Drivers for flood light keeps failing

qcroanoke

Sometimes I don't know if I'm the boxer or the bag
Location
Roanoke, VA.
Occupation
Engineering
The first thing I'd do is change that xfmr to 50 hz and see what happens. To me that's the obvious first step. You know that's not right.
 

FionaZuppa

Senior Member
Location
AZ
The first thing I'd do is change that xfmr to 50 hz and see what happens. To me that's the obvious first step. You know that's not right.
You can add a coil on one of the pri legs to correct impedance. But you are 100% right, that's the right thing to do, regardless of what's on the secondary as load.

I agree with this.
The strength of the magnetic field applied to the core is proportional to the volt-seconds at the primary winding. Obviously there will be 60/50 = 1.2 times more volt-seconds in a 50Hz positive or negative half-cycle than for a 60Hz one. If you integrate the sine wave voltages to get the volt-seconds, a 50Hz waveform will reach the same volt-seconds as a 60Hz half-cycle does in only 132° instead of 180°. So if the transformer was right on the edge of saturating with 60Hz applied, then with 50Hz it would saturate at 132° into the half-cycle and at that point the output voltage would fall to zero until the next half-cycle. Now sin(132°) = 0.745 and so in this case the voltage waveform on the transformer secondary would drop when the sine wave falls about 25% after reaching its peak value. In reality the transformer will have some design margin before it saturates and so the voltage should drop later in the waveform.

I would think an input rectifier in the driver should still tolerate this type of distorted waveform without damage, and possibly without much other effect if the rectifiers are only conducting near the voltage peaks anyway.

The only mechanism I can think of that might cause damage is if there were shunt capacitors at the driver input for EMI filtering or the like, and the fast drop (high dV/dT) in the applied voltage induced relatively high currents in such capacitors. But that's probably a stretch. The only way to know for sure would be to do a failure analysis.
1) The core may not saturate at all, really depends on how the sec is loaded.
2) I have seen many 120vac dc-dc pucks/wall-warts die when on non pure sine inverters. The input is just ugly and the cheapo dc-dc crud just cannot handle it. My guess is, these LED drivers in question are cheapo MIC items.

My approach might be, scope it to see & go from there.
Correcting input impedance is not hard to do, but who there can engineer it and install it?
Or, swap it out for a 50Hz xfrmr.
 
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Ainsley Whyte

Senior Member
Location
Jamaica
Occupation
Senior Electrical Engineer
I agree with this.
The strength of the magnetic field applied to the core is proportional to the volt-seconds at the primary winding. Obviously there will be 60/50 = 1.2 times more volt-seconds in a 50Hz positive or negative half-cycle than for a 60Hz one. If you integrate the sine wave voltages to get the volt-seconds, a 50Hz waveform will reach the same volt-seconds as a 60Hz half-cycle does in only 132° instead of 180°. So if the transformer was right on the edge of saturating with 60Hz applied, then with 50Hz it would saturate at 132° into the half-cycle and at that point the output voltage would fall to zero until the next half-cycle. Now sin(132°) = 0.745 and so in this case the voltage waveform on the transformer secondary would drop when the sine wave falls about 25% after reaching its peak value. In reality the transformer will have some design margin before it saturates and so the voltage should drop later in the waveform.

I would think an input rectifier in the driver should still tolerate this type of distorted waveform without damage, and possibly without much other effect if the rectifiers are only conducting near the voltage peaks anyway.

The only mechanism I can think of that might cause damage is if there were shunt capacitors at the driver input for EMI filtering or the like, and the fast drop (high dV/dT) in the applied voltage induced relatively high currents in such capacitors. But that's probably a stretch. The only way to know for sure would be to do a failure analysis.
There are shunt capacitors installed at the driver input would it be wise to disconnect it and monitor
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
There are shunt capacitors installed at the driver input would it be wise to disconnect it and monitor
At least for diagnostic purposes you could disconnect shunt capacitors at the input of a failed unit to see if it will work again. If it works then that shows it was the capacitor that failed.
If the capacitors were part of a listed assembly then I would not recommend to remove them permanently.
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
There are shunt capacitors installed at the driver input would it be wise to disconnect it and monitor
Is there a fuse or other component in series with the shunt capacitor? If so that could be be blown due to the excess current in a shunt capacitor from fast rising or falling voltage waveforms.
 

myspark

Senior Member
Location
SCV Ca, USA
Occupation
Retired EE
Hello there!


Here is an excerpt that describe the consequences of running a 60 hz transformer on a 50 hz power supply.
The deleterious effect of running a mismatched (hz vs copper loss) that can eventually cause failure of your aforementioned driver.
The rest of the paper will also explain how you can apply the appropriate remedy to prevent failures.
The equations are presented in such a way where--clearly-- there is not only the problem of saturation--it can also lead to breakdown of the insulation due to heat build-up.

Check this website and (hopefully) it will give you the idea of what's happening.

electricengineer13.com/60-hz-transformer/

The excerpt:

If a transformer of 60 Hz is to be operated at 50 Hz, its applied voltage must also be reduced by 1/6 or the peak flux in the core will be too high. This decrease in voltage applied with frequency is known as detonating.

Similarly, if this action does not cause insulation problems, a transformer of 50 Hz at 60 Hz can be operated at 20% high voltage.


Good luck .
Stay healthy, wealthy and wise while keeping distance.
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
1) The core may not saturate at all, really depends on how the sec is loaded.
The load current on the secondary will be reflected to the primary, and will be scaled by the inverse of the turns ratio. The load current has little effect on the magnetizing current which is what creates the magnetic field applied to the core, and which can make it saturate if the field is too strong.
 

Ainsley Whyte

Senior Member
Location
Jamaica
Occupation
Senior Electrical Engineer
Hello there!


Here is an excerpt that describe the consequences of running a 60 hz transformer on a 50 hz power supply.
The deleterious effect of running a mismatched (hz vs copper loss) that can eventually cause failure of your aforementioned driver.
The rest of the paper will also explain how you can apply the appropriate remedy to prevent failures.
The equations are presented in such a way where--clearly-- there is not only the problem of saturation--it can also lead to breakdown of the insulation due to heat build-up.

Check this website and (hopefully) it will give you the idea of what's happening.

electricengineer13.com/60-hz-transformer/

The excerpt:

If a transformer of 60 Hz is to be operated at 50 Hz, its applied voltage must also be reduced by 1/6 or the peak flux in the core will be too high. This decrease in voltage applied with frequency is known as detonating.

Similarly, if this action does not cause insulation problems, a transformer of 50 Hz at 60 Hz can be operated at 20% high voltage.


Good luck .
Stay healthy, wealthy and wise while keeping distance.
Excellent
 

Ainsley Whyte

Senior Member
Location
Jamaica
Occupation
Senior Electrical Engineer
I agree with this.
The strength of the magnetic field applied to the core is proportional to the volt-seconds at the primary winding. Obviously there will be 60/50 = 1.2 times more volt-seconds in a 50Hz positive or negative half-cycle than for a 60Hz one. If you integrate the sine wave voltages to get the volt-seconds, a 50Hz waveform will reach the same volt-seconds as a 60Hz half-cycle does in only 132° instead of 180°. So if the transformer was right on the edge of saturating with 60Hz applied, then with 50Hz it would saturate at 132° into the half-cycle and at that point the output voltage would fall to zero until the next half-cycle. Now sin(132°) = 0.745 and so in this case the voltage waveform on the transformer secondary would drop when the sine wave falls about 25% after reaching its peak value. In reality the transformer will have some design margin before it saturates and so the voltage should drop later in the waveform.

I would think an input rectifier in the driver should still tolerate this type of distorted waveform without damage, and possibly without much other effect if the rectifiers are only conducting near the voltage peaks anyway.

The only mechanism I can think of that might cause damage is if there were shunt capacitors at the driver input for EMI filtering or the like, and the fast drop (high dV/dT) in the applied voltage induced relatively high currents in such capacitors. But that's probably a stretch. The only way to know for sure would be to do a failure analysis.
think the Driver would have to be able to handle more than its rated voltage.
 

GoldDigger

Moderator
Staff member
I find it very hard to believe that the decrease in rated applied voltage as frequency decreases is called detonating. It might be a vaguely similar term morphed by auto-correct.
 

FionaZuppa

Senior Member
Location
AZ
The load current on the secondary will be reflected to the primary, and will be scaled by the inverse of the turns ratio. The load current has little effect on the magnetizing current which is what creates the magnetic field applied to the core, and which can make it saturate if the field is too strong.
An idle xfrmr may not be saturated, but add secondary load and it then may saturate.
Flux density B comes from magnetic field strength H. More H and you get more B. But there's a limit to B (where increasing H does not increase B), we call this saturation.
But, magnetic field strength H is directly proportional to the current flowing through the primary winding. When not saturated, pri current is proportional to sec current (VAin * 95% = VAout, roughly with good modern day xfrmrs).
 
I have been called to troubleshoot led flood lights for a big plant the drivers keeps failing over time very frequently voltage is 220 VAC. the lights are fed from a 220VAC secondary 37 KVA transformer 60HZ. In my country we used 50HZ could this be the reason why the drivers are failing so frequently ?
The most important thing I have found about troubleshooting is you need facts.

The 60 hz and 37 KVA suggest the transformer is 3 phase. How are the secondarys connected? I looked up power in Jamaica and it appears to be
Jamaica
110 V​
190 V​
50 Hz​
3, 4​
A/B​
It would appear that there is an inherent problem with voltage and Hz with most foreign (to Jamaica) equipment. I agree with others the transformer is highly suspect. Is this a new install or was the "old system" perhaps driven by electro magnetic ballasts. If so the old system might not care if both legs are significantly above ground potential.

Given the voltage tolerance of modern LED drivers I'm surprised a transformer is used. Are the lights designed for the European standard 230/400 volt systems? Also not knowing anything about 50 hz LED drivers, are they designed that one leg needs to be at near ground potential? I can tell you that US designed equipment (277volt) does, and when the manufacturer does not know you have a high resistance ground or floating neutral he will design the equipment with a 277 driver and an autotranny to lower the 480 to 277. This all fine until you have a ground fault.

You need more facts!
 

GoldDigger

Moderator
Staff member
An idle xfrmr may not be saturated, but add secondary load and it then may saturate.
Flux density B comes from magnetic field strength H. More H and you get more B. But there's a limit to B (where increasing H does not increase B), we call this saturation.
But, magnetic field strength H is directly proportional to the current flowing through the primary winding. When not saturated, pri current is proportional to sec current (VAin * 95% = VAout, roughly with good modern day xfrmrs).
You missed something FionaSuppa. The magnetizing current is roughly independent of load current because the increase in primary flux is exactly cancelled by the secondary flux. That is how transformers work. Motors draw more current with load. So do transformers, but the flux in their core stays pretty much the same.
 

gar

Senior Member
200703-1543 EDT

There have been many statements in this thread that show total ignorance on the part of the responder with respect to magnetic circuit theory, and ferro-magnetic characteristics.

For some magnetic material curves see "Electric and Magnetic Fields", Stephen S, Attwood, U of M, 1949, John Wiley and Sons, Chapter 13.

Back in the 1870s when Edison was developing an efficient dynamo he discovered ferro-magnetic core saturation.

Ordinary transformer iron has a moderately soft saturation characteristic as compared to a square loop material. Thus, as voltage to the primary is increased at a constant frequency the magnetizing current to an unloaded transformer is nonlinear with respect to excitation voltage, and rises faster than excitation voltage, but does not abruptly jump at some voltage.

Magnetizing current reaches its peak at a voltage zero crossing, and flux density is a maximum at the zero crossing.

Following is measured data from a Signal Transformer A41-175-24. Meaning a nominal 120 V input. 175 VA, and 24 V output at full load. Input current from the load at full load would be 175/120 = 1.46 A. Room temperature primary resistance is about 1,7 ohms. Assume this goes up by 30%, then primary power dissipation from the secondary load is about 1.46*1.46*2,2 = 4.68 W. Add about 0.15 to 1.46 for the magnetizing current effect, and we get 5.7 W. Multiply by 4 and the result is about 20 W, or an efficiency of about 175/195 = 90%.

Measurements to unloaded transformer ---
Input voltage from Variac dial, input current from RMS meter.

Input V, Out V, Out V/Out V at 120 input, Input V/120, Input I, Input I normalized to that at 120

20 .... 4.08 .... 0.15 ..... 0.17 .... 0.006 .... 0.03
40 .... 8.64 .... 0.31 .... 0.33 .... 0.014 .... 0.06
60 .... 13,8 .... 0.49 .... 0.50 .... 0.028 .... 0.12
80 .... 18.8 .... 0.67 .... 0.67 .... 0.045 .... 0.19
100 .. 23.7 .... 0.85 .... 0.83 .... 0.100 .... 0.42
120 .. 28.0 .... 1.00 .... 1.00 .... 0.240 .... 1.00
140 .. 32.2 .... 1.17 .... 1.17 .... 0.500 .... 2.08

This is enough for you to chew on now. Ordinary power transformers have substantial saturation effects, but not extremely sharp.

.

.
l
 

FionaZuppa

Senior Member
Location
AZ
The load current on the secondary will be reflected to the primary, and will be scaled by the inverse of the turns ratio. The load current has little effect on the magnetizing current which is what creates the magnetic field applied to the core, and which can make it saturate if the field is too strong.
True.

I was wrong in linking sec load amps to core flux.
 

myspark

Senior Member
Location
SCV Ca, USA
Occupation
Retired EE
. . . . It might be a vaguely similar term morphed by auto-correct.
. . . . . . I find it very hard to believe that the decrease in rated applied voltage as frequency decreases is called detonating. It might be a vaguely similar term morphed by auto-correct.

Maybe it was supposed to be "derating"?



…………………………..

Detone in AMPLIFYING technology parlance--refers to the degradation of sound--during the amplification process.
Amplifying sound signals involve increasing the amplitude of raw sound waves (unamplified) to a more audible or “pure” fidelity of the sound source.
This is done by transistors, transformers and capacitors etc.
Small transformers in amplifiers. . . along with age-old vacuum tubes are extensively used. In fact audiophiles prefer vacuum tubes when they have a choice over digital models.
The transition from high pitch to low pitch or vice versa--in analog systems are NOT “choppy” compared to transistorized models.

These small transformer components also suffer copper loss over a period of time and saturation is a common culprit. Filament heaters in vacuum tubes are energy hogs. . . they put out a lot of heat, thus putting these transformers to task.

Thus, when these transformers become saturated they no longer provide the high-fidelity sound that audiophiles are so keen on differentiating good sound from garbage.

Now, the word DETONATION as used in the above narrative --as derived from detone made its way to the English lexicon.
It has nothing to do with detonating explosives from IED roadside bombs.

Yeah, language evolve that sometimes people get caught unaware—that fire up their knowledge of Aunt Mabel’s grammar--acquired a hundred years ago.
 

GoldDigger

Moderator
Staff member
. . . . . . I find it very hard to believe that the decrease in rated applied voltage as frequency decreases is called detonating. It might be a vaguely similar term morphed by auto-correct.

Maybe it was supposed to be "derating"?



…………………………..

Detone in AMPLIFYING technology parlance--refers to the degradation of sound--during the amplification process.
Amplifying sound signals involve increasing the amplitude of raw sound waves (unamplified) to a more audible or “pure” fidelity of the sound source.
This is done by transistors, transformers and capacitors etc.
Small transformers in amplifiers. . . along with age-old vacuum tubes are extensively used. In fact audiophiles prefer vacuum tubes when they have a choice over digital models.
The transition from high pitch to low pitch or vice versa--in analog systems are NOT “choppy” compared to transistorized models.

These small transformer components also suffer copper loss over a period of time and saturation is a common culprit. Filament heaters in vacuum tubes are energy hogs. . . they put out a lot of heat, thus putting these transformers to task.

Thus, when these transformers become saturated they no longer provide the high-fidelity sound that audiophiles are so keen on differentiating good sound from garbage.

Now, the word DETONATION as used in the above narrative --as derived from detone made its way to the English lexicon.
It has nothing to do with detonating explosives from IED roadside bombs.

Yeah, language evolve that sometimes people get caught unaware—that fire up their knowledge of Aunt Mabel’s grammar--acquired a hundred years ago.
Thank you for that explanation. Given that usage, I still feel that the requirement to maintain a constant maximum V/F ratio should not, itself, be referred to as detonation. In speech, the distinction between the words should be crystal clear, since they are pronounced differently.

PS: In the above narrative I see the word detone, but detonation is only in your commentary.
 
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FionaZuppa

Senior Member
Location
AZ
Scope the secondary, it's probably clipping.
xfrmr maker should also have primary amps no-load #'s for both normal use (60Hz with matched pri voltage) and point of saturation. Amp clamp the primary and compare the #'s.
 
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