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

D

#### [email protected]

Jan 1, 1970
0
QUICK VERSION

Using LED calculators like http://www.bowdenshobbycircuits.info/led.htm andfor a given supply voltage, why would I not use more LED’s and have a smaller current limiting resistor?

DETAILED VERSION

Why – Mainly… I just want to experiment with LED lighting. But secondarily… 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 article http://www.bowdenshobbycircuits.info/page10.htm#ledlamp.gif

• 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 thebridge 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 inthe 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 (and 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.

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 2400 ohm, 2.263 Watts current limiting resistor. • However, if I use 44 LED’s, it suggests I can use a 240 ohm, 0.216 watt resistor. Thus, I would be putting out 75% more light for the same amount of electricity and using an easier to find quarter watt resistor. Is there some otheraspect about this circuit that suggests that is a bad idea? Thanks for your help. T #### Tom Biasi Jan 1, 1970 0 QUICK VERSION Using LED calculators like http://www.bowdenshobbycircuits.info/led.htm and for a given supply voltage, why would I not use more LED’s and have a smaller current limiting resistor? DETAILED VERSION Why – Mainly… I just want to experiment with LED lighting. But secondarily… 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 article http://www.bowdenshobbycircuits.info/page10.htm#ledlamp.gif • 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 (and 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. 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 2400 ohm, 2.263 Watts current limiting resistor.
• However, if I use 44 LED’s, it suggests I can use a 240 ohm, 0.216 watt resistor.

Thus, I would be putting out 75% more light for the same amount of electricity 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?

E

#### Ecnerwal

Jan 1, 1970
0
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:

http://www.supertex.com/pdf/datasheets/CL220x.pdf

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.

I

#### Inquisitor

Jan 1, 1970
0
Being new to electronics, I don't know all the key words to do the comprehensive searches.

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

Ecnerwal, Thanks for the links, the references and the new phrases. Between 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 120AC stepped down to less than 37DC. I'm guessing I'd need at least a transformer 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...

http://www.ebay.com/itm/New-3x1W-11...414?pt=LH_DefaultDomain_0&hash=item2c69b15936

T

#### Tom Biasi

Jan 1, 1970
0
Being new to electronics, I don't know all the key words to do the comprehensive searches.

Tom, I've seen that capacitor, but haven't yet found how to size it. It was 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

J

#### Jasen Betts

Jan 1, 1970
0
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.

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

K

#### [email protected]

Jan 1, 1970
0
120 VAC is 339.36 Volts, peak to peak.

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

T

#### Tom Biasi

Jan 1, 1970
0
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.
0.318 x Vmax of the input sinusoidal waveform or 0.45 x Vrms of the
input sinusoidal waveform.

T

#### Tom Biasi

Jan 1, 1970
0
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.

T

#### Tom Biasi

Jan 1, 1970
0
Show that I'm wrong. There were series sting TV sets built with a
"Dropping Diode". 120 VAC supply and 84 volt string of filaments. It
took a little calculation to see what was happening. The bad thing was
that the half wave AC played hell with the CRT filament. I modified
them by adding a separate filament transformer for the CRT.
I see what you are saying. But don't forget that RMS only applies to a
sine wave not 1/2 of a sine wave.
I just showed you what your "couple dozen others" were using.
Wire it up and see.

Tom

P

#### P E Schoen

Jan 1, 1970
0
"Michael A. Terrell" wrote in message
I've seen it 'wired up' in lots of variations. Also, I didn't say
it was RMS, I sad it was the 'Effective Voltage'. RMS can't
apply to anything except a 100% pure sine wave?

Here is an interesting calculator for RMS value:
http://www.daycounter.com/Calculators/RMS-Calculator.phtml

The RMS value of a half-wave rectified sine wave is 1/2 Vpeak, which is also
sqrt(2) * Vrms based on the unrectified waveform. It makes sense when you
consider what voltage is required for 1/2 power.

The effective voltage is by definition the same as RMS voltage, or the DC
voltage which will produce the same heating effect (wattage).

It gets interesting when you compute a portion of a sine wave with a
non-integral number of half-cycles (or quarter-cycles, actually). The
calculated value over time oscillates above and below the final value and
converges to it over many cycles. It equals the final value every 90
degrees. Thus, an RMS voltmeter using digital samples and a calculating
algorithm will be most stable if the measurement period is an integral
number of half-cycles. A period of 100 or 200 mSec is ideal for 50 and 60
Hz.

Phase Amplitude Square RMS
0.00 0.000 0.000 0.000
18.00 52.442 2750.155 52.442
36.00 99.750 9950.155 79.688
54.00 137.295 18849.845 102.551
72.00 161.400 26049.845 120.000
90.00 169.706 28800.000 131.453
108.00 161.400 26049.845 136.900
126.00 137.295 18849.845 136.957
144.00 99.750 9950.155 132.877
162.00 52.442 2750.155 126.491
180.00 0.000 0.000 120.000
198.00 52.442 2750.155 115.503
216.00 99.750 9950.155 114.273
234.00 137.295 18849.845 116.206
252.00 161.400 26049.845 120.000
270.00 169.706 28800.000 123.935
288.00 161.400 26049.845 126.602
306.00 137.295 18849.845 127.256
324.00 99.750 9950.155 125.886
342.00 52.442 2750.155 123.117
360.00 0.000 0.000 120.000

Here is an Excel spreadsheet you can play with:
http://enginuitysystems.com/files/SineWaveSample1.xls

Paul

J

#### Jasen Betts

Jan 1, 1970
0
I see what you are saying. But don't forget that RMS only applies to a
sine wave not 1/2 of a sine wave.
I just showed you what your "couple dozen others" were using.
Wire it up and see.

RMS can be applied to any waveform.

Square to voltage, compute the mean of this, and take the square root of that.

It works just fine for a half-sine wave, giving a result sqrt(0.5) of
the full wave.

T

#### Tom Biasi

Jan 1, 1970
0
RMS can be applied to any waveform.

Square to voltage, compute the mean of this, and take the square root of that.

It works just fine for a half-sine wave, giving a result sqrt(0.5) of
the full wave.
I wasn't implying it couldn't be calculated on any wave form, just
reminding Michael his was about a 1/2 sine wave at 50%.
Tom

T

#### Tom Biasi

Jan 1, 1970
0
50% power, which requires reducng the input to 70.7%. Mutiply
70.7%(I) * 70.7%(E) and you get 49.49%(P)
Wire it up and measure it with a true RMS meter.

Tom

T

#### Tom Biasi

Jan 1, 1970
0
I have. I have four working Fluke 8050A meters. I also have a
'Sensitive Research' brand bolometer type true RMS meter.
What did you see?

T

#### Tom Biasi

Jan 1, 1970
0
Exactly what I've been saying all along. 50% of the power of a full
sine wave delivered into a resistive load. I saw that 30 years ago.
You remove 50% of each complete cycle, and you remove 50% of the power.
Good. That's why I said way back on top, choose your own math.
I have used a diode in series with tube filaments for 50 years. I always
Have fun.
Tom

T

#### Tom Biasi

Jan 1, 1970
0
Then you are doing something wrong, like running it on 85 to 92
volts, or using a diode with a very high forward drop.
Not counting the forward conducting drop there would be 1/2 the RMS
across the diode.

T

#### Tom Biasi

Jan 1, 1970
0
Bullshit. You said you've been doing it for 50 years. What True RMS
Voltmeter did you have 50 years ago? You really need to study how the
diode works.
We were having a discussion and now you become insulting.
Have a nice Thanksgiving.

I

#### Inquisitor

Jan 1, 1970
0
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 constant current mode.

Here is a picture of the basic circuit off the Internet.
http://tech.garmf.com/Images/LM317LEDdriver.jpg

Here is an Internet calculator I used to size the resistor.
http://www.reuk.co.uk/LM317-Current-Calculator.htm

Here is a picture of my circuit (while running) and being lit by the bunch of LED’s.
http://tech.garmf.com/Images/LEDSetup.jpg

And here is a close-up of the circuit.
http://tech.garmf.com/Images/LM317.jpg

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 theLM317 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 over 40 LED’s is about 140V.

As I connect fewer and fewer LED’s the current goes up as expected. However, 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 for me to research.

P

#### P E Schoen

Jan 1, 1970
0
"Inquisitor" wrote in message

[snip]
As I connect fewer and fewer LED’s the current goes
up as expected. However, 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 for me to research.

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

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: http://www.mouser.com/ProductDetail...=sGAEpiMZZMsE420DPIasPj8rz0JawKoMhocZ5iO2q5I= http://www.mouser.com/ProductDetail/Clare/MXHV9910B/?qs=sGAEpiMZZMsGzNf1qgY4ZAg0jSvHVA1V 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:
http://www.supertex.com/pdf/datasheets/CL220x.pdf
http://www.mouser.com/ProductDetail...=sGAEpiMZZMsE420DPIasPiJw3Ed9o8pZbKrvGoXkOe8=

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