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Using 120VAC with LED’s – Why not use more LED’s versus large Current Limiting Resistor?
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dec...@att.net
Nov 17, 2012, 2:44:35 PM11/17/12
to
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?
Thanks for your help.
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?
Thanks for your help.
Tom Biasi
Nov 17, 2012, 3:22:23 PM11/17/12
to
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?
Ecnerwal
Nov 17, 2012, 5:36:14 PM11/17/12
to
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.
--
Cats, coffee, chocolate...vices to live by
Please don't feed the trolls. Killfile and ignore them so they will go away.
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.
--
Cats, coffee, chocolate...vices to live by
Please don't feed the trolls. Killfile and ignore them so they will go away.
Inquisitor
Nov 18, 2012, 9:36:11 AM11/18/12
to
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-110V-Home-Light-LED-Power-Supply-Driver-Electronic-Transformer-White-/190751791414?pt=LH_DefaultDomain_0&hash=item2c69b15936
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-110V-Home-Light-LED-Power-Supply-Driver-Electronic-Transformer-White-/190751791414?pt=LH_DefaultDomain_0&hash=item2c69b15936
Tom Biasi
Nov 18, 2012, 1:04:36 PM11/18/12
to
On 11/18/2012 9:36 AM, Inquisitor wrote:
> 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
> 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.
>
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
default
Nov 18, 2012, 3:49:21 PM11/18/12
to
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.
Jasen Betts
Nov 19, 2012, 6:11:33 AM11/19/12
to
On 2012-11-18, default <default> wrote:
> 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
> 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.
cheaper capacitor, like an ordinary 250V polyester
--
⚂⚃ 100% natural
--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---
Michael A. Terrell
Nov 19, 2012, 9:29:51 AM11/19/12
to
default
Nov 19, 2012, 9:50:28 AM11/19/12
to
was a standard 5% el-cheapo $2/100.
k...@att.bizzz
Nov 19, 2012, 12:47:33 PM11/19/12
to
more than half that. -ish.
.
Michael A. Terrell
Nov 19, 2012, 1:05:36 PM11/19/12
to
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.
Tom Biasi
Nov 19, 2012, 2:24:14 PM11/19/12
to
input sinusoidal waveform.
David Eather
Nov 19, 2012, 3:13:43 PM11/19/12
to
1. A small fluctuation in the mains supply will cause a large fluctuation
in the current through the LEDS - perhaps enough to destroy them.
2. The 3.5v @ 30mA is a fairly nominal figure. It will change with the
exact batch of LEDs, perhaps enough to destroy the LEDs
3. 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 and
use a resistor rated for twice the power calculated
>
> Thanks for your help.
Michael A. Terrell
Nov 19, 2012, 4:43:43 PM11/19/12
to
resistive load so it's equal to 70.7% * VAC, minus the diode drop.
Tom Biasi
Nov 19, 2012, 7:04:23 PM11/19/12
to
Michael A. Terrell
Nov 19, 2012, 7:31:28 PM11/19/12
to
"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.
Tom Biasi
Nov 19, 2012, 8:03:48 PM11/19/12
to
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
Michael A. Terrell
Nov 19, 2012, 8:14:04 PM11/19/12
to
was RMS, I sad it was the 'Effective Voltage'. RMS can't apply to
anything except a 100% pure sine wave? :)
P E Schoen
Nov 20, 2012, 1:00:51 AM11/20/12
to
"Michael A. Terrell" wrote in message
news:EsOdnddjTJj9RDfN...@earthlink.com...
> 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
news:EsOdnddjTJj9RDfN...@earthlink.com...
> 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? :)
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
Jasen Betts
Nov 20, 2012, 1:43:50 AM11/20/12
to
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.
Tom Biasi
Nov 20, 2012, 8:03:31 AM11/20/12
to
reminding Michael his was about a 1/2 sine wave at 50%.
Tom
David Eather
Nov 20, 2012, 10:48:00 AM11/20/12
to
On Tue, 20 Nov 2012 11:14:04 +1000, Michael A. Terrell
Not at all true. Every waveform has an RMS value.
Michael A. Terrell
Nov 20, 2012, 11:25:46 AM11/20/12
to
Michael A. Terrell
Nov 20, 2012, 11:28:28 AM11/20/12
to
70.7%(I) * 70.7%(E) and you get 49.49%(P)
Tom Biasi
Nov 20, 2012, 2:37:30 PM11/20/12
to
Wire it up and measure it with a true RMS meter.
Tom
Michael A. Terrell
Nov 20, 2012, 3:27:23 PM11/20/12
to
Tom Biasi wrote:
>
> On 11/20/2012 11:28 AM, Michael A. Terrell wrote:
> ? Tom Biasi wrote:
> ??
> ?? On 11/20/2012 1:43 AM, Jasen Betts wrote:
> ??? On 2012-11-20, Tom Biasi ?tomb...@optonline.net? wrote:
> ???? On 11/19/2012 7:31 PM, Michael A. Terrell wrote:
> ????? Tom Biasi wrote:
> ?????? On 11/19/2012 4:43 PM, Michael A. Terrell wrote:
> ??????? Tom Biasi wrote:
> ???????
> ??????? 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.
> ???????
> ?????? Choose your own math.
> ?????
> ?????
> ????? 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.
> ???
> ??? 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%.
> ?
> ?
> ? 50% power, which requires reducng the input to 70.7%. Mutiply
> ? 70.7%(I) * 70.7%(E) and you get 49.49%(P)
> ?
> Your math is fine Michael.
> Wire it up and measure it with a true RMS meter.
I have. I have four working Fluke 8050A meters. I also have a
> Wire it up and measure it with a true RMS meter.
'Sensitive Research' brand bolometer type true RMS meter.
Tom Biasi
Nov 20, 2012, 3:47:54 PM11/20/12
to
Michael A. Terrell
Nov 20, 2012, 6:00:45 PM11/20/12
to
Tom Biasi wrote:
>
> On 11/20/2012 3:27 PM, Michael A. Terrell wrote:
> ?
> ? Tom Biasi wrote:
> ??
> ?? ?
> ?? ? Tom Biasi wrote:
> ?? ??
> ?? ?? On 11/20/2012 1:43 AM, Jasen Betts wrote:
> ?? ??? On 2012-11-20, Tom Biasi ?tomb...@optonline.net? wrote:
> ?? ??
> ?? ?? On 11/20/2012 1:43 AM, Jasen Betts wrote:
> ?? ???? On 11/19/2012 7:31 PM, Michael A. Terrell wrote:
> ?? ????? Tom Biasi wrote:
> ?? ?????? On 11/19/2012 4:43 PM, Michael A. Terrell wrote:
> ?? ??????? Tom Biasi wrote:
> ?? ???????
> ?? ??????? 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.
> ?? ???????
> ?? ?????? Choose your own math.
> ?? ?????
> ?? ?????
> ?? ????? 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
> ?? ?????? Choose your own math.
> ?? ?????
> ?? ?????
> ?? ????? Show that I'm wrong. There were series sting TV sets built with a
> ?? ????? 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 see what you are saying. But don't forget that RMS only applies to a
> ?? ???? 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.
> ?? ??? 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
> ?? ???
> ?? ???
> ?? ?? 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%.
> ?? ?
> ?? ?
> ?? ? 50% power, which requires reducng the input to 70.7%. Mutiply
> ?? ? 70.7%(I) * 70.7%(E) and you get 49.49%(P)
> ?? ?
> ?? Your math is fine Michael.
> ?? Wire it up and measure it with a true RMS meter.
> ?
> ?
> ? 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?
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.
Tom Biasi
Nov 20, 2012, 7:53:24 PM11/20/12
to
I have used a diode in series with tube filaments for 50 years. I always
get about 60-65 volts.
Have fun.
Tom
Michael A. Terrell
Nov 20, 2012, 8:25:48 PM11/20/12
to
Tom Biasi wrote:
>
> On 11/20/2012 6:00 PM, Michael A. Terrell wrote:
> ?
> ? Tom Biasi wrote:
> ??
> ?? On 11/20/2012 3:27 PM, Michael A. Terrell wrote:
> ?? ?
> ?? ? Tom Biasi wrote:
> ?? ??
> ?? ?? On 11/20/2012 11:28 AM, Michael A. Terrell wrote:
> ?? ?? ?
> ?? ?
> ?? ? Tom Biasi wrote:
> ?? ??
> ?? ?? On 11/20/2012 11:28 AM, Michael A. Terrell wrote:
> ?? ?? ? Tom Biasi wrote:
> ?? ?? ??
> ?? ?? ?? On 11/20/2012 1:43 AM, Jasen Betts wrote:
> ?? ?? ??? On 2012-11-20, Tom Biasi ?tomb...@optonline.net? wrote:
> ?? ?? ??
> ?? ?? ?? On 11/20/2012 1:43 AM, Jasen Betts wrote:
> ?? ?? ???? On 11/19/2012 7:31 PM, Michael A. Terrell wrote:
> ?? ?? ????? Tom Biasi wrote:
> ?? ?? ?????? On 11/19/2012 4:43 PM, Michael A. Terrell wrote:
> ?? ?? ??????? Tom Biasi wrote:
> ?? ?? ???????
> ?? ?? ??????? 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.
> ?? ?? ???????
> ?? ?? ?????? Choose your own math.
> ?? ?? ?????
> ?? ?? ?????? Choose your own math.
> ?? ?? ?????
> ?? ?? ????? 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
> ?? ?? ????? Show that I'm wrong. There were series sting TV sets built with a
> ?? ?? ????? 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.
> ?? ?? ???
> ?? ?? ??? 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%.
> ?? ?? ?? I wasn't implying it couldn't be calculated on any wave form, just
> ?? ?? ?
> ?? ?? ?
> ?? ?? ? 50% power, which requires reducng the input to 70.7%. Mutiply
> ?? ?? ? 70.7%(I) * 70.7%(E) and you get 49.49%(P)
> ?? ?? ?
> ?? ?? Your math is fine Michael.
> ?? ?? Wire it up and measure it with a true RMS meter.
> ?? ?
> ?? ?
> ?? ? 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?
> ?
> ?
> ? 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
> get about 60-65 volts.
Then you are doing something wrong, like running it on 85 to 92
> I have used a diode in series with tube filaments for 50 years. I always
> get about 60-65 volts.
volts, or using a diode with a very high forward drop.
Tom Biasi
Nov 21, 2012, 7:49:43 AM11/21/12
to
across the diode.
Michael A. Terrell
Nov 21, 2012, 7:57:33 AM11/21/12
to
Voltmeter did you have 50 years ago? You really need to study how the
diode works.
Tom Biasi
Nov 21, 2012, 8:12:56 AM11/21/12
to
Have a nice Thanksgiving.
David Eather
Nov 21, 2012, 8:21:56 AM11/21/12
to
On Wed, 21 Nov 2012 23:21:10 +1000, David Eather <eat...@tpg.com.au> wrote:
> On Wed, 21 Nov 2012 02:25:46 +1000, Michael A. Terrell
> Ah. When I read it I was surprised at the name/mistake. My funny bone
> needs tweaking.
> On Wed, 21 Nov 2012 02:25:46 +1000, Michael A. Terrell
> needs tweaking.
Michael A. Terrell
Nov 21, 2012, 12:24:29 PM11/21/12
to
contains, rather than just posting nonsense. Sometimes it works. :)
Michael A. Terrell
Nov 21, 2012, 12:27:17 PM11/21/12
to
> > 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.
but you continue posting in ignorance and I have better things to do
with my little remaining time.
Tom Biasi
Nov 23, 2012, 3:11:02 PM11/23/12
to
On 11/21/2012 12:27 PM, Michael A. Terrell wrote:
>
> Tom Biasi wrote:
>>
>> On 11/21/2012 7:57 AM, Michael A. Terrell wrote:
>>>
>>> Tom Biasi wrote:
>>>>
>>>> On 11/20/2012 8:25 PM, Michael A. Terrell wrote:
>>>>>
>>>>> Tom Biasi wrote:
>>>>>
>>>> Not counting the forward conducting drop there would be 1/2 the RMS
>>>> across the diode.
>
>
>
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>
>
>>> 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.
>
>
> You too, but I'm done wth you. I gave you the math & the foremulas,
> but you continue posting in ignorance and I have better things to do
> with my little remaining time.
>
Hi Michael,
>
> Tom Biasi wrote:
>>
>> On 11/21/2012 7:57 AM, Michael A. Terrell wrote:
>>>
>>> Tom Biasi wrote:
>>>>
>>>> On 11/20/2012 8:25 PM, Michael A. Terrell wrote:
>>>>>
>>>>> Tom Biasi wrote:
>>>>>
>>>> Not counting the forward conducting drop there would be 1/2 the RMS
>>>> across the diode.
>
>
>
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>
>
>>> 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.
>
>
> You too, but I'm done wth you. I gave you the math & the foremulas,
> but you continue posting in ignorance and I have better things to do
> with my little remaining time.
>
I hope you had a nice thanksgiving.
I respect that you are done with me so I'll just answer your question.
Back then I worked for Ballantine Labs in Boonton, NJ. There were
several true RMS voltmeters at my disposal.
When applying your math and formulas one must consider we are not
working with AC anymore but unfiltered (or pulsating) DC. The peak is at
at a 50% duty cycle so the filaments would be subjected to that DC value
which would be Vpeak/pi. By your own admission several dozen others
tried to explain this to you.
I bid you well.
Tom
P E Schoen
Nov 23, 2012, 11:13:13 PM11/23/12
to
"Tom Biasi" wrote in message news:50afd828$0$9830$607e...@cv.net...
> I hope you had a nice thanksgiving.
> I respect that you are done with me so I'll just answer your question.
> Back then I worked for Ballantine Labs in Boonton, NJ. There were several
> true RMS voltmeters at my disposal.
> When applying your math and formulas one must consider we are not working
> with AC anymore but unfiltered (or pulsating) DC. The peak is
> at at a 50% duty cycle so the filaments would be subjected to that DC
> value which would be Vpeak/pi. By your own admission several dozen
> others tried to explain this to you.
> I bid you well.
Were the true-RMS voltmeters AC coupled? The specs of the Ballantine 323
show a low limit of 2 Hz and says nothing of DC:
http://www.ballantinelabs.com/323meter.htm
The older Model 320 only goes as low as 15 Hz and the manual specifically
states that it DOES NOT include any DC component. Thus you would need to
measure the true-RMS AC and then add the DC component using sqrt(AC^2+DC^2).
http://www.tubebooks.org/file_downloads/ballantine_320_true_rms.pdf
Apparently you did not work as a design engineer at Ballantine. Even a
technician should know the basics of true-RMS voltage measurement. Read my
previous post for more details.
Paul
> I hope you had a nice thanksgiving.
> I respect that you are done with me so I'll just answer your question.
> Back then I worked for Ballantine Labs in Boonton, NJ. There were several
> true RMS voltmeters at my disposal.
> When applying your math and formulas one must consider we are not working
> with AC anymore but unfiltered (or pulsating) DC. The peak is
> at at a 50% duty cycle so the filaments would be subjected to that DC
> value which would be Vpeak/pi. By your own admission several dozen
> others tried to explain this to you.
> I bid you well.
show a low limit of 2 Hz and says nothing of DC:
http://www.ballantinelabs.com/323meter.htm
The older Model 320 only goes as low as 15 Hz and the manual specifically
states that it DOES NOT include any DC component. Thus you would need to
measure the true-RMS AC and then add the DC component using sqrt(AC^2+DC^2).
http://www.tubebooks.org/file_downloads/ballantine_320_true_rms.pdf
Apparently you did not work as a design engineer at Ballantine. Even a
technician should know the basics of true-RMS voltage measurement. Read my
previous post for more details.
Paul
Tom Biasi
Nov 24, 2012, 9:25:23 AM11/24/12
to
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.
P E Schoen
Nov 24, 2012, 7:23:09 PM11/24/12
to
"Tom Biasi" wrote in message news:50b0d8a2$0$24755$607e...@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
> 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.
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
Tom Biasi
Nov 24, 2012, 7:32:38 PM11/24/12
to
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
Nov 25, 2012, 3:28:11 AM11/25/12
to
"Tom Biasi" wrote in message news:50b166f3$0$24745$607e...@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
> Paul,
> You can't really be serious. Dwelling on Michael's reference to the meters
> of 50 years ago is ludicrous.
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.
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
Nov 25, 2012, 3:32:23 AM11/25/12
to
Tom Biasi
Nov 25, 2012, 9:18:39 AM11/25/12
to
Inquisitor
Dec 1, 2012, 12:18:17 PM12/1/12
to
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.
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
Dec 1, 2012, 8:37:22 PM12/1/12
to
"Inquisitor" wrote in message
news:c0adbbff-76a3-403a...@googlegroups.com...
[snip]
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
news:c0adbbff-76a3-403a...@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
> 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.
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
Dec 1, 2012, 10:28:02 PM12/1/12
to
On 2012-12-01, Inquisitor <dec...@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
> 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.
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
--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---
Inquisitor
Dec 2, 2012, 10:24:51 AM12/2/12
to
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
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
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