Simple AC Powered LED

[carbonzit]

Well, how much reverse current are we talking about here; microamperes? Not much, at any rate, I'll wager.

 

[ronv]

This paper appears in: Reliability Physics Symposium (IRPS), 2010 IEEE International
Issue Date: 2-6 May 2010
On page(s): 522 - 527
Location: Anaheim, CA
ISSN: 1541-7026
Print ISBN: 978-1-4244-5430-3
References Cited: 11
INSPEC Accession Number: 11501038
Digital Object Identifier: 10.1109/IRPS.2010.5488776
Date of Current Version: 17 June 2010


Abstract
This paper describes an extensive analysis of the degradation of InGaN-based LEDs submitted to reverse-bias stress and Electrostatic Discharge events. Results described within the paper indicate that: (i) reverse-bias current flows through localized leakage paths, related to the presence of structural defects; (ii) the position of these paths can be identified by means of emission microscopy; (iii) reverse-bias stress can induce a degradation of the electrical characteristics of the devices (decrease in breakdown voltage); (iv) degradation is due to the injection of highly accelerated carriers through the active region of the LEDs, with the subsequent generation/propagation of point defects; (v) the localized leakage paths responsible for reverse-current conduction can constitute weak regions with respect to reverse-bias ESD events.

I didn't want to pay to read the whole article but the implication is clear.

So the real question is where does the led suck up the light from when it is reversed biased. Just a really small spot like a black hole or just a little bit from the whole room?

 

[audioguru]

Well, how much reverse current are we talking about here; microamperes? Not much, at any rate, I'll wager.

The reverse current is slightly less than the forward current (19.9mA instead of 20.0mA). But since the reverse voltage is much higher then the heating is much higher.

 

[carbonzit]

I didn't want to pay to read the whole article but the implication is clear.

Perhaps, but the $64,000 question is, how much? Without quantifying this, we're just taking shots in the dark. Again, what sorts of leakage currents are we talking about here?

I think my initial guess was at least in the ballpark; I just checked a datasheet for some LEDs I have (medium-high-intensity red), and it gives a Vr of 5V (minimum, typ. = 20V) at an Ir of 100µA. So really not very much. I suspect this is not being exceeded, or at least not very much, by my line-powered LED over in that other thread.

So the real question is where does the led suck up the light from when it is reversed biased. Just a really small spot like a black hole or just a little bit from the whole room?

So it's a darksucker for each 1/2 cycle, followed by being a lightsucker? But the darksucker must be just a little bit stronger ...

 

[DerStrom8]

Without quantifying this, we're just taking shots in the dark.

CZ, I'm afraid you're the only one taking shots in the dark. Just about everyone else knows exactly what they're talking about.

 

[ronsimpson]

I think my initial guess was at least in the ballpark; I just checked a datasheet for some LEDs I have (medium-high-intensity red), and it gives a Vr of 5V (minimum, typ. = 20V) at an Ir of 100µA. So really not very much. I suspect this is not being exceeded, or at least not very much, by my line-powered LED over in that other thread.

"Ir of 100µA. So really not very much" The data sheet might say 100uA (at 5V) but the current has to be 20mA (at 8v) or whatever the forward current is. You can't pass DC through a capacitor. The forward and reverse current must equal.
As been stated before, the part is built for 20mA at 1.5V, it was never built for 20mA at 8 volts.

 

[carbonzit]

"Ir of 100µA. So really not very much" The data sheet might say 100uA (at 5V) but the current has to be 20mA (at 8v) or whatever the forward current is. You can't pass DC through a capacitor. The forward and reverse current must equal.

Huh? Why must the forward and reverse current equal? It's a diode. Since when do the forward and reverse currents equal, say in a typical power supply diode?

On the reverse-bias cycle, there's much less current than on the forward-bias cycle. The capacitor cannot pass DC, and the LED cannot pass reverse current (beyond Ir).

 

[audioguru]

Perhaps, but the $64,000 question is, how much? Without quantifying this, we're just taking shots in the dark. Again, what sorts of leakage currents are we talking about here?

I think my initial guess was at least in the ballpark; I just checked a datasheet for some LEDs I have (medium-high-intensity red), and it gives a Vr of 5V (minimum, typ. = 20V) at an Ir of 100µA. So really not very much. I suspect this is not being exceeded, or at least not very much, by my line-powered LED over in that other thread.

Reverse breakdown is not a little leakage current. It is when the junction has avalanche breakdown like a zener diode with only the capacitor's reactance limiting the current.
The reverse breakdown is spec'd with the maximum amount of current at a certain voltage, maybe it is 100uA at 5V when it barely breaks down then it might be 100mA at 6V and might be 1A at 7V.

Philips Lumileds says that their LEDs are not designed to survive a reverse bias.
When this major manufacturer says DON'T Do It then don't.

 

[ronsimpson]

If you look back several people tried to tell you the current is the same.
Audio Guru and I have spiced it, not that we needed to.
"Why must the forward and reverse current equal?" "The capacitor cannot pass DC" quotes from you! If the + and - currents are the same for the cap then how can they be 1000:1 different for the diode.
Diodes can have huge reverse currents, just over voltage a diode and see what happens.....your are!

 

[RCinFLA]

The series capacitor is the 'ballast' for the LED. Like a fluorescent light ballast (which is usually an inductor), the high reactance of the capacitor creates an approximate current source with the high voltage AC mains input.

The LED reverse breakdown current will be close to forward current and the reverse current provides no illumination, just extra heating on the LED. Because the reverse breakdown is a higher voltage then the forward voltage there will be some DC bias developed on the capacitor. It will only be a few volts in DC offset in comparison to the large AC voltage drop. The higher the zener voltage happens to be for a given LED, the more power will be lost on the reverse current cycle.

Because the capacitor has relatively low loss, the AC load at the mains will have a very poor power factor. If the LED forward current is 20 mA's, the true power will be a little more then 140 mW's (assuming a 4v forward voltage drop, and 10 v reverse breakdown voltage). About 100 mW of dissipation will be heating the LED during reverse cycle (which produces no light), compared to 40 mW used in forward (light generating) cycle.

Apparent power at the AC mains will be about 2.4 VA assuming again 20 mA LED current and 120 vac mains.

Power factor will be true power / appearent power = 0.140 watts / 2.4 VA = 0.06 .

 

[ronv]

Main article: Electrical polarity of LEDs
As with all diodes, current flows easily from p-type to n-type material.[80] However, no current flows and no light is emitted if a small voltage is applied in the reverse direction. If the reverse voltage grows large enough to exceed the breakdown voltage, a large current flows and the LED may be damaged. If the reverse current is sufficiently limited to avoid damage, the reverse-conducting LED is a useful noise diode.

Try 4 caps in parallel. You should still be well below the maximum current in the forward direction. Let us know how it comes out.

 

[The Electrician]

I have a bag full of hundreds of surplus LEDs from years ago, so I decided to make some measurements. Using a Tektronix TPS2042 scope with 4 isolated channels and a clamp-on DC responding current probe, I captured some scope images.

I connected a .47 µF X2 capacitor in series with the LED and connected it to the grid. I tried this several times and each time the LED worked right away; I didn't have any initial failures to illuminate. I disconnected and reconnected the arrangement to the grid numerous times and with any given LED, after several re-connects, the LED would fail, usually shorted.

In these images, the grid voltage is shown in orange, the LED (and capacitor) current in purple and the voltage across the LED in green.

The first image shows one of the LEDs in action. The green trace shows the voltage across the LED. The forward voltage is about 4 volts, and the reverse (breakdown) voltage for this LED is about 35 volts. The current through the LED is shown by the purple trace and is about 25 mA in both forward and reverse.

The second image shows that same LED, but with the sweep speed changed, and another trace turned on showing the instantaneous product of the LED voltage and the LED current. This is the red trace, and what is shown is the instantaneous power dissipation in the LED. I offset the reference for the red trace to the bottom of the screen, so we could more clearly see the difference in dissipation in the forward and reverse directions.

Horizontally, from about 4 cm to about 7 cm, the green trace shows the voltage across the LED as it conducts in the forward direction; it's about 4 volts. The red trace is about 1/3 cm above the bottom of the screen during this time; this is the forward dissipation in the LED.

From about .5 cm to 4 cm, the LED is conducting in the reverse direction. The green trace shows the reverse breakdown voltage to be about 35 volts. The reverse current is about the same as the forward current, but with 35 volts across the LED when the reverse current is flowing, the instantaneous dissipation is much greater. The red trace shows this dissipation peaking out about 5 cm above the bottom of the screen. This peak dissipation in the reverse direction (35 volts times 25 mA = .875 watts) is about 15 times the dissipation in the forward direction. This heats the LED and produces no light; it's not a good thing!

The third image shows another LED which has an even higher reverse breakdown voltage of about 80 volts peak. Imagine what the reverse dissipation would be!

 

[colin55]

One point that has been overlooked by carbonzit is this:

The forward WATTAGE through the capacitor must equal the reverse WATTAGE (through the same capacitor), for a LED and capacitor on an AC line. Since the forward TIME and reverse TIME are about the same, this concludes that the current-values must also be the same.
In other words the amount energy DISCHARGED from a capacitor during one half of a cycle is the only amount energy that the capacitor will accept during the second portion (the CHARGING portion) of the cycle.
That's why you must discharge a capacitor during half the cycle so it can be charged during the second part of the cycle.
If you only discharge it 10%; only 10% of the possible energy will be available for the charging portion of the cycle.

 

[The Electrician]

I just checked a datasheet for some LEDs I have (medium-high-intensity red), and it gives a Vr of 5V (minimum, typ. = 20V) at an Ir of 100µA.

What this means is that if you apply 5V in the reverse direction, the current won't be more than 100µA. It isn't telling you what the reverse current will be if you apply more than 5V in reverse, or whether there will be damage if you do so.

What's going on here is this: in the U.S. where the grid voltage in your house is about 170 volts peak, the LED is guaranteed to break down in the reverse direction; no LED is going to have a reverse breakdown voltage of anything near 170 volts. An LED will break down and the current will be much higher than the 100 µA you get if you only apply 5V in reverse.

If you have a capacitor in series with a device such as a rectifier (or LED that doesn't break down in the reverse direction), and connect the combination across the grid, the device will conduct in the forward direction for a few cycles and charge up the capacitor with a DC voltage. When the DC voltage on the capacitor equals the peak grid voltage, no more current will flow (as long as the diode or LED doesn't break down in the reverse direction).

The only reason this line connected LED circuit works is because the LED breaks down in the reverse direction.

You can prove this to yourself by "creating" an LED that doesn't break down with 340 volts applied in the reverse direction. Connect a 1N4004 diode (or other rectifier diode with at least 400 volts reverse rating) in series with an LED, such that if you apply a DC current (use a 1k resistor in series with a 20 volt power supply) to the series combination of diode and LED, the 1N4004 will pass its forward current in the direction which will make the LED light up. When you connect a .47 µF capacitor in series with this diode/LED combination and connect to the grid, the 1N4004 will prevent the reverse breakdown of the LED. The combination will behave like an LED with a reverse breakdown rating of greater than 400 volts (or whatever the reverse voltage rating of the diode is); the combination won't conduct in the reverse direction in this grid connected circuit, and the LED won't light up.

Huh? Why must the forward and reverse current equal? It's a diode. Since when do the forward and reverse currents equal, say in a typical power supply diode?

If you apply a large enough voltage in reverse to a diode, you can make it break down, and the reverse current can be as large as, or larger than, the forward current.

That's what's happening with the LED in this case. There is so much reverse voltage available from the grid, that the LED breaks down in the reverse direction and conducts a current just as large as needs to be to keep the capacitor from conducting DC (which it can't do anyway); that current is exactly equal to the forward current. If it wasn't equal to the forward current, the capacitor would eventually charge with a DC voltage and the circuit would stop working. That's what happens when you put a 1N4004 in series with the LED. With the 1N4004 in series, you don't get reverse breakdown and you can't get a reverse current equal to the forward current; eventually the capacitor charges up with DC and things stop working.

Here are a couple more scope captures.

The first image has an additional trace; a blue trace shows the voltage across the capacitor. The zero volt reference is the center of the screen (vertically), so you can see that the voltage across the capacitor goes from about 170 volts positive to about 170 volts negative.

The second image shows what happens when you put a 1N4004 in series with the LED. It's a little hard to distinguish between the green and blue traces, but the voltage across the capacitor is the horizontal blue line at -170 volts. The capacitor has become charged with a DC voltage equal to the peak grid voltage. This voltage opposes the AC voltage from the grid so that the voltage applied to the LED isn't able to go sufficiently high in the forward direction to turn on the LED.

The voltage across the diode/LED is still the green trace; now it doesn't go sufficiently high in the forward direction to turn on the LED. This shows that the LED wouldn't light up if it didn't break down in the reverse direction.

 

[ronv]

Ahh, A picture is worth 1000 words.
While you have the scope out (Looks like a storage scope?) could you capture a high current surge when the turn on occurs towards the peak of the AC cycle?

 

[unclejed613]

ok, to limit the current to 20mA at 120V/60hz requires an Xc of 6kohms. this calculates to 422nF. closest lower value would be 330nF. it is advisable to put a reverse diode across the LED to prevent reverse breakdown of the junction of the LED. use a 1N4007 for this. the capacitor has the advantage of providing the voltage drop and current limiting without generating lots of heat as a resistor would (a resistor doing the same job would be dissipating 2.4 watts). the capacitor must be (as someone already mentioned) rated for "line side" use.

 

[alec_t]

put a reverse diode across the LED to prevent reverse breakdown of the junction of the LED. use a 1N4007 for this

The reverse diode doesn't need to be a high-voltage type such as the 1N4007. The maximum reverse voltage this diode experiences is the forward voltage of the LED (<5V). Using a second LED as the reverse diode is the preferred option, as the combination gives light on both half-cycles of the mains.

 

[carbonzit]

I have a bag full of hundreds of surplus LEDs from years ago, so I decided to make some measurements. Using a Tektronix TPS2042 scope with 4 isolated channels and a clamp-on DC responding current probe, I captured some scope images.

I connected a .47 µF X2 capacitor in series with the LED and connected it to the grid. I tried this several times and each time the LED worked right away; I didn't have any initial failures to illuminate. I disconnected and reconnected the arrangement to the grid numerous times and with any given LED, after several re-connects, the LED would fail, usually shorted.

In these images, the grid voltage is shown in orange, the LED (and capacitor) current in purple and the voltage across the LED in green. ... [snip]

First of all, thank you, Electrician, for taking the time and trouble to look at this problem seriously. It's been frustrating getting nothing but criticism from the electronic-nanny types here, which wouldn't bother me, except that nobody has bothered to provide any kind of analysis, which you did. Hat tip to you.

So I thought about this some more. Now, the reason a capacitor works at all is because of its reactance. In this case, the cap I grabbed turned out to be almost perfect for the job, as its reactance at 60 Hz is about 5.6KΩ, forming a voltage divider thus:

**broken link removed**

(I figured the LED's forward resistance roughly from a Vf of 2 and an If of 20mA)

So fine, we have the capacitor standing in for a resistance, acting both as a voltage divider and a current limiter. (Contrary to what is often stated here, the resistor does not function strictly as a current limiter.)

That's when the LED is forward biased. Then I thought about what happens on the negative half-cycle. Being a diode, it has much greater reverse resistance (after all, that's how PN junctions and thermionic valves work, to a first approximation, right?) So what we have is something like this:

**broken link removed**

So now the voltage divider ratio is reversed. I really have no idea what the reverse resistance of a LED is, but let's say for the sake of discussion that it's 50KΩ; OK, that means that ... oh, I see: there's now a lot of voltage across the LED.

So isn't that the real (and simple) explanation of why there's such a high reverse voltage across it? And of course, with a higher reverse voltage comes higher reverse current, being on the steep part of the V/I curve. (And there's also that business of possible erosion of the diode junction due to excessive reverse current, from that abstract someone posted up there.)

OK, got that. But I'm confused about one aspect of having the capacitor in the circuit:

**broken link removed**

During positive half-cycles, the capacitor will charge as shown on the left, correct? But on negative half-cycles, doesn't the stored charge counteract the line voltage, as shown on the right? Doesn't this lessen the reverse voltage across the LED?

I also still do not understand the statement that the forward and reverse currents through the capacitor must be equal. Why is that so? Let's look at an ideal diode, which has infinite resistance under reverse bias: in this case, there'd be zero reverse current. It's not as if there's some principle of conservation of energy at work here. Can you explain this better?

I can see that the right and proper way to light a LED this way is to use two, back-to-back, to allow the capacitor to evenly charge and discharge.

BTW, LED is still illuminated. I unplugged it and replugged it a bunch of times just to test it.

 

[Diver300]

During positive half-cycles, the capacitor will charge as shown on the left, correct?

Yes

But on negative half-cycles, doesn't the stored charge counteract the line voltage, as shown on the right? Doesn't this lessen the reverse voltage across the LED?

No!

It increases the voltage on the LED, just like putting two AA cells in series, + of one connected to - on the other, gives twice the voltage.

In your analysis, when the LED is reverse biased, the absolute value of its resistance increases, but it is still small compared with the reactance of the capacitor. As The Electrician measured, the current is controlled by the capacitor only, because its reactance is much larger than the resistance of the LED in either direction. The LED voltage is always a lot smaller than the supply voltage, so most of the supply voltage is across the capacitor.

If the diode were ideal, there would be forward current on the first cycle, as the capacitor charges up. On the reverse cycle, the diode would block the current and the capacitor would stay charged. On the next forward cycle, the capacitor would already be charged, so no current would flow.

So, if you were to put a 300 V or more diode in series with the LED, in the same direction, the LED would not light. It might flash once when it is turned on, and it might glow very dimly if there is any leakage in the capacitor.

The fact that the LED lights means that current is flowing into the capacitor every cycle, which means that current is flowing out of the capacitor every cycle as well.

 

[carbonzit]

The fact that the LED lights means that current is flowing into the capacitor every cycle, which means that current is flowing out of the capacitor every cycle as well.

Ah, so. So when those folks were saying "The currents through the capacitor in each direction must be equal", what they really meant is that in order for the LED to light in the forward direction, some current must flow through the capacitor in the reverse direction in order to discharge the capacitor: would you agree?

And I suppose that if you look at the current over time (di/dt), then the currents would have to be equal.

Just an editorial note: in case you don't see what I'm driving at, it's not sufficient to make an assertion ("the currents must be equal") without some explanation, even if it makes perfect sense to you. (Which is also a roundabout way of thanking The Electrician and Diver300 for their patience.)

Edit: Of course, I don't mean di/dt, I mean the opposite (the integral). But I don't know how to make that funny sign here yet. (Haven't learned TeX yet.)