Using 120VAC with LED?s ? Why not use more LED ?s versus large Current Limiting Resistor?

QUICK VERSION

Using LED calculators like

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and for a given supply voltage, why would I not use more LED?s and have a sm aller current limiting resistor? DETAILED VERSION

Why ? Mainly? I just want to experiment with LED lighting. But seconda rily? a 120VAC LED bulb is $25. With some $2 worth of electronics I can make a nice under cabinet light.

I?m not an EE. I don?t know what may be important information, so I? ve included everything I?ve done. I?ve basically started with this art icle

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? I?m using a free-be Harbor Freight volt meter ? My AC voltage is 121.4 VAC and pretty steady. ? Using pieces out of an old, burned out PC power supply, I?ve made the bridge out of (4) IN5406 diodes ? My measured DC out of the bridge is 108.6 VDC ? I?ve paralleled a 104J (400V) capacitor across the DC output found in the same power supply ? This brought it up to 161.2 VDC.

All this? ?I think? I understand.

Now, this is what I?m having difficulty understanding. In the example (a nd everyone else?s) the number of LED?s is quite a bit smaller than cou ld be supported by the voltage available. Thus the current limiting resist or is quite large in both ohms and power rating. In my limited understandi ng, I would think one would want to use more LED?s, thus reducing the amo unt of electricity wasted as heat in the resistor. As a secondary benefit, I?m finding it far easier to find 1/4 and 1/2 watt resistors instead of

2.5W resistors.

In my specific example, I used the ?LED Series Resistance Calculator?. The LED?s I have are White LED?s ($3.60 for 100 from China) so I have plenty to try and/or burn up :) They?re rated ~3.5V at 30 mA.

? If I use 25 LED's at 3.5V, 30mA, 161.2V, the calculator says I need a 2

400 ohm, 2.263 Watts current limiting resistor. ? However, if I use 44 LED?s, it suggests I can use a 240 ohm, 0.216 wa tt resistor. Thus, I would be putting out 75% more light for the same amount of electric ity and using an easier to find quarter watt resistor. Is there some other aspect about this circuit that suggests that is a bad idea?

Thanks for your help.

Reply to
dec720
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for a given supply voltage, why would I not use more LED?s and have a smaller current limiting resistor?

120VAC LED bulb is $25. With some $2 worth of electronics I can make a nice under cabinet light.

included everything I?ve done. I?ve basically started with this article

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out of (4) IN5406 diodes

same power supply

everyone else?s) the number of LED?s is quite a bit smaller than could be supported by the voltage available. Thus the current limiting resistor is quite large in both ohms and power rating. In my limited understanding, I would think one would want to use more LED?s, thus reducing the amount of electricity wasted as heat in the resistor. As a secondary benefit, I?m finding it far easier to find 1/4 and 1/2 watt resistors instead of 2.5W resistors.

LED?s I have are White LED?s ($3.60 for 100 from China) so I have plenty to try and/or burn up :) They?re rated ~3.5V at 30 mA.

ohm, 2.263 Watts current limiting resistor.

resistor.

and using an easier to find quarter watt resistor. Is there some other aspect about this circuit that suggests that is a bad idea?

Hi, Unless you own your own power station I can pretty much guarantee you that your voltage will not always be as you measure it now. If you get your LED total forward voltage close to the max available you may loose the conductance if the line voltage dips. Line spikes are another consideration since LEDs are not that forgiving of over voltage.

Did you note that Bowden's circuits often used a series capacitor?

Reply to
Tom Biasi

Why use a current limiting resistor at all?

LED operating voltage is not all that well specified in real life.

The parameter you really want to control is current. And operating voltage varies non-linearly with current.

If you are using a resistor to control the current, you need to size the resistor for the maximum voltage in, and the minimum LED voltage out, to make it "safe" - also limiting the current you can run when values are more "average" if a resistor is how you set your operating current.

Devices are made to light as many LEDs as possible* in a simple manner (others are made to do it in a complex manner, some of which may be more efficient.)

One of the simple ones is:

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You can also use an LM317 in floating current source mode. Really any current source you care to cook up - a resistor between a not entirely stable input voltage and a not terribly reliable operating point voltage is a fairly crappy current source, so you can do better in many ways.

So long as you are willing to eat the cost of a few parts if you let the magic smoke out of them, nothing like playing with a few different versions. You can always remove a few LEDs if you find that more is not better.

If you'd rather just build something that works, using a part designed to make it simple & reliable, such as the above, might make sense.

  • - I suppose the 5V the simple device typically costs means it can't really light as many as possible for a given voltage, but it's fairly close and much simpler than trying to maximize that figure while still holding everything else under control by other means.
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Reply to
Ecnerwal

Being new to electronics, I don't know all the key words to do the comprehe nsive searches.

Tom, I've seen that capacitor, but haven't yet found how to size it. It w as not intuitively obvious (to me) how it works. I'm guessing it works onl y because its on the AC side.

Ecnerwal, Thanks for the links, the references and the new phrases. Betwee n surfing the CL220 and the LM317, I found lots of good things to explore. It looks like the LM317 would be painful (for me) to figure how to get 120 AC stepped down to less than 37DC. I'm guessing I'd need at least a transf ormer and rectifying bridge. For my other projects (12 Volt DC automotive battery type) it will be a great addition to my "cook book".

My surfing brought up one end product, that I'm curious about. What might be in here that does this job... and so cheaply...

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m2c69b15936

Reply to
Inquisitor

comprehensive searches.

not intuitively obvious (to me) how it works. I'm guessing it works only because its on the AC side.

The series capacitor has a reactance at the line frequency. Reactance is like AC resistance. Since the voltage and current are not in phase through an ideal capacitor the device will not dissipate any power but there will be a voltage drop across it. Not quite like using a resistor because phase angles are involved. If you are interested you can pursue "Capacitive Reactance."

Tom

Reply to
Tom Biasi

a given supply voltage, why would I not use more LED?s and have a smaller current limiting resistor?

120VAC LED bulb is $25. With some $2 worth of electronics I can make a nice under cabinet light.

included everything I?ve done. I?ve basically started with this article

formatting link

out of (4) IN5406 diodes

same power supply

everyone else?s) the number of LED?s is quite a bit smaller than could be supported by the voltage available. Thus the current limiting resistor is quite large in both ohms and power rating. In my limited understanding, I would think one would want to use more LED?s, thus reducing the amount of electricity wasted as heat in the resistor. As a secondary benefit, I?m finding it far easier to find 1/4 and 1/2 watt resistors instead of 2.5W resistors.

LED?s I have are White LED?s ($3.60 for 100 from China) so I have plenty to try and/or burn up :) They?re rated ~3.5V at 30 mA.

ohm, 2.263 Watts current limiting resistor.

resistor.

and using an easier to find quarter watt resistor. Is there some other aspect about this circuit that suggests that is a bad idea?

Your power line voltage is given as a nominal value. 120 volts could mean anything from 105 to 126, your resistor has to be able to keep the leds happy in that range, with some derating for temperature too, if you plan to encounter some high ambient temps.

A capacitor makes a lot of sense to drop the voltage with minimal power loss. And then you still need some resistance for when the cap is discharged and hit with a instantaneous voltage of ~150 peak.

I usually also include a diac or zener to limit fast voltage spikes, a cap across the DC output might serve the same purpose.

I have a night light using a full wave bridge to power the leds. A .5 uf / 200 VAC cap in series with the AC input to the fwb along with a

100 ohm 1/4 watt series resistor. A diac across the 3 white leds limits voltage spikes. I accidentally shorted a cap when I was experimenting with it - all three leds shorted, the resistor opened, and it happened so fast I didn't see much of anything - didn't even singe the paint on the carbon film resistor.
Reply to
default

it may have been a fusible resistor, using one of them lets them use a cheaper capacitor, like an ordinary 250V polyester

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Reply to
Jasen Betts

120 VAC is 339.36 Volts, peak to peak.
Reply to
Michael A. Terrell

A "flame proof" resistor would certainly be a better choice, but this was a standard 5% el-cheapo $2/100.

Reply to
default

But unless you're putting it across 240V (both legs) it'll never see more than half that. -ish. .

Reply to
krw

I was just pointing out that in some circuits you can have that across a capacitor in a 120 VAC circuit. I had an argument with someone on another newsgroup who claimed he could use a single 1N34 diode in a tube radio to rectify the AC line. He also refused to beleive that a

1n4007 used in a series filament string wouldn't drop the effective voltage to 84.84 V minus the forward voltage drop. He a a couple dozen others insisted that it would drop to an effective voltage of 60 volts.
Reply to
Michael A. Terrell

0.318 x Vmax of the input sinusoidal waveform or 0.45 x Vrms of the input sinusoidal waveform.
Reply to
Tom Biasi

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It is a bad idea because

  1. A small fluctuation in the mains supply will cause a large fluctuatio= n =

in the current through the LEDS - perhaps enough to destroy them.

  1. The 3.5v @ 30mA is a fairly nominal figure. It will change with the =

exact batch of LEDs, perhaps enough to destroy the LEDs

  1. The voltage drop across the LED will also change with temperature. It= =

gets lower as the temperature increases so current increases, perhaps =

enough to destroy the LEDs.

The larger the resistor the less sensitive it will be to these effects.

As a note, if you want longevity don't run the LEDs at more than 20mA an= d =

use a resistor rated for twice the power calculated

Reply to
David Eather

One half of the AC waveform = 50% power, not 50% voltage into a resistive load so it's equal to 70.7% * VAC, minus the diode drop.

Reply to
Michael A. Terrell

Choose your own math.

Reply to
Tom Biasi

First off? Thank you for all of you that are being helpful. I now understand the points that you?ve made about my original question. Since then, I have dug into a more active circuit using LM317 chip in cons tant current mode.

Here is a picture of the basic circuit off the Internet.

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Here is an Internet calculator I used to size the resistor.

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Here is a picture of my circuit (while running) and being lit by the bunch of LED?s.

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And here is a close-up of the circuit.

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I used calculator above which indicated I need the 56 ohm resistor to give me an output current of 22.3 ma. With an input of 120.1 VAC, I measure 107 .1 VDC out of the full bridge. I also measured 106.6 VDC coming out of the LM317 with no load connected. When connecting to the load of 40 LED?s, I get a current of 10 ma. I kind of expected this since the voltage drop o ver 40 LED?s is about 140V.

As I connect fewer and fewer LED?s the current goes up as expected. Howe ver, I kind of expected the LM317 to start kicking in and keep the current around 22.3 ma. In the picture above, you can see that at 36 LED?s, the current was showing near 30 ma.

Can you tell me what I?m missing? Some key words would be very helpful f or me to research.

Thanks for all your help.

Reply to
Inquisitor
[snip]

With no capacitance in the circuit, the voltages will be pulsing with peaks of 170 V. The LM317 may be unstable without a capacitor on the input. The 56 ohm resistor should limit the output to 22 mA peak, but the 36 LEDs will

clip at 126 V which means the regulator will see about 170-126=54 volts, which is beyond the absolute maximum voltage rating of 40V. The LEDs may be seeing a very high pulse current during the time the device is overvoltaged, and it is a wonder that catastrophic failure has not occurred. You may want to add a capacitor to the bridge, which will be a steady 160-170 VDC, then a resistor and a 35 volt zener across the LM317 so the differential will be limited. The resistor should be chosen to allow about 25 mA at 35 volts, or about 1.4 kOhms 2 watts.

You could also do this without the capacitor, and it will be more efficient, but the LEDs will be subjected to a pulsing waveform and the average current will be less than the peak as determined by the 56 ohm resistor.

The "best" way to do this is with a little switchmode driver which can be obtained for less than $1 and it will work from 20VDC to 400VDC:

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1qgY4ZAg0jSvHVA1V

The first is in a little TO-92 package and fixed at 50mA, while the second is an SOIC-8 and has variable PWM dimming. All you need are a few external components.

For $2 you can get a TO-220 device with a fixed 20mA output that needs only a 10nF capacitor and works from 5V to 220V:

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Paul

Reply to
P E Schoen

pro

LED?s.

WHOA! You can't use an LM317 with more than 40V.

me to research.

search this: LM317 datasheet

On the natinnal semiconductor data sheet the 40V limit is at the top of page 4 An example current regulator circuit can de seen on page 18, it's basicallythe same as yours.

The 40V limit means the voltage difference between your LEDs and the input should never exceed 40V. regular 110V AC can have DC peaks up around 170V somewhere so your LED string is constrained to be one that drops 130V or more else you risk damaging the LM317 and incorrect operation.

don't forget the fuse.

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Reply to
Jasen Betts

I?ll be looking into your other component suggestions, but for now, I?d like to understand what?s going on with this circuit if for no other rea sons, than educational purposes?

PART I

Ok? I think reading your two?s posts, you?re basically saying the sam e thing. I knew that the LM317 was limited to a 40V differential. The pro blem was? I should have known better about the output of the bridge. I w as just reading the DC on the meter and not rationalizing the 107 V wasn? t really DC.

OK? so I found a 200V, 10uf capacitor lying around and added it across th e DC output of the bridge. I?m now reading 165VDC. So? even if I?m seeing 165V, I should assume it still has an AC component and is really pea king between 160 and 170 volts. I don?t have an oscilloscope, so I?m g uessing it would looks something like this?

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So is the following a valid rationalization? If I take the LOWEST voltage and divide by the nominal LED voltage drop, I would get a number of LED?s (160 / 3.5 = 45). This should be a kind of threshold for the LM317. At this value, the LM317 is always restricting t he current to the LED?s and I should see the 22.3 ma?

And further? that at the high peak 170V, would only be using 170 - 45*3.5 = 12.5V of the 40V differential allowed on the LM317?

And that? the AC power could go all the way up to (45*3.5 + 40) = 197 V p-p before the LM317 would be compromised?

PART II

I started reading about the zener dode you are suggesting. If I understand correctly, this protects the LM317 from seeing more than it?s rated 40 V olt differential. Does it serve any other functionality? Considering that its rated for 1.5 amps, what kind of failure mode should I expect from run ning the LM317 with too much voltage through it, but only at 30ma?

Thanks

Reply to
Inquisitor

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