Is stepping down 110V mains to drive LEDs with a cap efficient? - Page 1
Forum section: electronics / projects

author: jarrodhroberson
author_rank: Contributor
timestamp: February 09, 2015, 12:00:15 am
content: I was reading this article ->http://www.instructables.com/id/Using-AC-with-LEDs-Part-2-and-make-this-handy-/and I know that just plopping a resistor in there will just generate heat and pretty much cancel out any cost savings of using LEDs.Does that CAP solution fair any better in not wasting power/money?

author: IanB
author_rank: Super Contributor
timestamp: February 09, 2015, 12:07:18 am
content: Let me quote from your link:QuoteYes, it does work the way they say. Capacitors can limit the current without generating heat.

author: jarrodhroberson
author_rank: Contributor
timestamp: February 09, 2015, 12:23:44 am
content: Thanks for clarifying, the"without the (or as much) heat"confuses the issue.That semantic implies that there is some heat/waste just not as much but it does not specify an actual value or percentage.So it does not generate any heat, does any significant percentage of the power go to waste?Does this explain how you get LED night light bulbs in tiny packages that replace the same size incandescent bulb package and they can still claim the extreme cost savings?

author: IanB
author_rank: Super Contributor
timestamp: February 09, 2015, 12:45:08 am
content: A capacitor by itself would in theory not consume any power or generate any heat. However, in the example they did combine the capacitor with a lower value resistor and the resistor will still waste some power, just not as much as before. Putting a resistor in there is helpful to moderate the peak inrush current through the capacitor.Using a capacitor to limit the current will lead to a bad power factor (something the electricity company doesn't like), but for very small things like LEDs the power factor problem is inconsequential.

author: mariush
author_rank: Super Contributor
timestamp: February 09, 2015, 12:56:54 am
content: Pretty much the same thing is explained here:

author: jarrodhroberson
author_rank: Contributor
timestamp: February 09, 2015, 01:06:40 am
content: Thanks for the video, that makes it all make sense!

author: dannyf
author_rank: Super Contributor
timestamp: February 09, 2015, 01:13:36 am
content: QuoteIn the simplest sense, yes, as an ideal capacitor is "lossless".No in the sense of VA vs. Watt. Google "apparent power" for more.

author: Psi
author_rank: Super Contributor
timestamp: February 09, 2015, 04:05:43 am
content: It gets a bit weird when you use this kind of power supply to run ICs and switch stuff. You end up burning off the energy you don't use but only up to the max current which can flow through the cap.So it can be efficient or not depending on your cap value and how much energy your using at a specific time

author: DanielS
author_rank: Frequent Contributor
timestamp: February 09, 2015, 04:24:40 am
content: Quote from: Psi on February 09, 2015, 04:05:43 amThe energy stored in the cap gets returned to the grid on the next polarity change, so the only losses there are conduction. Unless you meant unused power as in putting a shunt regulator in the load to prevent rectified voltage from exceeding the load's operating voltage when it is not drawing sufficient current to keep voltage down through normal operation.

author: Psi
author_rank: Super Contributor
timestamp: February 09, 2015, 04:31:49 am
content: yep, i meant with most cap psu implementations there's a zener/bridge.Either the current flows through your load or through the zener.