Using 120VAC with LED’s – Why not use more LED ’s versus lar

[Tom Biasi]

On 11/24/2012 7:23 PM, P E Schoen wrote:

"Tom Biasi" wrote in message news:50b0d8a2$0$24755$607ed4bc@cv.net...

I was just answering the question of what volt meters I had available.
It's totally irrelevant to what was being discussed.

Michael used the 1/2 power to calculate what voltage would produce
it, but the voltage would be sine-wave AC not 1/2 rectified DC.

It matters not if the waveform is sinusoidal, square, half-wave
rectified, either polarity DC, or any combination thereof. A proper
true-RMS meter will read any of them as the equivalent heating effect.
However, components other than ideal resistors are affected by the
harmonic content, duty cycle, peak voltages, and crest factor. Taken to
extremes, there will be a measurable difference because of the effects
of time, although the total amount of heat generated will be the same.

Since you mentioned the Ballantine meters it would be assumed that they
were what you used for your measurements as you posted previously for a
half-wave rectified voltage. The fact that they are capacitively coupled
is highly relevant.

Paul
Paul,

You can't really be serious. Dwelling on Michael's reference to the
meters of 50 years ago is ludicrous.
The bottom line is this: If I have a 120 volt RMS sine wave and I power
a resistive load through a diode, what voltage will I see across the
resistive load?
Just give me your answer and I'll go away.

Tom

 

[P E Schoen]

"Tom Biasi" wrote in message news:50b166f3$0$24745$607ed4bc@cv.net...

Paul,
You can't really be serious. Dwelling on Michael's reference to the meters
of 50 years ago is ludicrous.

I have been designing industrial test equipment for almost 40 years, using a
variety of means to read true RMS. My Ortmaster uses DC coupling and
computes the RMS value from a sequence of ADC samples. It reads exactly the
same using a sine wave or DC, or any combination thereof. I have also worked
with true-RMS meters which used a vacuum bulb heater and thermocouple, also
DC coupled, and that technology was at least 50 years old. One of my first
analog designs used an AD533 RMS IC:
http://www.analog.com/static/imported-files/data_sheets_obsolete/OBSOLETE%20WATERMARK/AD533.pdf

The bottom line is this: If I have a 120 volt RMS sine wave and I
power a resistive load through a diode, what voltage will I see
across the resistive load?
Just give me your answer and I'll go away.

http://enginuitysystems.com/pix/120AC_Half_Wave.png

For 119.29 VAC through a silicon rectifier, the voltage across 100 ohms is
83.4 VRMS and 53 V Average.

It does go counter to intuition.

Paul

 

[P E Schoen]

Actually it was the AD536:
http://www.analog.com/static/imported-files/data_sheets/AD536A.pdf

Paul

 

[Tom Biasi]

On 11/25/2012 3:28 AM, P E Schoen wrote:

"Tom Biasi" wrote in message news:50b166f3$0$24745$607ed4bc@cv.net...

Paul,
You can't really be serious. Dwelling on Michael's reference to the
meters of 50 years ago is ludicrous.

I have been designing industrial test equipment for almost 40 years,
using a variety of means to read true RMS. My Ortmaster uses DC coupling
and computes the RMS value from a sequence of ADC samples. It reads
exactly the same using a sine wave or DC, or any combination thereof. I
have also worked with true-RMS meters which used a vacuum bulb heater
and thermocouple, also DC coupled, and that technology was at least 50
years old. One of my first analog designs used an AD533 RMS IC:
http://www.analog.com/static/imported-files/data_sheets_obsolete/OBSOLETE%20WATERMARK/AD533.pdf


The bottom line is this: If I have a 120 volt RMS sine wave and I
power a resistive load through a diode, what voltage will I see
across the resistive load?
Just give me your answer and I'll go away.

http://enginuitysystems.com/pix/120AC_Half_Wave.png

For 119.29 VAC through a silicon rectifier, the voltage across 100 ohms
is 83.4 VRMS and 53 V Average.

It does go counter to intuition.

Paul
Thanks. The 53 V. average was my entire point.

 

[Inquisitor]

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 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 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.

Thanks for all your help.

 

[P E Schoen]

"Inquisitor" wrote in message
news:c0adbbff-76a3-403a-915e-6cbe300eeac7@googlegroups.com...

[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.

Thanks for all your help.

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:
http://www.mouser.com/ProductDetail/Supertex/HV9922N3-G/?qs=sGAEpiMZZMsE420DPIasPj8rz0JawKoMhocZ5iO2q5I%3d
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/Supertex/CL220N5-G/?qs=sGAEpiMZZMsE420DPIasPiJw3Ed9o8pZbKrvGoXkOe8%3d

Paul

 

[Jasen Betts]

On 2012-12-01, Inquisitor <dec720@att.net> wrote:

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

pro

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 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 over 40 LED’s is about 140V.

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

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.

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.

--
⚂⚃ 100% natural

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[Inquisitor]

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 reasons, than educational purposes…

PART I

Ok… I think reading your two’s posts, you’re basically saying the same thing. I knew that the LM317 was limited to a 40V differential. The problem was… I should have known better about the output of the bridge. I was 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 the 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 peaking between 160 and 170 volts. I don’t have an oscilloscope, so I’m guessing it would looks something like this…

http://tech.garmf.com/Images/Voltages.png

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 the 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 Vp-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 Volt differential. Does it serve any other functionality? Considering that its rated for 1.5 amps, what kind of failure mode should I expect from running the LM317 with too much voltage through it, but only at 30ma?

Thanks