Induced Voltage

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jbond

Member
I have 5 - 120VAC LEDs that are used for status indications. The LEDs are connected to a 500 ft, 25-conductor cable. The cable is used for 120VAC control. There is some induced voltage present between 4-10VAC that is partially lighting the LEDs.

Any ideas to fix this problem? Resistor in series? Incandescent bulbs?
 

physis

Senior Member
Use a bleeding resister. What do have for unwanted current, 4 milliamps? 30 Ohms 'll kill that.

What in the world are you doing driving LED's with 120 AC?

There's more to this, huh.

Edit: The resistance should be in parallel.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
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Electrical Contractor
The resistance should be in parallel.
Agreed, but ahead of each LED's individual current-limiting resistor if you can access that point. Otherwise, the parallel resistor might change the voltage across the LED too much.
 

physis

Senior Member
Agreed, but ahead of each LED's individual current-limiting resistor if you can access that point. Otherwise, the parallel resistor might change the voltage across the LED too much.

Yeah, OK, but what about a diode Larry? :grin:

Edit: Change the voltage across 120 VAC too much. :grin:

I'm just messing with you Larry, there's more to this thing than's been mentioned yet.
 
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jbond

Member
I haven't had a chance to measure the resistance.

I didn't ask why they are using 120 VAC LEDs. I came in to help commission the job and ran into the problem.

Thanks for your help guys.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090307-2227 EST

jbond:

I suspect capacitive coupling, not inductive (induced). Try 5000 ohms 5W across the pilot light. Maybe you can use a larger resistor and thus a lower power rating, but try this first to see if it solves the problem.

.
 

jbond

Member
Gar,

Why do you suspect that it is capacitive coupling? and what is the difference between capacitive coupling and induced voltage? This is new to me and something I have not seen before.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090308-2100 EST

jbond:

Two conductors spaced by an insulator form a capacitor. Two wires in a cable have a spacing that is not too great, and between them is insulating material. This material will have a dielectric constant greater than 1. Even if the dielectric constant was 1 because the wires were in a vacuum there is capacitance.

As an example 2 conductor #12 Romex has about 20 pfd capacitance per foot, based on a short sample of some old wire. Assuming your cable had the same capacitance, then 500 ft is 10,000 pfd or 0.01 mfd. Your cable might have a greater coupling capacitance.

At 60 Hz 0.01 mfd has a capacitive reactance of 265,000 ohms. The maximum current from this at 120 V would be about 0.5 MA. If the LED in the pilot lamp is one that has high brightness, means little current to produce useable light, then you are likely to see some residual light.

If the input wire to the pilot lamp is connected to nothing (supposed to be off), then you do not have a loop circuit for magnetic flux lines to cut and therefore no magnetic field induced current. There will be an induced voltage but no current. Even if you had a closed circuit I doubt you would induce 1 V in the loop and this would never light the LED with the amount of series resistance that is in the pilot lamp.

The use of the wording "induced voltage" would normally apply to a voltage in a conductor generated from a changing magnetic field. For example, connect a coil of wire to a sensitive meter and move a strong magnet near the coil. With a small 1200 turn coil about 1.5 x 1.5 inches outside and a strong magnetic I can get 5 to 10 MV deflection on a Simpson 270 on the 250 MV (50 microamp) range.

.
 

physis

Senior Member
I haven't had a chance to measure the resistance.

I didn't ask why they are using 120 VAC LEDs.

They're not. LED's operate on about three volts DC between 5 and 15 milliamps typically.

The LED's are actually driven by some intermediate circuitry that's only reading the 120 lines.

Has anybody tried replacing whatever it is that posesses the LED's?

I can't even imagine a situation where inductive or capacitive reactance on a 120 VAC signal input would cause a correctly working device designed for that, to act stupid.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090313-1213 EST

physis:

An LED pilot light assembly for AC will consist of an LED, a means to protect the LED from reverse voltage, and a current limiter. This might be a single LED at the output of a bridge rectifier, and a series resistance.

With a particular LED that I use as an indicator for 9 V I use a 25,000 series resistor and get plenty of light for indicator purposes. Scale 9 V to 90 V and the resistance is 250,000 ohms. The peak current from 120 V with this resistance is 120*1.414/250,000 = 0.7 MA.

Do a little calculation and see how much capacitance will produce this current at 120 V and 60 Hz.

.
 

physis

Senior Member
physis:

An LED pilot light assembly for AC will consist of an LED, a means to protect the LED from reverse voltage, and a current limiter. This might be a single LED at the output of a bridge rectifier, and a series resistance.

With a particular LED that I use as an indicator for 9 V I use a 25,000 series resistor and get plenty of light for indicator purposes. Scale 9 V to 90 V and the resistance is 250,000 ohms. The peak current from 120 V with this resistance is 120*1.414/250,000 = 0.7 MA.

Do a little calculation and see how much capacitance will produce this current at 120 V and 60 Hz.

.

With a particular LED that I use as an indicator for 9 V I use a 25,000 series resistor and get plenty of light for indicator purposes.

I don't care what you say, you can't see an LED driven by 360 Microamps.

The peak current from 120 V with this resistance is
120*1.414/250,000 = 0.7 MA.

I have no idea what you're even talking about aside from 120 V.

What in the world are you doing multiplying that by the square root of 2 and then dividing that by 250k?

Edit: I'm sorry, I wasn't thinking about bringing the 120 back to peak. Mostly because there's really no need here.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090313-1709 EST

physis:

With 25.5 K and 2 V I get illumination of a Stanley EFR5366X LED and much brightness at 10 V. At 255 K I can see illumination at 2 V and probably adequate light at 10 V for a satisfactory indicator.

I have used Stanley LEDs for their high brightness for a long time. Maybe 25 years.

The reason for the peak voltage of the sine wave is to get the peak current. Just a useful measure.

Also note that this particular LED has a threshold close to the 2 V level.

These LEDs are neat because of the low current requirements and thus lower power rating for the current limiting resistor.

.
 

glene77is

Senior Member
Location
Memphis, TN
090307-2227 EST
I suspect capacitive coupling, not inductive (induced).
Try 5000 ohms 5W across the pilot light.
.

Gar,
Good suggestion.
I have used 5K Ohm (draws aprox. 40mA) in parallel
to draw down phantom voltage, prior to igniting the LED's.

I know you will not laugh at this: :smile:
I have also setup a series of LED's fed via a capacitor.
The cap was calculated to have the 60Hz impedance
appropriate for limiting the current to 20mA driven by 120Vac.
The forward voltage for the LED's was about 1.9V,
and the reverse voltage for the LED's was about 3.5V.
I had the current limited to a level that applied 2.2V to the LED's,
and the higher reverse voltage protected the LED's.
Silly stuff one does on a slow day.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090314-0838 EST

glene77is:

For comparison at 70 deg. F. On the Stanley EFR5366X, sample of one, my forward drop at the threshold of noticeable light is about 1.4 to 1.5 V, I went up to 27 V in the reverse direction without breakdown. Reverse leakage at 27 V was less than 0.1 microamps. In the forward direction at 1 MA the drop is 1.80 V, and at 20 MA 2.08 V.

Did your LED have an actual reverse breakdown of 3.5 V?

.
 
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physis

Senior Member
I have used Stanley LEDs for their high brightness for a long time. Maybe 25 years.

I've been using stuff like this forever, I've never even heard of a stanley LED, I thought the galium arsnide was fancy. The specs you're posting are rather extreme for an LED.

The reason for the peak voltage of the sine wave is to get the peak current. Just a useful measure.

I dig that but only if you're fully going DC supply from the AC. You know, full wave recification and filter caps, but that's hardly necessary. LED's act a lot like light bulbs, they can deal with voltage very well, it's current that kills them, and they mostly care about wattage over time.

Also note that this particular LED has a threshold close to the 2 V level.

Not many LED's light up very good at 2 volts.

These LEDs are neat because of the low current requirements and thus lower power rating for the current limiting resistor.

If you can light this thing up at 360 uAmps, I'm impressed.

Aside from all that, I'm still having a problem with reactive coupling getting through the circuitry and being able to drive indicators. I don't care how you put it, the reactive current should either be insignificant or prevented from having an effect.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090314-1638 EST

physis:

If you are interested in experimenting with Stanley LEDs here is a source
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=404-1090-ND

The datasheet for the Stanley FR5366 is here
http://www.stanley-components.com/en/search/search_product.cfm#FFFFFF
Put in FR5366 in the search box, and pick datasheet.

It would be more appropriate to classify LEDs as as a constant voltage device and thus not very tolerant of voltage, and therefore need to be driven by current controlled sources. Same problems with neon bulbs and fluorescent bulbs. Actually tungsten filament lamps are not very tolerant of voltage. Life decreases very rapidly with increasing voltage.

Probably the majority of LEDs light up very well at 2 V.

In a previous post I estimated the capacitance for a 500 ft cable. My estimate should be within one order of magnitude up or down for virtually any cable without shielding between the wires of interest.

On a green LED that I quickly checked it required about 1.4 V to get 0.1 microamp forward current.

.
 
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