1 of my drivers is hot to touch.
- Thread starter diggs99
- Start date
diggs99
Well-Known Member
I am of no help but just stopped in to wish you the best man. You’ll get it figured out.
Thanks bud, appreciate that.
Got some good people guiding me, no doubt ill get it figured eventually.
ChiefRunningPhist
Well-Known Member
I'd test voltage of both DC sides first. But when you swapped drivers the whiny one was still whiny on the WAGO setup (and high heat when cranked)? And the quiet driver was still quiet on the terminal block setup (and normal heat when cranked)? If that's the case you got a bad driver.
But if you want to test the current then you'll want to wire your meter in before the terminal block. So break the loop by pulling your positive driver lead from the terminal block it's hooked to, and then pluging the positive driver lead that you just unhooked from the terminal block and plug it into the 10A hole on the multimeter. Then use another wire to connect the "negative" hole or "COM" on the multimeter to the terminal block that the positive driver lead was hooked up to.
So current flows out the positive driver lead into the 10A hole of multimeter, then through multimeter, then out the "negative" hole of multimeter, then through the wire connecting the negative hole on the multimeter and the positive terminal block, then through the positive terminal block, then though all the wires hooked up to the terminal blocks then to your lights, then back around through the negative light wires to the negative terminal block to the negative driver lead where the loop is finally completed.
But if you want to test the current then you'll want to wire your meter in before the terminal block. So break the loop by pulling your positive driver lead from the terminal block it's hooked to, and then pluging the positive driver lead that you just unhooked from the terminal block and plug it into the 10A hole on the multimeter. Then use another wire to connect the "negative" hole or "COM" on the multimeter to the terminal block that the positive driver lead was hooked up to.
So current flows out the positive driver lead into the 10A hole of multimeter, then through multimeter, then out the "negative" hole of multimeter, then through the wire connecting the negative hole on the multimeter and the positive terminal block, then through the positive terminal block, then though all the wires hooked up to the terminal blocks then to your lights, then back around through the negative light wires to the negative terminal block to the negative driver lead where the loop is finally completed.
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diggs99
Well-Known Member
Thanks 1212, that was also the conclusion i came too, has to be the driver, otherwise it would have worked fine when connected to the other fixture and if the wiring was bad on the first fixture, it should have made the good driver run hot too
Thanks Chief,
Yes, this is exactly how it went. I never even bothered sending you the wiring pics after i switched drivers around and seen the result.
The Bad driver showed the same symptoms when connected to either fixture. That leads me to also believe the wiring and blocks are all fine, its gotta be the driver right?
Ok this makes clear sense. yes i still want to test the current and learn how to use the multimeter. I think any diy guy should know this stuff for troubleshooting at the very least.
You guys dont know how much i appreciate you taking the time to help as usual.
Thank you
ChiefRunningPhist
Well-Known Member
If your driver is hot it means it's dissipating power. I'm starting to think it's from bad secondary side switching, but just a guess. Not worried about the terminal blocks anymore though.
Check the drivers to make sure both are rated for 120VAC input or whatever you're pulling from the wall. In the states it's 120VAC, other countries is 240VAC sometimes (lots of times actually). But if they are both rated for 120VAC or whatever you're pulling then I'd go get a refund on the driver.
When you do your multimeter testing, if you get a current and voltage measurement from the DC side you can calculate the "wattage out" (Watts = Volts × Amps), and then if you have a kill-a-watt meter, plug the driver into it to see what the "wattage in" from the AC side looks like. You'll be able to calculate just how much your hot driver is wasting. A 480h should be about 94% effecienct, so (wattage out) ÷ (wattage in) = ~0.94. The left over 6% will be the warmth you feel from a normal driver like your quiet WAGO driver, on the hot driver, it should be much less effecient, when you do (wattage out) ÷ (wattage in) on the hot driver, you might only be 0.8 - 0.85, so the left over portion might be 15-20% or 3× as much.
Attached are some charts I use for quick reference, the electrical equation color wheel at the bottom of the electrical graphic shows the power, current, voltage, and resistance relationships. You can see what the effeciency of the 480h should be at for various AC input voltages and load %'s on the effeciency graphics, 1 is CC, 1 is CV.
EDIT:
If you try to test power on the AC side with the multimeter you have to also account for PF, its not just V·A. Thats why I recommend the kill-a-watt meter unless your multi meter tests for PF, mine doesn't.
Check the drivers to make sure both are rated for 120VAC input or whatever you're pulling from the wall. In the states it's 120VAC, other countries is 240VAC sometimes (lots of times actually). But if they are both rated for 120VAC or whatever you're pulling then I'd go get a refund on the driver.
When you do your multimeter testing, if you get a current and voltage measurement from the DC side you can calculate the "wattage out" (Watts = Volts × Amps), and then if you have a kill-a-watt meter, plug the driver into it to see what the "wattage in" from the AC side looks like. You'll be able to calculate just how much your hot driver is wasting. A 480h should be about 94% effecienct, so (wattage out) ÷ (wattage in) = ~0.94. The left over 6% will be the warmth you feel from a normal driver like your quiet WAGO driver, on the hot driver, it should be much less effecient, when you do (wattage out) ÷ (wattage in) on the hot driver, you might only be 0.8 - 0.85, so the left over portion might be 15-20% or 3× as much.
Attached are some charts I use for quick reference, the electrical equation color wheel at the bottom of the electrical graphic shows the power, current, voltage, and resistance relationships. You can see what the effeciency of the 480h should be at for various AC input voltages and load %'s on the effeciency graphics, 1 is CC, 1 is CV.
EDIT:
If you try to test power on the AC side with the multimeter you have to also account for PF, its not just V·A. Thats why I recommend the kill-a-watt meter unless your multi meter tests for PF, mine doesn't.
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whytewidow
Well-Known Member
If the current bounces as the driver gets hotter its the driver. More than likely it's a capacitor inside the driver. Kicking the relay in and out. That's the noise your hearing.
diggs99
Well-Known Member
@ChiefRunningPhist
When you say use another wire to plug into negative on multimeter and the terminal block, is 18awg fine or should I use 14 awg or does it not matter?
When you say use another wire to plug into negative on multimeter and the terminal block, is 18awg fine or should I use 14 awg or does it not matter?
Axion42
Well-Known Member
For the test it does not matter
ChiefRunningPhist
Well-Known Member
According to the chart the 18ga stranded wire is rated for 3.5A max, and if it's a 480W driver and pushing 48V, then 10A would be flowing from driver to multi meter to terminal block (watts = V·A; 480W = 48V × A). The chart also says there's about 22Ω of resistance in 1000ft or 0.022Ω/ft, so I think an amperage of 10 through .022Ω (12in wire) is not to much to worry about, V = A·Ω (same as V=I·R just using different abreviations), so 10A·0.022Ω = 0.22V, and Watts = V·A, so 0.22V × 10A = 2.2W dissipated across a 1ft wire. It may get warm after a while, but I think for a minute or so you'd be Ok.
I've seen several charts for current ratings and they all give different numbers, so just watch for heat and if it gets too warm you know it's not big enough.
EDIT:
Remember the multimeter is rated for only 10A, so you don't want to exceed that.
Current is measured in Amps: A, I
Voltage is measured in Volts: V
Power is measured in Watts: W, P
Resistance is measured in Ohms: Ω, R
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diggs99
Well-Known Member
Thanks for yet another very informative post bud.
So should i leave the fixture as it is right now, before i check?
or turn both pots down and then once im hooked to the meter begin turning up?
i have the 480h-48a drivers.
diggs99
Well-Known Member
And the voltage pot? i orginally turned that up to max, but have since 1/4 turned it back on suggestion from whytewidow
should i leave it where its at aswell?
Sorry if some of these questions seem dumb or redundant , i just want to make sure i know what im doing before i start playing around with things.
should i leave it where its at aswell?
Sorry if some of these questions seem dumb or redundant , i just want to make sure i know what im doing before i start playing around with things.
ChiefRunningPhist
Well-Known Member
Start with Vo max, Io min for a driver with the built in POTs, max is CW (righty tighty), min is CCW (lefty loosey). If using a driver with dimming leads then turn the dimming leads all the way down, which ever way down is for your dimming leads POT.
After that, you can plug the driver in and you can ramp up the current to whatever amperage you're looking for by watching the multimeter and turning the built in Io POT, or the dimming leads POT.
Once the current has been measured to your desired level, you'll want to take note of the measurement, (take precaution not to bump or move the Io POT, or dimming leads POT between these next steps), then un-plug the driver, remove the multimeter from the loop and then reconnect the wires that went into the multimeter to each other so the loop is closed again and current could possibly flow if the driver were still plugged in, but its not because you unplugged it.
Then once the multimeter is removed and the cct is mended back to a closed loop, you'll want to measure the voltage (change holes, and rotate multimeter dial) so you can determine the power (Watts = V·A), and how much resistance is in your lights and wiring, or the "load" of the cct (Ω = V/A).
You can then compare this info with the other driver. Put the other driver through the same paces but turn the current up to the same as the whiny driver you just tested, then once current is the same (make sure to not bump POT after you've dialed it in, if you bump it then your voltage reading will then be corresponding to a different amperage that we don't know), unplug driver, remove multimeter from loop, mend loop, then plug driver back in, then check the voltage, if there is a V difference between the 2 drivers when both drivers push the same A, then there will also be a power difference, or wattage difference.
This would mean 1 of your loads has less Ω then the other or requires less Vf than the other, if the loads came from the same place and are identical it would be plausible that any V difference or increase in Ω from one load over the other would be due to the only thing different, which would be the WAGOs or terminal blocks. This is why I was suggesting testing voltage earlier.
I was wondering if the whiny driver had to produce more V than it was rated for to flow the same A that the other setup could flow, and only because I was guessing that the terminal blocks had a bit more resistance in them, like maybe they were made from a non effecienct alloy that ended up being higher Ω and would then require more V from the driver to flow the same A, compared to a driver flowing current through an effecient Cu conductor with virtually 0Ω. This scare was put to rest when you found the whine dependent to the driver and not the load. I was having a hard time accepting such a huge piece of metal (terminal block conductor) could be dropping V by that much, but your current/voltage test could tell us just how much more effecient one is over the other.
After that, you can plug the driver in and you can ramp up the current to whatever amperage you're looking for by watching the multimeter and turning the built in Io POT, or the dimming leads POT.
Once the current has been measured to your desired level, you'll want to take note of the measurement, (take precaution not to bump or move the Io POT, or dimming leads POT between these next steps), then un-plug the driver, remove the multimeter from the loop and then reconnect the wires that went into the multimeter to each other so the loop is closed again and current could possibly flow if the driver were still plugged in, but its not because you unplugged it.
Then once the multimeter is removed and the cct is mended back to a closed loop, you'll want to measure the voltage (change holes, and rotate multimeter dial) so you can determine the power (Watts = V·A), and how much resistance is in your lights and wiring, or the "load" of the cct (Ω = V/A).
You can then compare this info with the other driver. Put the other driver through the same paces but turn the current up to the same as the whiny driver you just tested, then once current is the same (make sure to not bump POT after you've dialed it in, if you bump it then your voltage reading will then be corresponding to a different amperage that we don't know), unplug driver, remove multimeter from loop, mend loop, then plug driver back in, then check the voltage, if there is a V difference between the 2 drivers when both drivers push the same A, then there will also be a power difference, or wattage difference.
This would mean 1 of your loads has less Ω then the other or requires less Vf than the other, if the loads came from the same place and are identical it would be plausible that any V difference or increase in Ω from one load over the other would be due to the only thing different, which would be the WAGOs or terminal blocks. This is why I was suggesting testing voltage earlier.
I was wondering if the whiny driver had to produce more V than it was rated for to flow the same A that the other setup could flow, and only because I was guessing that the terminal blocks had a bit more resistance in them, like maybe they were made from a non effecienct alloy that ended up being higher Ω and would then require more V from the driver to flow the same A, compared to a driver flowing current through an effecient Cu conductor with virtually 0Ω. This scare was put to rest when you found the whine dependent to the driver and not the load. I was having a hard time accepting such a huge piece of metal (terminal block conductor) could be dropping V by that much, but your current/voltage test could tell us just how much more effecient one is over the other.
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ChiefRunningPhist
Well-Known Member
Courtesy @CobKits ...
https://www.google.com/url?sa=t&sou...FjAAegQIARAB&usg=AOvVaw3dKaii_ART_IuoOlW6yf55
Looks like you're right there within tested specs, a bit more on either end but nothing to worry about.
https://www.google.com/url?sa=t&sou...FjAAegQIARAB&usg=AOvVaw3dKaii_ART_IuoOlW6yf55
Looks like you're right there within tested specs, a bit more on either end but nothing to worry about.
