hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
12V lighting track ghost voltage?
- Thread starter tortuga
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220421-2339 EDT
My post #36 needs a little additional information.
Note: that my Simpson 270 reads 1.51 ratio of square to sine wave compared to either the Beckman 4410 or Fluke 87, where they read about 1.37 for the ratio, thus my 270 is a ratio of square wave to sine wave greater than 1.414 . This ratio needs to be divided by 1.11 to get the correct ratio for sq-wave to sine wave. And when 1.11 is used for correction the ratio is about 1.36 .
This1.36 ratio then is close to the ratios read by the two RMS meters.
Why is the ratio coming out at 1.36 instead of 1.414? Probably because of the high frequency rolloff of all these meters, and it not being really quite a square wave that the meter is reading ( measuring ).
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My post #36 needs a little additional information.
Note: that my Simpson 270 reads 1.51 ratio of square to sine wave compared to either the Beckman 4410 or Fluke 87, where they read about 1.37 for the ratio, thus my 270 is a ratio of square wave to sine wave greater than 1.414 . This ratio needs to be divided by 1.11 to get the correct ratio for sq-wave to sine wave. And when 1.11 is used for correction the ratio is about 1.36 .
This1.36 ratio then is close to the ratios read by the two RMS meters.
Why is the ratio coming out at 1.36 instead of 1.414? Probably because of the high frequency rolloff of all these meters, and it not being really quite a square wave that the meter is reading ( measuring ).
.
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220422-0840 EDT
tortuga:
I am going to try to get back to your basic problem. Tungsten incandescent bulbs burning out in a short time.
We know that an incandescent bulb's brightness increases as we increase power input.
In your case, because bulbs are burning out quickly, we can assume that more than the nominal voltage, current, and thus power are being applied to the bulbs.
To get a frame of reference we can apply the correct nominal voltage to a good bulb. I suggest a 12 V car battery that has been fully charged, and then let to rest for possibly 12 hours ( not at all critical amount of time, but long enough that the battery gets to a nominal rest value. This should be slightly over 12 V. ) .
You want to have a means to measure this bulb intensity. I won't describe that now. This means you also need to measure input power to the bulb, and therefore RMS voltage across the bulb.
Next you connect the light to the track via some wires, and measure its light intensity. We are estimating that this brightness will be quite a bit higher because of the short lamp life you have experienced. Now that you have exposed wires to the bulb, again try to measure voltage across the bulb. Try both AC and DC measurements. What do you see?
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tortuga:
I am going to try to get back to your basic problem. Tungsten incandescent bulbs burning out in a short time.
We know that an incandescent bulb's brightness increases as we increase power input.
In your case, because bulbs are burning out quickly, we can assume that more than the nominal voltage, current, and thus power are being applied to the bulbs.
To get a frame of reference we can apply the correct nominal voltage to a good bulb. I suggest a 12 V car battery that has been fully charged, and then let to rest for possibly 12 hours ( not at all critical amount of time, but long enough that the battery gets to a nominal rest value. This should be slightly over 12 V. ) .
You want to have a means to measure this bulb intensity. I won't describe that now. This means you also need to measure input power to the bulb, and therefore RMS voltage across the bulb.
Next you connect the light to the track via some wires, and measure its light intensity. We are estimating that this brightness will be quite a bit higher because of the short lamp life you have experienced. Now that you have exposed wires to the bulb, again try to measure voltage across the bulb. Try both AC and DC measurements. What do you see?
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LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
How about measuring the current, and calculating the voltage using the load power rating?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220422-2014 EDT
LarryFine:
Measuring current would useful in troubleshooting this problem, but voltage is better for predicting lamp life.
A tungsten filament bulb has a resistance characteristic that falls somewhere between being a constant current load and a linear resistance. Thus, voltage is a better measure of lifetime than current.
Whereas an LED is more of a constant load voltage than a constant load current. Therefore, it is more important to measure current to an LED for lamp life.
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LarryFine:
Measuring current would useful in troubleshooting this problem, but voltage is better for predicting lamp life.
A tungsten filament bulb has a resistance characteristic that falls somewhere between being a constant current load and a linear resistance. Thus, voltage is a better measure of lifetime than current.
Whereas an LED is more of a constant load voltage than a constant load current. Therefore, it is more important to measure current to an LED for lamp life.
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LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
And what I said was that the meter uses the same method to measure current as it does voltage. It still actually measures voltage but across a known resistance in series with the load. So, if it can't measure voltage, trying to measure the current isn't going to work either.
-Hal
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
And I'm still sticking to my story, the way to figure this all out is to look at the output of that power supply with a scope. You can measure the voltage and frequency and see what it looks like. I have a feeling that whatever it is putting out isn't playing nice with those halogen lamps which is why they don't last long. They are used to a clean 12VAC.
-Hal
-Hal
ELA
Senior Member
- Occupation
- Electrical Test Engineer
Here is a link that shows a track light transformer operating at > 20Khz.
https://img.acuitybrands.com/public...rsion=01/20/2021+05:03:47&DOC_Type=SPEC_SHEET
Also at this site a recommendation to use a Fluke 289 or better meter to read the high frequency output. Fluke 289 has AC RMS specs up to 100Khz.
https://www.totaltracklighting.com/...sformer-12vac-elv-dimmable-ta75wa12-0007.html
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220426-1504 EDT
If one can not get to the pins of the light bulb, then measuring current is better than nothing.
However, power dissipation tends to go up at a faster rate than current.
Following I present some data from a 75 W tungsten filament bulb. This shows power going up faster than current for increasing voltage. What is also interesting is that both power and voltage are about linear in the region I measured. Of course we know that that voltage is linear with itself.
Data from a 75 W 120 V tungsten filament bulb.
090 V .... 0.53 A .... 47.7 W
100 V .... 0.56 A .... 56.0 W
110 V .... 0.59 A .... 64.9 W
120 V .... 0.62 A .... 74.4 W
130 V .... 0.64 A .... 83.2 W
140 V .... 0.67 A .... 93.8 W
A projection of the power curve to 0 current comes out at about 47 V. That is not really what happens. So somewhere the curve has to curve toward 0 current at 0 volts.
Lamp voltage is a better indicator, more sensitive, of lamp life than current.
I believe that a Simpson 260 or 270 could be one of the best starting points for analyzing this problem. However, neither of these meters will directly measure AC current.
.
If one can not get to the pins of the light bulb, then measuring current is better than nothing.
However, power dissipation tends to go up at a faster rate than current.
Following I present some data from a 75 W tungsten filament bulb. This shows power going up faster than current for increasing voltage. What is also interesting is that both power and voltage are about linear in the region I measured. Of course we know that that voltage is linear with itself.
Data from a 75 W 120 V tungsten filament bulb.
090 V .... 0.53 A .... 47.7 W
100 V .... 0.56 A .... 56.0 W
110 V .... 0.59 A .... 64.9 W
120 V .... 0.62 A .... 74.4 W
130 V .... 0.64 A .... 83.2 W
140 V .... 0.67 A .... 93.8 W
A projection of the power curve to 0 current comes out at about 47 V. That is not really what happens. So somewhere the curve has to curve toward 0 current at 0 volts.
Lamp voltage is a better indicator, more sensitive, of lamp life than current.
I believe that a Simpson 260 or 270 could be one of the best starting points for analyzing this problem. However, neither of these meters will directly measure AC current.
.
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
That's insane! They want you to spend upwards of $600 just to check some customers track lighting powered by their crapola. Total overkill unless you regularly work with VFDs and the like.
I have a better idea. Just make up a 12V lamp with some test leads. Clip it across the power supply output and if it doesn't light or is dim the unit is defective. Simple, done! We don't care about the frequency or waveform or voltage or what the weather is that day...
And if the lamp life isn't what you would expect that's just the way it is. Change the power supply to a normal12VAC transformer.
-Hal
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220426-1943 EDT
hbiss:
The problem is not low voltage to the bulbs, it is excessive voltage. Bulbs do not burn out on low voltage, rather their life is greatly extend.
Post #1 does not imply that voltage is really low, just some measurement problem makes the apparent voltage reading look low, rather it implies an inability to get a correct voltage reading.
So the real problem is how to actually measure the voltage across the bulb terminals.
We need to hear back from tortuga.
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hbiss:
The problem is not low voltage to the bulbs, it is excessive voltage. Bulbs do not burn out on low voltage, rather their life is greatly extend.
Post #1 does not imply that voltage is really low, just some measurement problem makes the apparent voltage reading look low, rather it implies an inability to get a correct voltage reading.
So the real problem is how to actually measure the voltage across the bulb terminals.
We need to hear back from tortuga.
.
OK so I still dont have access to the defective track but I was able to borrow a similar replacement power supply and the scope.
Here is the supply specs:
https://www.bulbsdepot.com/media/pdf/lightech/LET-75.pdf
As I said I dont really know anything about scopes other than pressing the auto button.
Here it is with no load:
And here with a 12V 20W Halogen MR 16:
And with the LED MR 16
With the LED and the Halogen the scope looks exactly like halogen waveform.
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I tried to measure voltage with a Brymen BM786 and got basically nothing on AC or DC, just like my regular meter, and thats with the 20watt halogen lit.
I did not have access to a Fluke 289
(I also measured the AC input voltage at 118.3 VAC)
I did not have access to a Fluke 289
(I also measured the AC input voltage at 118.3 VAC)
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hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
I can tell you know nothing about 'scopes. 
Those pictures are useless. I might suggest that you learn with a basic analog scope then go on to digital. I see you have a meter from EEV Blog. He has a pretty good tutorial and the same scope you have. Take a look. Hint: practice with the calibrate signal available on those two terminals bottom right.
-Hal
Those pictures are useless. I might suggest that you learn with a basic analog scope then go on to digital. I see you have a meter from EEV Blog. He has a pretty good tutorial and the same scope you have. Take a look. Hint: practice with the calibrate signal available on those two terminals bottom right.-Hal
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220427-2346 EDT
tortuga:
Your scope traces for the halogen are quite useful.
What we see here is a square wave with a quite varying frequency, and amplitude, around a 40 microsecond to 14 microsecond period, and a peak to peak amplitude of +/- 32 V down to +/-15 V. That means the equivalent ( in terms of power ) DC voltage to the bulb is ranging from 7.5 to 15 V. This is probably too high for good lamp life. This signal can be readily measured with a Simpson 260 / 270 in the AC Output mode of the meter.
There are ways for you to use the scope to get a better look at the AC waveform. For the moment I will not discuss this. Try to find a Simpson 260 and a small bridge rectifier, 1N4148 diodes. Connect the AC signal to the bridge AC input, and the Simpson in DC mode 50 V range to the bridge output. Since the signal is somewhat of a sq-wave the DC volt reading will be close the peak value of the sq-wave. But since the sq-wave amplitude is varying ( i suspect at 120 Hz ) I would add shunt capacitance, like a 25 mfd 25 V rating across the meter input. This should reasonably hold the peak AC voltage.
The reason for converting your AC signal to DC and filtering, actually peak holding, is that the meter in AC mode will not give you an adequately reading without a number of gyrations that are hard to understand, and you would not get the highest level either.
There is much more for you to learn about your scope.
.
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tortuga:
Your scope traces for the halogen are quite useful.
What we see here is a square wave with a quite varying frequency, and amplitude, around a 40 microsecond to 14 microsecond period, and a peak to peak amplitude of +/- 32 V down to +/-15 V. That means the equivalent ( in terms of power ) DC voltage to the bulb is ranging from 7.5 to 15 V. This is probably too high for good lamp life. This signal can be readily measured with a Simpson 260 / 270 in the AC Output mode of the meter.
There are ways for you to use the scope to get a better look at the AC waveform. For the moment I will not discuss this. Try to find a Simpson 260 and a small bridge rectifier, 1N4148 diodes. Connect the AC signal to the bridge AC input, and the Simpson in DC mode 50 V range to the bridge output. Since the signal is somewhat of a sq-wave the DC volt reading will be close the peak value of the sq-wave. But since the sq-wave amplitude is varying ( i suspect at 120 Hz ) I would add shunt capacitance, like a 25 mfd 25 V rating across the meter input. This should reasonably hold the peak AC voltage.
The reason for converting your AC signal to DC and filtering, actually peak holding, is that the meter in AC mode will not give you an adequately reading without a number of gyrations that are hard to understand, and you would not get the highest level either.
There is much more for you to learn about your scope.
.
.
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
220427-0821 EDT
tortuga:
Turn off the vertical column displays on the left and right sides of the display when not needed.
Put the scope on 2 mS per major division for horizontal display. This is a good time base for 60 Hz waveforms. I suspect that you may actually have a source power supply with full wave rectification. This would mean 120 Hz was the base frequency, and possibly 1 mS per major division could be fine. Note that 60 Hz is 16.7 milliseconds per cycle.
Next set trigger source to line. This makes the horizontal sweep trigger from the scope AC line input. Then select single cycle triggering, and get a single cycle sweep, and display it for us.
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tortuga:
Turn off the vertical column displays on the left and right sides of the display when not needed.
Put the scope on 2 mS per major division for horizontal display. This is a good time base for 60 Hz waveforms. I suspect that you may actually have a source power supply with full wave rectification. This would mean 120 Hz was the base frequency, and possibly 1 mS per major division could be fine. Note that 60 Hz is 16.7 milliseconds per cycle.
Next set trigger source to line. This makes the horizontal sweep trigger from the scope AC line input. Then select single cycle triggering, and get a single cycle sweep, and display it for us.
.
synchro
Senior Member
- Location
- Chicago, IL
- Occupation
- EE
I agree with gar's analysis of the 'scope waveforms.
Having the amplitude of the output square wave be in proportion to the instantaneous magnitude of the AC voltage is actually a good thing. That's because it can reduce the harmonic distortion of the load current that is at multiples of 60Hz. This is in comparison to methods that rectify the AC input and feed a capacitor, thereby drawing current only near the peaks of the AC waveform. That results in a poorer distortion power factor.
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