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

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.

 

[Tom Biasi]

On 11/17/2012 2:44 PM, dec720@att.net wrote:

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.

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?

 

[Ecnerwal]

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.

 

[Inquisitor]

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 Biasi]

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
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

 

On Sat, 17 Nov 2012 11:44:35 -0800 (PST), dec720@att.net wrote:

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.

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.

 

[Jasen Betts]

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
cheaper capacitor, like an ordinary 250V polyester

--
⚂⚃ 100% natural

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Michael A. Terrell]

Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester

120 VAC is 339.36 Volts, peak to peak.

 

On 19 Nov 2012 11:11:33 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

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
cheaper capacitor, like an ordinary 250V polyester

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

 

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.
..

 

[Michael A. Terrell]

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.

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.

 

[Tom Biasi]

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.

 

[David Eather]

On Sun, 18 Nov 2012 05:44:35 +1000, <dec720@att.net> wrote:

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?

It is a bad idea because

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]

Tom Biasi wrote:

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.

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.

 

[Tom Biasi]

On 11/19/2012 4:43 PM, Michael A. Terrell wrote:

Tom Biasi wrote:

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.


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.

 

[Michael A. Terrell]

Tom Biasi wrote:

On 11/19/2012 4:43 PM, Michael A. Terrell wrote:

Tom Biasi wrote:

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.


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.

 

[Tom Biasi]

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:

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.


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.

Tom

 

[Michael A. Terrell]

Tom Biasi 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:

On 11/19/2012 1:05 PM, Michael A. Terrell wrote:

krw@att.bizzz wrote:

On Mon, 19 Nov 2012 09:29:51 -0500, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Jasen Betts wrote:

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
cheaper capacitor, like an ordinary 250V polyester



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.


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.


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.

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?

 

[P E Schoen]

"Michael A. Terrell" wrote in message
news:EsOdnddjTJj9RDfNnZ2dnUVZ_tadnZ2d@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

 

[Jasen Betts]

On 2012-11-20, Tom Biasi <tombiasi@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.


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