, 390 kohm, R2 is quarter watt, 300 ohm. The capacitor C1 is rectangular in shape, not electrolytic. Thank you.

 

It uses a Half cycle method, by charging the battery voltage up on the reverse cycle via R1,D1,3 and lighting up the led.

 

It uses a Half cycle method, by charging the battery voltage up on the reverse cycle via R1,D1,3, on the opposite cycle it lights up the led, via D2,3

Why would a designer choose this strategy? I mean, why not just use a different charger voltage in the first place?

 

Why would a designer choose this strategy? I mean, why not just use a different charger voltage in the first place?

Dunno, maybe the TS has drawn it wrong.... Seems stupid to charge a non electrolytic cap up on one cycle and do nowt with it, it's too small to be used as a Voltage doubler method ???

 

To try to understand this circuit, I whipped up an LTspice simulation. It's not making sense to me.

 

I broke apart a cheap flashlight to look at how it works. It is a rechargeable flashlight you can just stick into a wall supply, but you're supposed to do this only when the flashlight is in "OFF" state. By mistake it was being charged in "ON" state and seemed to stop working, hence the breaking.
I have drawn the circuit of the flashlight when it's charging:
View attachment 124673
Can you please explain how it is working? There doesn't seem to be stepping down, rectification, so how does it convert the AC to DC and charge the battery?

The battery is pretty big in size. Also, R1 in the circuit is quarter watt, 390 kohm, R2 is quarter watt, 300 ohm. The capacitor C1 is rectangular in shape, not electrolytic. Thank you.

If you posted that circuit as a project; the moderators would close the thread.

There's no mains isolation, I'm guessing you wouldn't find a wattless dropper downstream of a step down transformer.................

 

It uses a Half cycle method, by charging the battery voltage up on the reverse cycle via R1,D1,3 and lighting up the led.

Thank you for your response. Could you explain this "half-cycle method" in a little more detail?

 

It uses One Half of a sinewave voltage, so uses one Diode instead of using a Full wave bridge rectifier, its a cheaper method to drop Half of the Voltage.

 

If you posted that circuit as a project; the moderators would close the thread.

There's no mains isolation, I'm guessing you wouldn't find a wattless dropper downstream of a step down transformer.................

It's not a project. I would've imagined using a transformer, rectifier and regulator before charging a battery. I saw this circuit inside of a REALLY REALLY shabby flashlight, the "Made in You-Know-Where" kind, no disrespect. So, I don't know if this is a scam circuit that isn't meant to work, or it's some workable circuit that just overlooks aspects for smaller budgets.

It uses One Half of a sinewave voltage, so uses one Diode instead of using a Full wave bridge rectifier, its a cheaper method to drop Half of the Voltage.

Then what does the cap, D2 and D4 do? Thank you.

Dunno, maybe the TS has drawn it wrong.... Seems stupid to charge a non electrolytic cap up on one cycle and do nowt with it, it's too small to be used as a Voltage doubler method ???

So the cap is useless? Is the "voltage doubler" method used for charging batteries?

Why would a designer choose this strategy? I mean, why not just use a different charger voltage in the first place?

Thank you for your response. What's a charger voltage?

 

Then what does the cap, D2 and D4 do? Thank you.

Good question, unless you have drawn it wrong, i would say nothing,, although it looks like a simple voltage doubler, but i assume your circuit is powered by a transformer??

Can you post pictures of it?

 

Good question, unless you have drawn it wrong, i would say nothing,, although it looks like a simple voltage doubler, but i assume your circuit is powered by a transformer??

Can you post pictures of it?

There was no transformer inside the flashlight. I will post detailed pictures of it. It may take a while. Please be patient. Thanks.

 

There's no mains isolation, I'm guessing you wouldn't find a wattless dropper downstream of a step down transformer.................

Huh. I assumed there was a step-down transformer. In the simulation, you can eliminate the capacitor and it doesn't matter much. So I'm still confused about this. It uses 4 diodes, why not just make a bridge?

 

Thank you for your response. What's a charger voltage?

I meant the voltage from an external step-down transformer, a wall wart. Are you saying this thing plugs in directly to mains?

 

So the cap is useless? Is the "voltage doubler" method used for charging batteries?

The charge pump voltage doubler also works as a constant current supply.

 

Good question, unless you have drawn it wrong, i would say nothing,, although it looks like a simple voltage doubler, but i assume your circuit is powered by a transformer??

Can you post pictures of it?

The red and black leads go to a single LED. This is connected in series with the battery, a half watt, 1 ohm resistor and switch.
The two green leads go to two round prongs that can be inserted into the wall supply, this is the "AC in" in my drawing of the charging circuit.
The green and black leads go to an array of smaller light sources. Again, this is in series with the battery, the half watt 1 ohm resistor and switch. The switch has 3 modes, "SINGLE LIGHT", "OFF", "LIGHT ARRAY". The yellow and black leads go to the battery.

 

I meant the voltage from an external step-down transformer, a wall wart. Are you saying this thing plugs in directly to mains?

No wall wart, it's plugged directly into the mains. I am confused too...is this a scam?

 

In that case the Capacitor C1 is being use as a Transformerless Voltage Dropper, with the 390K resistor R1 discharging it, Very Dangerous to work on!!

 

No wall wart, it's plugged directly into the mains. I am confused too...is this a scam?

No, not necessarily. It's a valid technique.

 

Hello,

A wattless voltage dropper is a technique used in cheap stuff like your led tourch.
The circuit is not allowed here on this forum out of safety.
The User Agreement has a list of restricted topics.

Bertus