Look at the schematic...

[Clarkdale44]

**broken link removed**

What's the use of 5th diode in this schematic..(which is in parallel with cap)?
This is a circuit to drive led from mains AC.

Sorry if this question is too simple or obvious.. I just want to know what that diode's doing there.

 

[Veraxis]

Is there any chance that it is a zener diode for clamping the DC voltage at the output? it uses a 25V capacitor, so if the LEDs went open circuit, the capacitor would get overloaded normally, but the zener would clamp the voltage down.

 

[Nigel Goodwin]

It IS a zener diode, it gives a sensible DC voltage, and current is then set by the 47 ohm.

 

[Clarkdale44]

Is there any chance that it is a zener diode for clamping the DC voltage at the output?

It is not zener.... it is clearly written top right "1n4007 x5", 4 used for bridge rectifier and last one is parallel with capacitor.

It IS a zener diode

It is not zener..

 

[GromTag]

some kind of frequency tank?

 

[GromTag]

Concept of thought, that 2.5V 100uF capacitor will encounter up to full mains voltage if output becomes disconnected. That aside it is configured as a frequency filter, a smoothing function.

:Edited: "may" encounter to will encounter mains voltage. - if not all ready known -

 

[Clarkdale44]

2.5V 100uF

It is 25v 100uf

 

[jpanhalt]

The "image address" **broken link removed** goes to an interesting page...

Maybe it would help to tell us where you got the design?

Maybe it was intended to limit the voltage to a forward biased diode, but was drawn backward?

John

 

[Clarkdale44]

Maybe it would help to tell us where you got the design?

I thought it wouldn't necessary to share the source...
I got it from youtube video, the guy's schematic looked interesting so...

 

[GromTag]

Was thinking about the capacitors voltage on 2 points, 1 the low state voltage of around 23V when driving low loads 1K range and the avg LED not favoring the lower voltage..

 

[jpanhalt]

An accompanying YouTube has the polarities marked correctly:

I think it is just another meaningless YouTube circuit (sorry, my snippet was before author drew other supply line).

John

 

[Veraxis]

Hmm, I suspect then that the fifth 1N4007 is a mistake. A rectifier diode wouldn't really do anything in that circuit. This circuit was likely reverse engineered from the inside of an ordinary "capacitive dropper" type LED light bulb, and the person doing the reverse engineering didn't realize that the fifth diode was different from the others.

Given that the zener is only for clamping the voltage in case of the LEDs going open circuit, it would operate normally as long as the LEDs were connected. However, if something ever went wrong, the capacitor would get overvolted and probably go up in smoke or pop like a firecracker.

It's also worth pointing out that you probably shouldn't handle that circuit board the way the person in the video does. There is no mains isolation in a capacitive dropper circuit, so that circuit board is effectively live at mains voltage!

 

[Nigel Goodwin]

It is not zener.... it is clearly written top right "1n4007 x5", 4 used for bridge rectifier and last one is parallel with capacitor.

It is not zener..

I still disagree, it SHOULD be a zener, and having now heard the origin of this hand drawn circuit it's likely that it's been drawn out wrongly - or it was 'designed' (and I use the term VERY loosely) by someone with no knowledge of what they were doing.

 

[Clarkdale44]

I guess you both seem right... With zener there instead of 1n4007 can it help protect capacitor in case something goes wrong?

 

[Veraxis]

Yes, replacing the fifth diode with a zener would work better, although for the values shown you would probably need a fairly high power zener (>=2W probably) to dissipate the current. 1uF is quite high for a 220V mains capacitive dropper. The person in the video doesn't specify the voltage drop across that LED board, but I count 17 LEDs, so my guess is that board is more likely to be a 5V board meant to be run from a USB, which would have all of the LEDs in parallel and would be very inefficient (having them in series would have a Vf~=51V, which would be too high for the capacitor and I think would have a current of around 50-60mA). Having all the LEDs in parallel would also have very poor mains power factor.

Capacitive droppers are more efficient and have better (relatively) power factor when run at lower current (i.e. lower value on the mains-side film cap) through a higher voltage series string of LEDs. Of course, you would also need to use a higher voltage DC filter capacitor and zener, so some LED bulbs just omit the zener and just use a 400V-rated capacitor (the extra-cheap ones just omit the capacitor entirely and are super flickery). For a 120V (170Vpeak) mains voltage, one could use a lower voltage cap but you would need a higher value on the film cap for the same current.

Quite frankly, this isn't the best or most efficient example of a capacitive dropper circuit, so you may want to re-work the values if you plan to build one.

 

[Nigel Goodwin]

I guess you both seem right... With zener there instead of 1n4007 can it help protect capacitor in case something goes wrong?

With the zener there it's a transformer-less PSU, with just a diode (that has no function whatsoever) it's a disaster waiting to happen.