LEDs from 120VAC and understanding diode usage - Page 1
Forum section: electronics / beginners

author: GnatGoSplat
author_rank: Regular Contributor
timestamp: August 03, 2012, 06:34:58 pm
content: I was curious about how people run LEDs from a 120VAC source, so I consulted Google and I don't completely understand the possible solutions.The most popular involves using an anti-parallel diode. The author of this page goes into great detail:http://www.boat-project.com/electro/panellight.htmI'm not clear on how the anti-parallel diode actually works. When current flows in reverse (LED is reverse biased), is the anti-parallel diode acting as a sink, and that's why the LED sees none of the reverse current?Why would you use the anti-parallel diode instead of putting the diode in series and using it as a half-wave rectifier as suggested here:http://www.dutchforce.com/~eforum/index.php?showtopic=27520&st=0&pid=229984&#entry229984Is one solution more correct or preferable than the other? The latter with series rectifier is how my mind first envisioned the solution because I thought the diode would block reverse current and therefore doesn't have to act as a sink like an anti-parallel and wastes no current when LED is reversed biased, but the former using anti-parallel is by far more commonly suggested which makes me wonder if there is an inherent flaw in the latter suggestion.

author: IanB
author_rank: Super Contributor
timestamp: August 03, 2012, 06:39:59 pm
content: Quote from: GnatGoSplat on August 03, 2012, 06:34:58 pmThe linked boat project article explains in great detail how it actually works and what the anti-parallel diode is for and why it is there.There is nothing I can add that will explain it better than that article does. Have you read the article all the way through?(In case you missed the relevant part it is to avoid blowing up the LED by applying 120 V or more across it when it is reverse biased.)

author: GnatGoSplat
author_rank: Regular Contributor
timestamp: August 03, 2012, 06:44:04 pm
content: Yes, I did read it, but I think I lack some basic understanding because I'm not sure why a forward-biased diode in anti-parallel would prevent any current from reaching the LED. Only thing I can think of is the anti-parallel is acting as a current sink and loading down the circuit enough that no reverse voltage reaches the LED. If that's what it's doing, then I guess I do understand it. I think I'm just not confident on whether I am understanding it or not.Also, I'm definitely not understanding why it's better to use the anti-parallel diode over using a series diode as a half-wave rectifier.

author: IanB
author_rank: Super Contributor
timestamp: August 03, 2012, 06:49:36 pm
content: The LED has a reverse breakdown voltage that is quite low, such as about 5 V. So we don't want to apply more than 5 V across it in case we damage it.If we put two diodes in series in half wave rectifying mode, then the reverse voltage will be divided in two across the two diodes. So the LED could still get half of 120 V across it when the polarity reverses.On the other hand, if the anti-parallel diode is used the maximum reverse voltage seen by the LED will be the forward bias voltage of the anti-parallel diode. The LED will be quite safe.

author: SeanB
author_rank: Super Contributor
timestamp: August 03, 2012, 07:01:32 pm
content: Depends on what you are doing with the LED. If it is a simple indicator there is a current limiting resistor that limits the current to around 2mA, and the reverse diode is there to protect the LED from high reverse voltage, as they typically break down at around 5V reverse voltage. If you want less flicker you use 2 LED chips in anti parallel in the same package, so that each one lights alternately. Otherwise you use a bridge rectifier to drive it with pulsed DC.If you want higher current you will generally use a capacitor in series in place of the resistor, as this reduces the heat produced in the resistor, by virtue of it's reactance. This is used in cheaper lamps along with using a number of chips in series, generally 10-20 LED's in series to get a voltage of around 35V along with a capacitor of around 330nF to do the current limiting, along with a bridge rectifier and very little else.If you want higher power you will use a dedicated inverter and a transformer to generate the required voltage at around 1A to drive the high power LED units.

author: GnatGoSplat
author_rank: Regular Contributor
timestamp: August 03, 2012, 07:09:34 pm
content: I see, thanks for the explanation, Ian. I had actually wondered if that was the case (reverse voltage divided across the two diodes).Sean, I did read a little about using a capacitor to reduce current. Using a capacitor in that way is pretty new to me, I assume it's far more efficient due to less energy released as heat?

author: SeanB
author_rank: Super Contributor
timestamp: August 03, 2012, 07:30:02 pm
content: Yes it is for efficiency. Using a resistor to provide 20mA to a LED means you dissipate around 2.4W of heat in the resistor. With the capacitor the losses are almost nothing. There are drawbacks in that you need added series resistance and voltage clamps as the capacitor is very good at feeding all the noise and impulses through to the LED.