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Evan
05-02-2008, 08:16 AM
All the usual disclaimers are in the link at the bottom. If you die don't call me to complain.

There have been quite a few threads over the last few years about using LEDs for work lights. LED lighting is something I have been playing with as well.

I finally found information on how to properly power white LEDs from the AC line directly. Note that this method will produce 120 hertz flicker (100 hertz in those places with poor frequency regulation :D).

http://vts.bc.ca/pics3/ledpower.jpg

Dawai
05-02-2008, 11:03 AM
Hi..

Wouldn't a capacitor reduce flicker? That would make a RC network, I still have not seen a place to purchase them cheaply.. I can buy a flashlight cheaper than I saw the LED's..

David

Lew Hartswick
05-02-2008, 11:13 AM
Yep. I just ran the reactance and it comes out about 18 ma.
For the electronicaly challenged:
1
X(sub)c = ______________
2 Pi F C

:-) Gad trying to do a math expression in ASCII is a riot.
Wonder how this is going to display?
...lew...
(edited) FORGET IT . I gues there is no way to make it,the spaces are ignored.

lazlo
05-02-2008, 11:18 AM
:-) Gad trying to do a math expression in ASCII is a riot.
Wonder how this is going to display?

I had the same problem when I posted the equations for the skin depth. Oddly, if you open the quote window, you can see your formula with the correct spacing, just as you intended :)

By the way, there's a flood of Chinese LED-lighted consumer products at your local Sears/Walmart/Home Depot... For \$20 you can buy a 20 LED work light, rip out the guts, and have an AC-power string of white LED's for far less than you could buy the parts.

dp
05-02-2008, 12:07 PM
White LED's have a forward voltage drop of 3.0 - 3.5 v, so when you put 12 in series you have to back out that voltage. The current is actually less than 15ma assuming impedance of 5.6k for the capacitor plus the 1k resistor.

It is not mentioned in the schematic but it is critical to the success of these circuits that diodes be placed back to back exactly as shown. Otherwise the full line voltage will be imposed on the diode or diode string in the reverse direction and they cannot handle that kind of reverse voltage. Putting them back to back provides a voltage clamping function for the non-conducting path.

Evan
05-02-2008, 12:07 PM
Try a dollar store. I recently picked up a dozen cheezy fiber optic decorations for a buck each that contain three super high intensity leds, red, green and blue. I'm in the process of making a special effects picture frame. If it works the way I hope I'll post it. If it doesn't then I will have a full spectrum adjustable night light.:D

tmarks11
05-02-2008, 12:20 PM
Yep
1
X(sub)c = ______________
2 Pi F C
Your trying to be too fancy. Easier to state it as:

Xc=1/(2*Pi*f*C) = 1/(2*3.14*60*0.47e-6) = 5644 ohms

I personally think this is asking for trouble. Just throw a \$5 wall wort transformer into the circuit, and it becomes easy to make without risking setting something on fire.

Anybody every overload an LED before? It blows up in a spectacular fashion at about 30V dc... scatterring shrapnel everywhere.

Evan
05-02-2008, 12:25 PM
FORGET IT . I gues there is no way to make it,the spaces are ignored.
Use the 'code' tag and the 'System' font.

Xc = -( ___1___ )
2 Pi F C

Evan
05-02-2008, 12:27 PM
I personally think this is asking for trouble. Just throw a \$5 wall wort transformer into the circuit, and it becomes easy to make without risking setting something on fire.
I take it you don't use LED Christmas lights either? This is a safer circuit.

I'm having a hard time visualizing shrapnel from something that weighs 0.4 grams, including leads...

05-02-2008, 01:19 PM
Hm Why not have two strings one with cap to provide a phase shift and another without but with extra dropping resistance. That way you'd get "overlap" to eliminate flicker.

Also wouldn't some of the LED's be brighter than others? How well do they match?

Dawai
05-02-2008, 01:21 PM
The lil lady across the road used to have my dad come over to plug her iron up... she had saw a spark once and it scared her so bad she never really had her house wired correctly. I can still remember seeing coal oil lamps glowing over there.

it had these twist electrical switches I wanted so bad.. I offered to rewire the house *labor for the electrical service in it.. a 120 volt meter base with twist switch on it. All the light switches and exposed flex.. old black wires hanging on the fixtures..

She was scared of electricity too. either that or she had the hots for my dad..

BobWarfield
05-02-2008, 01:28 PM
Awesome!

Keep tinkering fellas, when you've got the perfect circuit figured out, I want to put together one of these:

http://www.thewarfields.com/cnccookbook/img/OthersProjects/MillRingLight.jpg

You just can't have too much light directly where the machine is doing its thing.

Best,

BW

dp
05-02-2008, 01:43 PM
Note that this method will produce 120 hertz flicker (100 hertz in those places with poor frequency regulation :D).

These circuits will actually produce two 60 Hz flicker sources, 180 degrees out of phase. In some situations this would be very noticible - a light rod, for example, where the LED's are not phycially interleaved.

rantbot
05-02-2008, 02:01 PM
What does the cap do in this circuit that a 5.6k resistor wouldn't do?

Evan
05-02-2008, 02:18 PM
The cap doesn't act as a voltage divider.

Evan
05-02-2008, 02:25 PM
These circuits will actually produce two 60 Hz flicker sources, 180 degrees out of phase.

I'm going to experiment using my Audiolumnitron to see what approaches might help to minimize flicker. One thing I thought of doing is to make a reflector that is painted with luminescent paint. The new paints are extremely bright for the first few milliseconds and should go a long way to reduce the flicker.

Also wouldn't some of the LED's be brighter than others? How well do they match?

This type of series circuit ensures that all the LEDs have the same current. While they may not be perfectly matched in brightness it isn't possible for one to hog current as it is in a parallel circuit.

rantbot
05-02-2008, 03:21 PM
The cap doesn't act as a voltage divider.
Neither does the resistor. In these circuits, both are simple series impedances. One has a reactive component, but that's inconsequential here, as the circuit is powered at a single fixed frequency.

dp
05-02-2008, 03:51 PM
The cap doesn't act as a voltage divider.

All series circuits are voltage dividers. If you measure the voltage across each device the sum of those will equal the line voltage. And the voltage across each device will be different from the other devices (simplifying the diode string to a single device). Measuring the voltage using one side of the line as a reference, the series circuit is now a tapped series divider.

There's been an unmentioned wrinkle that is not real important to this circuit and that is the power factor of the circuit needs to include the reactive component, the capacitor, which introduces the j operator into the math. Add to that the fact that a scope across the back to back diodes will paint a very squared off wave form and that will create radio noise. A closer look at the cross-over points should reveal the square-law region of conductance of the diodes.

There's no reason a full wave bridge rectifier can't be installed in the circuit. Doing so immediately improves the ripple frequency, all the diodes can be run in series and not back to back. The capacitor would be connected in series with the bridge input, and the resistor and diodes would be wired in series with the bridge output.

Now you can consider using an active three-wire voltage regulator or capacitor mulitplier using a FET or bipolar transistor to drive the diodes with a very clean DC that will remove the flicker entirely.

http://sound.westhost.com/project15.htm

But it's also no longer a very simple circuit :)

wshelley
05-02-2008, 04:12 PM
Bob's picture is almost identical to a led light I saw at the local hardware store for use on a patio umbrella. Didn't grab one since it was \$20 and the Jeep was on empty :rolleyes: In any case, online they are selling for \$14ish. Has anyone tried one on a mill yet?

http://www.wardscorner.net/images/led light.jpg
Ward

2ManyHobbies
05-02-2008, 05:11 PM
A full wave bridge will give you a 120 Hz flicker on half of the LEDs and the other half will be dead, thought adding a capacitor would easily reduce the remaining flicker at that point.

What about a resonant tank tuned to ~4x line frequency? Besides *really* wanting line isolation at that point, you'd have two sets of flicker running 240 Hz 180 degrees out of phase.

Of course, for added cool points for spindle lighting, nothing beats wireless lighting if one where to add permanent magnets to the spindle and coils in the ring of LEDs. :p

japcas
05-02-2008, 06:47 PM
Bob's picture is almost identical to a led light I saw at the local hardware store for use on a patio umbrella. Didn't grab one since it was \$20 and the Jeep was on empty :rolleyes: In any case, online they are selling for \$14ish. Has anyone tried one on a mill yet?

http://www.wardscorner.net/images/led light.jpg
Ward

I bought a light just like the one pictured above for \$6 at Walmart including batteries. I machined a brass bushing that just slips over then lens on my Nikon Coolpix 4600. I have to hold it on as I made it a slip fit on the lens because I don't think the extra weight would be good for the servo motor that zooms the lens. You need to preset the white balance on the camera before taking pictures with it, otherwise the pics come out real white. But when done right, it helps to take macro shots without a tripod because the camera is able to use a faster shutter speed than without the light. And for macro shots the the flash ruins the shot in most cases. Here are some pics of it.
http://i32.photobucket.com/albums/d39/japcas/DSC_2344.jpg
http://i32.photobucket.com/albums/d39/japcas/DSC_2345.jpg

dp
05-02-2008, 08:03 PM
A full wave bridge will give you a 120 Hz flicker on half of the LEDs and the other half will be dead, thought adding a capacitor would easily reduce the remaining flicker at that point.

You don't wire the LED's back to back in a DC circuit - they are all wired in one long string.

Evan
05-02-2008, 08:08 PM
I should have said the cap isn't a pure resistance. The reactance is sensitive to frequency and waveform. Any time you are rectifying you introduce high frequency harmonics that tend to bypass the capacitor and in that sense it isn't a divider since ω is frequency dependent. Trying to measure this with most meters will give a false reading.

I know there are plenty of more complex methods of running LEDs, I'm using some of them. I like the total simplicity and it's a circuit that those with limited electronics knowledge can handle. The only real advice I would offer on the safety aspect is that if you aren't sure how to wire anything that uses 120 vac then don't try this either.

dp
05-02-2008, 08:15 PM
I know there are plenty of more complex methods of running LEDs, I'm using some of them. I like the total simplicity and it's a circuit that those with limited electronics knowledge can handle. The only real advice I would offer on the safety aspect is that if you aren't sure how to wire anything that uses 120 vac then don't try this either.

It is probably worth mentioning that if for any reason one of the two diode paths should open up (blown diode, bad solder joint, spiteful supreme being...), the other path will blow like a popcorn phart.

05-02-2008, 10:44 PM
There is one problem with that circuit, do not submit it to UL for aprovel. :) They would go ballistic. What “IF” the capacitor shorts out! The full line voltage is now across the LED’s and they will be exploding like little bombs, the wiring will melt down, the entire circuit will burst into a roaring fire and if someone were to grab this burning mass to move it, they could be electrocuted from contacting raw house power. And of course it will ignite whatever it is next to. That in turn will burn down your house. Oh it is just way to scary to use.

On the other hand I have seen it used in commercial products, which were not UL approved. This is a nice simple cheep way to reduce line current without using a transformer. For the faint hearted you can always add a small fuse.

steve45
05-03-2008, 12:26 AM
Those lights look a lot like the one from Northern Tool: http://www.northerntool.com/webapp/wcs/stores/servlet/product_6970_200342912_200342912

The Northern light doesn't split, and it has more LEDs.

I LIKE the ideas for mounting, I may have to buy a couple more!!!

steve45
05-03-2008, 12:27 AM
On a related topic, I would like to replace some indoor flood lights with LEDs. C.C. Radio has some that they say will put out enough light (their prices are pretty high), but they say they won't work well with dimmers.

Any thoughts on using LEDs with a dimmer?

Evan
05-03-2008, 01:02 AM
There is one problem with that circuit, do not submit it to UL for aprovel
At least it has the capacitor for current limiting. LED Christmas lights don't even have that. Also, 20 ma at 120 vac is only borderline able to cause death from electrocution and won't cause electric burns. It takes around 100 ma to stop the heart. The only common way that 20 ma current could kill is by paralyzing the diaphragm muscles.

The circuit above will work fine with a dimmer. LED bulb replacements won't because they use solid state micro ASICs to control them and they depend on having a sine wave as input.

ASIC=Application Specific Integrated Circuit

dp
05-03-2008, 01:09 AM
On a related topic, I would like to replace some indoor flood lights with LEDs. C.C. Radio has some that they say will put out enough light (their prices are pretty high), but they say they won't work well with dimmers.

Any thoughts on using LEDs with a dimmer?

Any of Evan's circuits will work fine with a standard lamp dimmer. The control will not likely be close to linear, but the LED's will definitely dim in a controllable manor.

dp
05-03-2008, 01:14 AM
The circuit above will work fine with a dimmer. LED bulb replacements won't because they use solid state micro ASICs to control them and they depend on having a sine wave as input.

ASIC=Application Specific Integrated Circuit

My wife bought several solar powered yard lights that are clear plastic dragon flies, humming birds, etc with embedded LEDs. They're just fascinating because they have ASIC controllers and shift over quite a color range using just red, green, and blue LED's. A quick look showed only a single pair of wired going to the lamps, so there's got to be some back to back wiring and PWM going on. The colors blend all through the rainbow. And amazingly, they're still going in the morning just as the sun is coming up.

Evan
05-03-2008, 01:24 AM
I dropped in to Dollarama today and bought a bunch more of the RGB LED light toys. They have a tiny bump mounted controller that has about 10 modes of operation. Each time you press the on switch it goes into another mode. I have been trying to find a data sheet that matches the modes for these because it may well have more modes or a way to control it further especially since there is an unused solder pad on the ASIC. Many of these controllers have a very simple serial bus input for programming.

Evan
05-03-2008, 01:40 AM
Check out these:

http://www.dealextreme.com/products.dx/category.907

darryl
05-03-2008, 02:30 AM
Well, here I go again. I just lost the entire repy I had typed up to a single misplaced keystroke on the keyboard. I hate that, and it's so easy to do with the little finger.

Anyway, I'm also a fan of using an elegantly simple circuit to do the job. In this case, because the load is fixed, the series capacitor is a decent way to power the string. The leds will still flicker, and they will make some rfi. My preference would be to use one of those round blob 1 amp bridge rectifers with a filter capacitor, still use the series capacitor, but then power the led string through a three terminal regulator configured as a current source. This is the simpest circuit that a regulator can be configured as, and it completely eliminates flicker in the leds. The filter capacitor need not be a large value, as a ripple voltage of up to 25 or so volts can be tolerated by the following current regulator. The series resistor value would then be chosen to put the lowest voltage value seen by the filter capacitor to a point above the voltage requrement of the led string, plus the minimum voltage requirement for the regulator (which is about the same as one led voltage drop, 3v or so).

Without the series capacitor and resistor, the circuit could still be made to function using the regulator chip, but the led string would have to have a fixed number of leds, probably 42-44. Not much room to change this. This circuit could be as simple as one rectifer, one filter cap, one three terminal regulator, and one resistor- plus the led string. Ok, add a fuse, or two (one on each ac line). If you use a non polarized wall plug, you won't know which line is live, so both fused would be required for protection.

Usual disclaimers, all points in these circuits are capable of delivering a shock, etc.

tmarks11
05-03-2008, 09:23 AM
I'm having a hard time visualizing shrapnel from something that weighs 0.4 grams, including leads...
Yeah, I was surprised too, and so was my 5yo who was helping me :eek: ...

I had some LEDs hooked up to a variable DC power supply for testing... I was installing them into one of my son's trains. Turned on the power supply without rotating the voltage down to zero. The top of the LED blew off and bounced off the wall.

J Tiers
05-03-2008, 10:06 AM
The capacitor dropping reactance is NOT A PROBLEM.

If you do NOT like it then you MUST throw away any of these highly dangerous items:

Both those commonly use a series capacitor as a "wattless" series impedance.

BOTH those products are commonly UL recognized.

The two things that I WOULD suggest changing about the circuit are:

1) Change the capacitor to a 250VAC UL recognized component, rated as an "X" capacitor. They are extremely common, cheap, but are made NOT to fail as mentioned..... One of that type is made for use on AC voltage up to 250V nominal.

2) because nothing is perfect, stick a low amperage fuse in series. Alternately, you can put a fuse resistor in series in place of the metal oxide one shown, although that isn't as nice, and gets you involved in details that are not user-friendly.

The RFI can be reduced or avoided for the most part, by putting a suitably rated capacitor of around 0.01 uF across the diode string

As noted, you can use all the LEDs efficiently by using a bridge rectifier between the cap and the LEDs, and using the LEDs in one series string.

Dawai
05-03-2008, 10:31 AM
Charging a capacitor through a diode-resistor.. it has decay time relating to charge..

Used to make time delay relays like that, a capacitor charged though diode and resistor.. the resistor stops the diode from burn-out.. that was before the advancement of electronic time delays when we dealt with pneumatic timers.. what a pain in the electricians butt.. adjusting the lil needle valve to time/barometric pressure/humidity/temperature/phase of moon.

Should give flicker free light. Wire the led's in strings equal to voltage stored. Ya'll work it out and repost it.. I am up to my armpits here..

IF low voltage makes the LED's eat their rear out, install a zener, or a zener regulator circuit on the back side of another resistor..

Any of ya real machinists want to run away from home for a week? I got a extra bedroom.

Evan
05-03-2008, 11:49 AM
David,

LEDs are current operated devices. The voltage only needs to be high enough for the LED to forward conduct. You will notice that the circuits above don't depend on the number of LEDs in the circuit. What LEDs need is a controlled amount of current because once they begin conducting they will hog as much as they can limited only by the supply capability and the internal resistance of the LED.

Cheap LED flashlights take advantage of the fact that the internal resistance of the batteries acts as a current limiter. In many cases if you were to take a LED flashight head and run it from a line powered 4.5 vdc supply with no current limiting you will over drive the LEDs and greatly shorten the lifespan. I know because I have done it.

The above circuits take advantage of the fact that at a particular AC frequency a particular value of capacitor will pass a certain amount of current regardless of what the load is. To run the LEDs on DC requires that the current be limited in some manner. A zener diode won't do that as it is a voltage regulating device, not a current limiter.

dp
05-03-2008, 02:06 PM
The above circuits take advantage of the fact that at a particular AC frequency a particular value of capacitor will pass a certain amount of current regardless of what the load is. To run the LEDs on DC requires that the current be limited in some manner. A zener diode won't do that as it is a voltage regulating device, not a current limiter.

The capacitor uses Ohm's law just as a resistor does. The 5600 Ohm impedance can pass a max of V/5600 amps, or appx 20 ma at 120 VAC, 60 Hz. It can pass less current as other series resistances reduce the voltage applied across the capacitor.

All the circuits you provided are current limited but not current regulated.

Evan
05-03-2008, 03:43 PM
The capacitor uses Ohm's law just as a resistor does.

All the circuits you provided are current limited but not current regulated.
Those statements are contradictory. In each circuit the effective series resistance of the load is different but the current remains the same. This does not occur with a pure resistor instead of a capacitor.

By definition a current limiter is a current regulator.

Paul Alciatore
05-03-2008, 04:16 PM
At least it has the capacitor for current limiting. LED Christmas lights don't even have that. Also, 20 ma at 120 vac is only borderline able to cause death from electrocution and won't cause electric burns. It takes around 100 ma to stop the heart. The only common way that 20 ma current could kill is by paralyzing the diaphragm muscles.

........

This arguement does not hold water. If a fault conditioin like a shorted capacitor exists, then depending on which part of the LED string you come in contact with and how much damage occurs to that LED string, you could easily be in contact with the raw 115Volts AC from the socket. In that case, the amount of current passing through your body will have no relationship to the design parameters of the LED circuit. It will depend only on the path the current takes from the AC lines to ground - through your body. This could be a high or a low impedance path and the current could possibly exceed the dangerous level through your heart.

I got bit by about 50,000 Volts once. It went in one hand and out the bottom of my foot. It went through my shoe and around the plastic probe in my hand. I found the burn holes in both places. Absolutely no fun. But it was a high resistance path so the current was obviously low. I was quite shook up for several hours, but walked away from it. With better contact I might not have. Under good conditions, even a "low Voltage" like 115 Volts can be fatal.

If anyone tries any of these circuits, I would advise adding a fuse and possibly using it only on a GFI outlet. If installed on a machine, please insure that the safety ground is connected to that machine - as it should be in any case. SAFETY FIRST, please.

dp
05-03-2008, 04:34 PM
Those statements are contradictory. In each circuit the effective series resistance of the load is different but the current remains the same. This does not occur with a pure resistor instead of a capacitor.

By definition a current limiter is a current regulator.

Consider a constant current source. It is dynamic and works to ensure that within it's range the current is constant regardless of the load. NiCad battery chargers being an example, incandescant lamps being another. These are current regulators. The current will remain constant even with a zero resistance load.

Compare that to a current limiter which is nothing more than a resistor. The current through it varies with the load except that the current is limited to a short circuit value (the load resistance drops to zero). That is the upper current limit. No regulation involved.

The impedance of the capacitor does not change with load - it is fixed by the frequency of the source voltage. In the example that impedance is 5600 Ohms 0 A, 20 mA, or 20 A.

dp
05-03-2008, 07:09 PM
It is probably worth mentioning that if for any reason one of the two diode paths should open up (blown diode, bad solder joint, spiteful supreme being...), the other path will blow like a popcorn phart.

Another thought occured to me regarding the loss of one path in the back to back diode wiring. The remaining circuit is now a voltage doubling half-wave rectifier, so 220 (the line voltage plus the voltage of the fully charged capacitor) is available to the remaining diode string and series resistor for the non-conducting half of the cycle. If leakage in the diodes is low then the actual voltage applied will be twice the peak voltage, not RMS.

Evan
05-03-2008, 09:13 PM
This arguement does not hold water. If a fault conditioin like a shorted capacitor exists, then depending on which part of the LED string you come in contact with and how much damage occurs to that LED string, you could easily be in contact with the raw 115Volts AC from the socket. In that case, the amount of current passing through your body will have no relationship to the design parameters of the LED circuit. It will depend only on the path the current takes from the AC lines to ground - through your body. This could be a high or a low impedance path and the current could possibly exceed the dangerous level through your heart.

Why all the paranoia about the circuits? It's no different than any other 120 vac powered lamp. I have yet to see a lamp with a fuse anywhere. LEDs fail in a similar manner in that they usually fail open. They have an inherent fuse inside in the form of the "p" connection lead which is a very fine wire.

As far as my "argument" it holds water better than your average incandescent lamp in which the current available is far more than enough to fibrillate your heart, failed or not.

dp,

The term regulator is not very finely defined. It is any device that exercises some measure of control over something. In particular, the capacitor circuit is able to provide the same current through as few as 2 to as many as 80 LEDs, something a single value of resistor cannot do.

dp
05-03-2008, 09:53 PM
dp,

The term regulator is not very finely defined. It is any device that exercises some measure of control over something. In particular, the capacitor circuit is able to provide the same current through as few as 2 to as many as 80 LEDs, something a single value of resistor cannot do.

Kirchhoff's second law is pretty firm on this. Your circuit simplifies to a series of three resistors for purposes of a Kirchhoff analysis. The capacitor, diode strings, and resistor can all be considered resistance. You know what happens when you wire diodes in series - the forward voltage drop is additive. For white led's that is about 3.0 vdc/device. That is 18 volts for 6 devices in series vs 9 volts for 3 devices in series. The current difference is not much but is measureable. And certainly calculable.

Assuming an impedance of 5600 Ohms for the capacitor and 1 k for the resistor we have A = 120/6600, or 18.2 milliamps. That is the most current the system will deliver with no LED's. Add in 3 LED's now:

3 * 3.0 = 9 volts forward drop. A = 120 - 9 /6600, or 16.8 mA. Add 3 more: A = 120 - 18 / 6600, or 15.5 mA.

With 56 diodes in series the current will drop to zero because the cumulative forward voltage drop will equal the source peak voltage.

I agree with the semantics issue. I also agree the danger is overstated but I think there's some rib poking at the bottom of that. The failure mode is rather spectacular compared to a Christmas light string, though.

Usually at this point in a thread somebody drags out the "somebody is wrong on the Internet" cartoon :)

Evan
05-03-2008, 10:26 PM
A white gallium nitride LED operating at around 20 ma has a forward drop of about 4 volts.

It allows you to calculate the real resistance required to produce a desired current through whatever LED voltage drop you wish to specify. For a single white LED operating on 120 volts it spits out 6800 ohms for a 4 volt forward drop and 20 ma. That's in close agreement with the above #1 circuit.

However, for six in a row with a drop of 24 volts it gives a resistance of only 4800 ohms and for 12 LEDs with a drop of 48 volts it gives 3600 ohms. That isn't in agreement with the above circuits.

Explanation?

dp
05-03-2008, 10:38 PM
A white gallium nitride LED operating at around 20 ma has a forward drop of about 4 volts.

It allows you to calculate the real resistance required to produce a desired current through whatever LED voltage drop you wish to specify. For a single white LED operating on 120 volts it spits out 6800 ohms for a 4 volt forward drop and 20 ma. That's in close agreement with the above #1 circuit.

However, for six in a row with a drop of 24 volts it gives a resistance of only 4800 ohms and for 12 LEDs with a drop of 48 volts it gives 3600 ohms. That isn't in agreement with the above circuits.

Explanation?
I used to have a calculator like that.

Edit: ok - I did the math and the calculations from your calculator are exactly right on for determining what resistance value you need to put in series with the diode/diode string to maintain 20 mA current. In other words, 12 LED's require a series resistance of 3600 Ohms to limit the current to 20 mA when powered with 120 V.

All that is happening is the Vfd of the diode is being negated which leaves the balance of the voltage to be dealt with using a resistance (or impedance). So 120 - 48 = 72 volts applied across the series resistor. E/I = R so to get 20 mA we use a 3600 Ohm resistor. Because it is a series circuit the current in the LED is also 20 mA. That is 9.6 watts of heat which is pretty warm. I'd use silver solder.

Edit: It also means the circuit is wrong and misleading, but I knew that when I first saw it. It's not wrong by an important amount, though.

gunbuilder
05-03-2008, 10:44 PM
I got bit by about 50,000 Volts once. It went in one hand and out the bottom of my foot. It went through my shoe and around the plastic probe in my hand. I found the burn holes in both places. Absolutely no fun. But it was a high resistance path so the current was obviously low. I was quite shook up for several hours, but walked away from it. With better contact I might not have. Under good conditions, even a "low Voltage" like 115 Volts can be fatal.

I don't know the voltage but I got hit several times from testing AEDs. Some moron set the test rack up so the load resistors could touch the metal rack. I was following ESD protocol and had my wrist strap and ankle grounds on so I was well grounded. I will tell you that smarted. Oh the "fun" job of a test technician.

Thanks,
Paul

dp
05-03-2008, 11:08 PM
I don't know the voltage but I got hit several times from testing AEDs. Some moron set the test rack up so the load resistors could touch the metal rack. I was following ESD protocol and had my wrist strap and ankle grounds on so I was well grounded. I will tell you that smarted. Oh the "fun" job of a test technician.

Thanks,
Paul

I got hit with a 25kv jolt from a low impedance source. I was working on a Decca radar on a container ship in Long Beach harbor in California. I had to swap out the magnetron and the drill is to shut down the TR unit then use the supplied wand to bleed off the B+ on the maggie. The ground wire was broken on the wand unknown to me, so the initial spark went through me. I recoiled quickly (and uncontrollably) and watched a blue spark follow that wand quite a ways before it went out.

I was able to continue on and got the radar working, but I made my own grounding tool after that.

darryl
05-03-2008, 11:56 PM
It's been a while now, but I recall reading about shock, voltage and current. What I came to 'know' was that a very low current passing through you is dangerous because you tend to reciol, and thus can smash your elbow on something, etc. Other than that, you get a harmless shock. When the current rises to the low milliamp range, roughly between 3 and 20 ma, that can set your heart fibrillating and you can die. With a higher current, your heart is clamped, so the potential is there for less damage, and you can live through it. Once the current is high enough, or persists for a longer time, you will get cooked and die from that. Basically they were saying that if you can let go quickly enough, or become disconnected quickly enough, you chances of survival are greater if a slightly higher current is passing through your body.

Of course it is not that cut and dried. It depends on the path through your body, etc, but it was interesting to read that, and it seems to make some sense.

J Tiers
05-04-2008, 12:13 AM
Why all the paranoia about the circuits? It's no different than any other 120 vac powered lamp. I have yet to see a lamp with a fuse anywhere. LEDs fail in a similar manner in that they usually fail open. They have an inherent fuse inside in the form of the "p" connection lead which is a very fine wire.

As far as my "argument" it holds water better than your average incandescent lamp in which the current available is far more than enough to fibrillate your heart, failed or not.

dp,

The term regulator is not very finely defined. It is any device that exercises some measure of control over something. In particular, the capacitor circuit is able to provide the same current through as few as 2 to as many as 80 LEDs, something a single value of resistor cannot do.

Several interesting misconceptions...........

1) "paranoia"? Nope, the realization that there is nothing limiting the flow of current aside from the capacitor dielectric..... which, in a standard type capacitor, is thin, and not that reliable against say, a surge voltage. (But the special "X" type caps I suggested are pretty reliable, the "Y" type cap even more so).

A resistor, by contrast, almost cannot fail shorted. It is very reliable. The capacitor is NOT, and CAN fail shorted. (a couple constructional details affect that).

EITHER should have some sort of a fuse, and you would be unable to obtain the right to UL/CE marking for that device without it or an equivalent feature which limits current.

2) A resistor of the same general impedance will act the same way as that capacitor. Aside from a phase angle difference, there is no difference, either is so many ohms impedance at 60 Hz.

There is NO magic about a capacitor that makes it act differently as far as impedance, aside from the phase angle deal, and the fact it only works on AC.

3) LEDs are not reliable devices, and have not generally got approved insulation qualities between the leads and the outside of the case. So they should be guarded, because the whole shebang is at mains voltage.

I;ve been through this with the UL folks, and I know.

4) Lamps don't have fuses, both due to tradition, and because regular lamps are capable of blowing the breaker/fuse for the branch circuit if shorted. And the construction discourages shorts. The lamp itself (incandescent) is quite reliable, generally opening under any abuse such as high voltage.

Evan
05-04-2008, 04:19 AM
ok - I did the math and the calculations from your calculator are exactly right on for determining what resistance value you need to put in series with the diode/diode string to maintain 20 mA current. In other words, 12 LED's require a series resistance of 3600 Ohms to limit the current to 20 mA when powered with 120 V.

All that is happening is the Vfd of the diode is being negated which leaves the balance of the voltage to be dealt with using a resistance (or impedance). So 120 - 48 = 72 volts applied across the series resistor. E/I = R so to get 20 mA we use a 3600 Ohm resistor. Because it is a series circuit the current in the LED is also 20 mA. That is 9.6 watts of heat which is pretty warm. I'd use silver solder.
So that begs the question of where is all that heat when the capacitor is being used? And, why don't capacitors have a wattage rating?
------------------------

A resistor, by contrast, almost cannot fail shorted. It is very reliable. The capacitor is NOT, and CAN fail shorted. (a couple constructional details affect that).
Um, yes, constructional details indeed. A .47 mfd 250v capacitor isn't a failure prone electrolytic. Most usually it will be a poly or even ceramic capacitor. I have never seen one fail in a short condition other than being hit by lightning in which case the capacitor disappeared (failed short then failed extremely open).

There is NO magic about a capacitor that makes it act differently as far as impedance, aside from the phase angle deal, and the fact it only works on AC.

And the fact that capacitive reactance, while measured in ohms, has nothing to do with resistance.

Same question. Where is the heat?

LEDs are not reliable devices, and have not generally got approved insulation qualities between the leads and the outside of the case. So they should be guarded, because the whole shebang is at mains voltage.
So, where does that leave us with LED Christmas lights? UL approved, no fuse.

The lamp itself (incandescent) is quite reliable, generally opening under any abuse such as high voltage.
All the ones I have owned have failed. One, a properly installed heat lamp in the bathroom, exploded one day and rained glass and hot metal down on the floor outside the tub when my wife was in the shower. She was not impressed.

LEDs also tend to fail open.

J Tiers
05-04-2008, 10:36 AM
So that begs the question of where is all that heat when the capacitor is being used? And, why don't capacitors have a wattage rating?
------------------------

Capacitors DO have a "wattage" rating...... but it isn't strictly what you think. it is a CURRENT rating, and the limit is due to the inherent "residual" resistance of the part, which is generally 50 to 1000 times smaller than the reactance at 120Hz. That ratio is generally given as one of the specs of the part, for capacitors intended for applications where that is important.. The heat dissipation in the ESR limits the allowable current.

There is very little heat compared to a similar resistive impedance, because most of the capacitive impedance is "reactive", it is a feature of the "math" of the part (that the voltage across it is proportional to the integral of the current through it), and not strictly a dissipative resistance.

The above-mentioned "ESR" does dissipate heat.

Um, yes, constructional details indeed. A .47 mfd 250v capacitor isn't a failure prone electrolytic. Most usually it will be a poly or even ceramic capacitor. I have never seen one fail in a short condition other than being hit by lightning in which case the capacitor disappeared (failed short then failed extremely open).

As I mentioned, there are parts which ARE considered reliable enough to be connected from line to chassis of electronic equipment.

And there are parts which are NOT considered reliable enough to be so connected.

YOU CHOSE THE "NOT RELIABLE" TYPE PART. it would be so simple to use the right type, but noooooooooo......... that one is plenty good enough for you.

Incidentally, the standard test for any component not considered reliable by the test agency, is to short and open the part and operate the unit.

If either condition produces a hazard, which may include arcs, a condition which leaves live parts exposed after the test, or other "hazards", then the part is not accepted in that application.

A "recognized" part is not subject to those tests.

And the fact that capacitive reactance, while measured in ohms, has nothing to do with resistance.

Same question. Where is the heat?

See above, do the math

So, where does that leave us with LED Christmas lights? UL approved, no fuse.

NO FUSE THAT YOU CAN SEE DOES NOT MEAN THERE IS NO FUSE. The fuse may be permanently molded into the plug, since a regular incandescent type generally has a fuse in the plug (replaceable). The fuse may be a recognized fusible resistor, or any of a number of other things you would not recognize as a fuse.

THEN AGAIN, THE UL MARK MAY BE FRAUDULENT. CANADA WAS SHIPPED A CONSIDERABLE NUMBER OF CHINESE COUNTERFEIT CHRISTMAS LIGHTS WITH FAKE UL MARKS..........THOSE MAY HAVE NO FUSE, AND BE INCENDIARY DEVICES IN FACT.

All the ones I have owned have failed. One, a properly installed heat lamp in the bathroom, exploded one day and rained glass and hot metal down on the floor outside the tub when my wife was in the shower. She was not impressed.

"RELIABLE" in this case does NOT mean that it won't fail. It means that it won't burn your house down when it does. I assume your house did not burn down as a result of the failure, so the light evidently did its job. BTW.........THE FAILURE MAY HAVE BEEN DUE TO FAULTY INSTALLATION, ALLOWING COLD WATER TO DRIP ON THE HOT LIGHT.

LEDs also tend to fail open.

And the significance of this is?

Lots of thing fail open, and maybe draw hot, hissing arcs as they do, potentially igniting the surrounding materials.

There may be UL LEDS, made of non-flammable material, containing the arcing, etc. I suspect there are, since there are now mains-connectable LED lights.

Do you think the cheap chinese Christmas lights always use the best types?

If you do, I have a bunch of really good investments you absolutely have to get into. Double your money in 2 weeks, no questions. Just send me a check and your contact information.......................

Dunno why I have bothered with this. Evan is an amusing and decent sort, but once he is committed to a position, he is "never wrong".

Evan
05-04-2008, 11:13 AM
Just what position am I committed to?

I have taken apart the Christmas lights to extract the LEDs for other uses. There aren't any fuses or fusible links to be found anywhere.

Capacitors DO have a "wattage" rating...... but it isn't strictly what you think. it is a CURRENT rating, and the limit is due to the inherent "residual" resistance of the part, which is generally 50 to 1000 times smaller than the reactance at 120Hz.

A current rating isn't a wattage rating Jerry. You know that.

You and DP keep skating around not wanting to admit one key point. Capacitive reactance isn't the same as resistance even though the same unit is used to measure it. Capacitive reactance can be expressed in terms of effective resistance measured in ohms. It isn't resistance in any way though. The capacitor charges and stores electrons. When it is "full" no more can pass through. During the half cycle as the capacitor charges the effective resistance continuously varies from the instantaneous equivalent of a direct short to a the equivalent of an open circuit. It doesn't actually ever equal the effective resistance value in ohms except for a single instantaneous moment in time during each half cycle.

The reason no significant energy is dissipated as it is with a resistor is because the capacitor isn't a resistor and is not limiting the circuit current via resistance. Once the capacitor is charged on the half cycle no more current can flow until the current direction reverses on the next half cycle. Then the energy stored in the capacitor is returned to the power line and the capacitor begins to charge in the other direction.

While a capacitor does have true resistance to current flow as you pointed out it is only a very small fraction of the reactance.

The bottom line is that the capacitor is not the same as a resistor. The current that flows through a resistor is mainly voltage and load dependent. The current that flows through a capacitor is mainly frequency and load dependent. Those are very different modes of operation.

dp
05-04-2008, 11:34 AM
So that begs the question of where is all that heat when the capacitor is being used? And, why don't capacitors have a wattage rating?
------------------------

Ideal capacitors have no resistance so wattage is not a factor. In reality, capacitors have ESR (equivalent series resistance) and it is a factor in some high current circuits. Skin effect in RF circuits is important for capacitors, and inductance found in some capacitor configurations is also a factor for RF circuits.

Um, yes, constructional details indeed. A .47 mfd 250v capacitor isn't a failure prone electrolytic. Most usually it will be a poly or even ceramic capacitor. I have never seen one fail in a short condition other than being hit by lightning in which case the capacitor disappeared (failed short then failed extremely open).

Mylar caps and very definitely the old paper caps are vulnerable to perferation with high frequency pulses. This can short them out. I used to write a lot of Hi-Pot testing procedures for MIL-HDBK-217 and we would test capacitors to the failure point. They definitely short out.

And the fact that capacitive reactance, while measured in ohms, has nothing to do with resistance.

Reactance is an imaginary number (j operator).

Same question. Where is the heat?

Depends on the ESR. See above. Also have a look that this page:

Evan
05-04-2008, 11:36 AM
Didn't I just address those items?

J Tiers
05-04-2008, 11:39 AM
Just what position am I committed to?

among others, apparently:

1) that you don't need a fuse because LEDs are perfectly safe for use on mains power since they fail open.

2) that any old capacitors are perfectly OK to use on mains power with no fuse.

3) that capacitors limit current in a way that a resistor cannot, setting a current regardless of load or voltage.

I have taken apart the Christmas lights to extract the LEDs for other uses. There aren't any fuses or fusible links to be found anywhere.

Maybe you need to check the UL/CSA numbers..............

What other components are present other than the LEDS themselves?

A current rating isn't a wattage rating Jerry. You know that.

But it "implies" wattage at any particular voltage and frequency, since it is a limit on the allowable current through the part.

A resistor wattage rating is ALSO an effective "implied" current rating due to I^2*R...........................

You and DP keep skating around not wanting to admit one key point. Capacitive reactance isn't the same as resistance even though the same unit is used to measure it.
The bottom line is that the capacitor is not the same as a resistor. The current that flows through a resistor is mainly voltage and load dependent. The current that flows through a capacitor is mainly frequency and load dependent. Those are very different modes of operation.

Oh. we are not skating...........

We might accuse YOU of skating a bit, you have changed your point around......

Your original point was that the capacitor had unique features that made it limit current in a way that the resistor cannot.

such as:

In each circuit the effective series resistance of the load is different but the current remains the same. This does not occur with a pure resistor instead of a capacitor.

The above circuits take advantage of the fact that at a particular AC frequency a particular value of capacitor will pass a certain amount of current regardless of what the load is.

Neither of those statements is in fact true, and that is what we objected to.

Lew Hartswick
05-04-2008, 12:09 PM
Evan I sympthize with you. The Nit pickers rule the web. :-)
...lew...(retired electronics engineer)

dp
05-04-2008, 01:24 PM
There is yet another subtlety in this circuit - at the end of the first complete half cycle the charge across the capacitor is 170 v minus the voltage drop across the diodes and resistor. That is added to the line voltage when the cycle reverses, so the peak voltage applied to the diodes and resistor is actually quite high. We've been tossing around RMS figures so far and that is not the most accurate way to represent the circuit.

But Kirchhoff's 2nd law still trumps.

J Tiers
05-04-2008, 03:33 PM
Evan I sympthize with you. The Nit pickers rule the web. :-)
...lew...(retired electronics engineer)

REJECTED

No, the main issues are safety related, AS YOU VERY WELL KNOW.

Sympathize if you wish........ when someone electrocutes himself or starts a fire due to poor design.

Just like the unfused, unprotected EDM that was in HSM some years ago......... and was defended as "not that bad" :rolleyes: :rolleyes: :rolleyes: :rolleyes:

The other points are just wrong and confusing to someone who, unlike yourself, does NOT know better.

IF SOMEONE STATED, AND STICKS TO, A POSITION WHICH IS WRONG AND MAY LEAD OTHERS INTO A DANGEROUS OR TROUBLESOME POSITION.....

OTHERS ARE OBLIGATED TO CORRECT THAT.

to do otherwise is to become complicit in causing their potential death or injury in the one case.....

Or just complicit in the spewing of false technical data everywhere on the web..... which someone may actually believe and try to use.....

I couldn't get away with that at work........ Why should someone "get a pass" because "it's only on the internet"?

YOU, Mr Hartswick, are now an official, complicit, part of the "dumbing-down" of the internet.

dp......... the capacitor charge isn't really the issue, it comes up as a current, since the diodes etc will conduct enough to hold voltage down to their forward voltage. But one half cycle of current may be higher than expected. Not a particular problem, I think.

Evan
05-05-2008, 08:20 AM
that any old capacitors are perfectly OK to use on mains power with no fuse.
Exactly when did I say or imply that? When did I object to your suggestion that a particular type be used? That's what all this nitpicking revolves around, isn't it?

Swarf&Sparks
05-05-2008, 08:28 AM
An incandescent lamp makes a good current regulator.

Oh, that kinda defeats the purpose, eh? :D

Looks fine to me Evan. If they can't take care, they shouldn't be there!

dp
05-05-2008, 10:08 AM
An incandescent lamp makes a good current regulator.

Oh, that kinda defeats the purpose, eh? :D

You bet - they're quite good at it, actually, for such a simple device. The filament resistance goes up with heat, and heat goes up with voltage. The current also goes up but only slightly.

Here's an interesting use of the principle including a constant current curve chart:

http://www.passdiy.com/projects/zenlite1.htm

I'm sorry to see the tone of the thread go south. I enjoyed thinking about Evan's circuit while spewing chips in the shop. For such a simple circuit it has some very complex and interesting characteristics.

darryl
05-05-2008, 05:57 PM
Come to think of it, the students in my industrial power class in highschool built a power supply based on a series capacitor direct from the ac line. I was one of the students, not the perpetrator of this highly unsafe device. It was for running a radio normally meant to run on batteries. I guess I didn't think of it at the time, or didn't know any better, but that was obviously a dangerous project to be having the students build and use. Maybe we were supposed to get a shock out of it to build our awareness of the dangers of electricity. I know that one of the guys forgot to include the series resistor to the neon bulb, and when he first plugged it in, there was a 'significant event'.

J Tiers
05-05-2008, 11:18 PM
Exactly when did I say or imply that? When did I object to your suggestion that a particular type be used? That's what all this nitpicking revolves around, isn't it?

No nitpicking involved, and THAT was not necessarily the issue, if you agree a better part exists for the purpose. It's news to me that you agree, but OK.

Generally I think there isn't anything basically awful , it's fine...... as I said...... with a couple issues. I don't agree with the "its a deathtrap" folks, with proper insulation and so forth, a properly rated series capacitor is fine.

I suggested a fuse, and a change of capacitor, and expressed some doubt that LEDs were properly rated for direct contact to people while on a non-isolated mains line...... For UL or CE there should generally be 2 layers of insulation, equivalent spacing, OR one layer and a grounded chassis. Dunno if the LEDs in your christmas lights have that or not.........

And a couple of us had an issue with the "magic" current regulation of a capacitor............................... (the bit about how "a resistor can't do that")

The response, and other statements were as I have quoted above.... I think it is quite self-explanatory.

Not MY nitpicking, but I HAVE been paid to assure that the (sometimes quite nitpicky) rules made by others were adhered to..... So I can claim some specialized knowledge in the matter.

In the end, those rules, and others, are what distinguishes good stuff from the chinese stuff we keep reading about in the paper. The poisons or contamination in food and medicine, the fake UL tags, the unsafe counterfeit safety equipment, etc, that is what the various rules keep out of the stuff you use and trust.

By the way, you asked where the heat is................from the capacitor.........

The answer is "in the generator". A reactive load returns power to the source without dissipating it. The generator has to "eat" it, which heats up the generator, and explains much of why why the power co hates reactive loads.

In your case, the capacitor probably just offsets some microscopic part of the overall inductive load on the mains, but that is via the same effect that would cause heating without an inductive load.

dp
05-05-2008, 11:46 PM
And a couple of us had an issue with the "magic" current regulation of a capacitor...............................
Unless there are some physics I don't understand, magic may be required. With a small number of LED's in series there is not a large impact on current in the circuit by changing the number of diodes. There does come a time though when conduction angle comes into play. The LED's only conduct when the voltage across them exceeds the cumulative forward voltage drop. Throw a little trig at the problem and you'll see that with a not very large number of diodes, the conduction time drops to less than half of a half-cycle. From that point and above they are off most of the time. So both the peak current and the conduction time are reduced as diodes are added. I don't see how that circuit could support 80 LED's (I presume two sets of 40 back to back) without a serious impact on both average current and peak current.

I think though that the series capacitor does provide a measure of safety not found without it. It increases the source impedance of the nuclear power station output to substantially 5600 ohms, and that can't hurt anything. It's equivalent to an isolation transformer in this regard.
Unlike the transformer it does require a carefully wired power cord to ensure the capacitor is on the hot side of the outlet though it is trivial to simply use two capacitors rather than one.

Paul Alciatore
05-06-2008, 12:13 AM
Why all the paranoia about the circuits? It's no different than any other 120 vac powered lamp. I have yet to see a lamp with a fuse anywhere. LEDs fail in a similar manner in that they usually fail open. They have an inherent fuse inside in the form of the "p" connection lead which is a very fine wire.

As far as my "argument" it holds water better than your average incandescent lamp in which the current available is far more than enough to fibrillate your heart, failed or not.

......

Evan,

Sorry if I offended. I was concerned because this is a machinist board, not an electronics or electrical one. Some members here have a very good grasp on electrical/electronic matters and some do not. I would not like to see any one get hurt. To me, your remark seemed to say or imply that the circuit was perhaps safer than it really is. I just wanted everyone to understand that there are risks in experimenting with this kind of thing.

As for regular lamps, I would agree that they are not the safest construction. A standard medium screw base socket has a paper insulator that is all that separates the outer screw socket from the outside world which is often a metal lamp. These paper insulators age and often crack apart with the heat of the bulbs. This is why such lamps are almost always rated for 40 Watts or LESS. But everyone puts 60, 75, 100, and even 150 Watt bulbs in them. I can only wonder how many are shocked by them.

As for some of the other remarks here, a LED may have a very fragile connection within it, but the length of that connection may be very short. Fuses have Voltage ratings as well as current. The Voltage rating is for the Voltage it can withstand (is guaranteed to withstand) after blowing. A higher Voltage may arc over and get you. An LED acting as a fuse would also have a Voltage rating and it may not be very high. 115 VAC actually has a peak Voltage of about 160 Volts or greater. This could jump a small gap. Again, I am trying to make everyone aware that you can not count on these kinds of things for protection. Fuses are fuses and LEDs are LEDs.

I am not at all sure how the capacitor functions in the LED circuits. Yes it will have an AC impedance. Any impedance will help to limit the current. But, a current that is limited by a constant impedance, be it purely resistive or capacitive reactance, will be dependent on the Voltage drop in the rest of the circuit: that is, across the LEDs. So it does not make any sense to me for the same value of capacitor or resistor to be used with the same source Voltage and different numbers of LEDs in series. A capacitor is not an active component so it's impedance will be the same unless the frequency changes. I don't have time to try to figure it out or to experiment and measure. I leave that to someone else. I have not yet seen a good explanation here.

A parting thought: some of the jury rigged electrical circuits I have seen have scared the heck out of me. In my opinion, you can't be too careful. The emergency rooms and fire departments are overworked already.

Evan
05-06-2008, 07:11 AM
This isn't a new idea and it isn't my idea. Capacitor input lamp ballasts have been around for a long time. It's the reason that very old fluorescent lamp ballasts contain PCBs since they used oil filled capacitors.

J Tiers
05-06-2008, 08:10 AM
This isn't a new idea and it isn't my idea.

Correct. And it is a good idea. Fairly "green" as well, since the capacitive reactance offsets the typical mains inductive reactance in some small measure, and avoids a certain amount of resistive losses that would simply waste power.

There is NO PROBLEM with the general capacitor-limited circuit when proper parts etc are used. In some ways it can be SAFER than a resistor, since the heat is largely avoided (at least locally).

Properly done, it is perfectly acceptable.

I have repaired a DeWalt battery charger (for a portable drill). There was a large series capacitor, a fuse, and some charging circuitry. The battery plugged directly into the charger base, putting everything at mains voltage. NO isolating transformer.

That item is perfectly fine, carries UL etc.

IIRC, the LED indicators on it were behind "light pipes" that avoided having the LEDs exposed directly, but I do not know if that was due to safety, or simply assembly convenience.

BTW for the "miss Emily" worry-wart types who will tell me not to attempt to repair such things; a) I am capable of designing the product. b) the problem was a fuse, which blew due to the owner using the charger powered from a "modified square wave" inverter. The waveform from such an inverter is NOT suited to the capacitive dropping technique. Too much high frequency.

Peter Sanders
05-07-2008, 01:56 AM
Yep. I just ran the reactance and it comes out about 18 ma.
For the electronicaly challenged:
1
X(sub)c = ______________
2 Pi F C

:-) Gad trying to do a math expression in ASCII is a riot.
Wonder how this is going to display?
...lew...
(edited) FORGET IT . I gues there is no way to make it,the spaces are ignored.
I copied this ....

1
XC = --------
2*pi*f*C
from here...

http://www.geocities.com/siliconvalley/2072/elecrri.htm

I wonder if it works? Uh-oh, it does not work.

There is always screen capture (image) although that's a pain because you need somewhere (online) to store the image as this forum does not allow attachments.

http://www.multiline.com.au/%7Epsanders/images/Image1.jpg
Print Screen and MS Paint :eek:

MSPaint is your friend ?(shudder :eek: )

I use Paint Shop Pro X2, but even Irfanview can do better cropping and pasting :)

wmgeorge
05-07-2008, 10:09 PM
I copied this ....

1
XC = --------
2*pi*f*C
from here...

http://www.geocities.com/siliconvalley/2072/elecrri.htm

I wonder if it works? Uh-oh, it does not work.

There is always screen capture (image) although that's a pain because you need somewhere (online) to store the image as this forum does not allow attachments.

http://www.multiline.com.au/%7Epsanders/images/Image1.jpg
Print Screen and MS Paint :eek:

MSPaint is your friend ?(shudder :eek: )

I use Paint Shop Pro X2, but even Irfanview can do better cropping and pasting :)

I guess what my concern is, all the imported electrical and electronics from China. How do we really know if the UL and Canadian approvals are real? You gotta wonder, Christmas lights for \$4.95, and you want safe too?