EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: SimpleOne on September 30, 2020, 12:10:57 pm
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Hi,
I picked up a second hand isolation transformer to power DUT's. I think I have it wired correctly, but as this is new to me, I wanted to esquire with better minds about a strange voltage level that I see on the output side of the transformer between earth and both the line terminal and neutral terminal. I'm seeing an indicated ~80v AC between earth and line, and between earth and neutral on the output side.
Any idea what that 80v is about?
I'm not sure if this 80v is 'real' or if it is some sort of induced artifact due to the transformer core?
The background details:
It's a Xentek 5418 (but that won't help you much, not much information on the net about it).
The transformer's input side can take either 115v or 230v. It is wired as per it's diagram for 230v input.
The earth is bonded on both sides, so there is continuity between the input side earth, the chassis housing and the output side earth.
The output side is wired as per it's diagram for 230v output.
There does indeed appear to be isolation between the input line/neutral and the output side line/neutral.
Voltage between line and neutral on the output side is correct, there is ~236v between line and neutral on the output side for about the same input voltage.
Across a few observations when measured on the input side, there is the expected ~236v between line and earth, and ~0 to ~1v between neutral and earth.
Is this something to worry about or just ignore as an artifact in the measuring instrument (a 121GW DMM)?
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The transformer is from your decription probably a center tap transformer. The 80V must be if the transformer is wired for 110V because at 236v becuse it should be half the voltage about 118V.
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hi,
if you measure with digital dmm, use low impedance mode, otherwise those 80V means nothing
or use analog voltmeter and tell us the result
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+1
It's most probably just capacitive coupling from primary to secondary. Quite normal, this will go away as soon as you use a low impedance measurement.
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What you want to measure is the leakage current, not the voltage. The voltage will depend a lot on what you are using to measure. A typical isolation transformer may have from 20 to 200uA of leakage current. If you can't measure uAAC, put a 1K resistor where you want to measure and then test the voltage across the resistor--1mV = 1uA.
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It can't be centre tapped, because the voltages don't add up to 236V
It's leakage current. Assuming the input impedance of the meter is about 10M you have around 8µA of leakage, which is tiny and nothing to worry about. It'll just about make a neon lamp, or high efficiency white/green/blue LED glow in a dark room. If you test the LED, then add a diode in reverse parallel, to protect it against reverse voltage.
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Okay, thanks all. I suspected leakage of one form or another. The capacitance value that is stamped on it is 0.001 pF.
I don't have any specific low impedance measuring gear, but I do have a 1 omh and a 10 omh power resistor somewhere that I can use as a shunt. Failing that, I'll hook up a brave, compulsorily volunteered, LED and see what we get.
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I don't have any specific low impedance measuring gear, but I do have a 1 omh and a 10 omh power resistor somewhere that I can use as a shunt.
I wouldn't use that low a resistance - and definitely not a high power one - just in case there is real power behind it.
I'd try this first:
... put a 1K resistor where you want to measure and then test the voltage across the resistor--1mV = 1uA.
... and make sure it's a low power resistor. 1/4W or some such. If there is real power behind this voltage, a low power resistor will fail - but the destruction will be contained to that resistor and be far less dramatic than the result of using a high power resistor.
The reason why a 1K resistor would be OK for this test is, that with capacitive leakage, the impedance is in the order of megohms and 1K is a nice value to load down the circuit. You don't really want a low value resistor and you certainly do not need a 0.0000000000001 ohm shunt like you would like to use in a high current power supply. Accuracy is not particularly important here - you just want to confirm which ball park you're playing in.
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Thanks for the great information Brumby, much appreciated!
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I don't have any specific low impedance measuring gear, but I do have a 1 omh and a 10 omh power resistor somewhere that I can use as a shunt.
I wouldn't use that low a resistance - and definitely not a high power one - just in case there is real power behind it.
I'd try this first:
... put a 1K resistor where you want to measure and then test the voltage across the resistor--1mV = 1uA.
... and make sure it's a low power resistor. 1/4W or some such. If there is real power behind this voltage, a low power resistor will fail - but the destruction will be contained to that resistor and be far less dramatic than the result of using a high power resistor.
The reason why a 1K resistor would be OK for this test is, that with capacitive leakage, the impedance is in the order of megohms and 1K is a nice value to load down the circuit. You don't really want a low value resistor and you certainly do not need a 0.0000000000001 ohm shunt like you would like to use in a high current power supply. Accuracy is not particularly important here - you just want to confirm which ball park you're playing in.
There's no guarantee any resistor will fail open circuit, unless it's specifically designed to do so, i.e. it's a fusible resistor. In practise that's what normally happens to low value resistors, but it isn't a certainty and shouldn't be relied upon for safety.
If there's real power behind it, then bad things can happen if a resistor of too lower power rating is connected across it. Fortunately, the fact that the voltages when measured to earth don't add up to the total secondary voltage, due to loading from the meter's high impedance indicates it's definitely high impedance, so this isn't anything to worry about. Use a 100k 0.6W resistor, which won't smoke, even if connected across the mains. A leakage current of around 8µA will only give 8mV arcoss a 1k resistor, which isn't enough to reliably measure with most multimeters anyway. 100k, will give a much more sensible value of around 800mV.
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Why should you use a complicated measurement setup?
Connect a bog standard (incandescent) light bulb from L or N to earth at the output side and see if it lights up.
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There's no guarantee any resistor will fail open circuit, unless it's specifically designed to do so, i.e. it's a fusible resistor. In practise that's what normally happens to low value resistors, but it isn't a certainty and shouldn't be relied upon for safety.
My experience comes from the "normal" example - which, I admit, only covers the one failure mode. Like most safety policies, the most dramatic failures can be far more unlikely - but their severity warrant provisions be put in place to cover them.
I shall always defer to the safest approach available.
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Why should you use a complicated measurement setup?
Connect a bog standard (incandescent) light bulb from L or N to earth at the output side and see if it lights up.
That's too simple. :-DD
:palm: (Why didn't I think of that?)
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And if the transformer will have leakage in the mA range, light bulb will not glow, but you will ... after touching the output.
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And if the transformer will have leakage in the mA range, light bulb will not glow, but you will ... after touching the output.
Exactly. That could be used, as a quick test first, to gauge whether it's safe to connect a resisitor, but is not good enough to prove the current is too low to not shock.
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Okay, thanks all. I suspected leakage of one form or another. The capacitance value that is stamped on it is 0.001 pF.
I has to be a lot more than that; 100s of picofarads is typical.
I don't have any specific low impedance measuring gear, but I do have a 1 omh and a 10 omh power resistor somewhere that I can use as a shunt. Failing that, I'll hook up a brave, compulsorily volunteered, LED and see what we get.
a 22k 1/2 watt or 47k 1/4 watt resistor works well.
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I don't have any specific low impedance measuring gear, but I do have a 1 omh and a 10 omh power resistor somewhere that I can use as a shunt. Failing that, I'll hook up a brave, compulsorily volunteered, LED and see what we get.
a 22k 1/2 watt or 47k 1/4 watt resistor works well.
Those values are only safe to connect across 120V mains. On 230VAC they'll start to emit smoke fairly quickly. One of my first childhood experiments with the mains was a 22k resistor and a red LED. The LED survived, even though I didn't put a diode in series or parallel with it, as it non-destructively avalanched over on the reverse cycles, but the resistor emitted magic smoke and died fairly quickly. Fortunately it failed safe: open circuit and no explosion, or fire. You need at least 100k 0.6W, or 220k ¼W for 230V mains and even that's pushing it to the limit: fine for a quick test, but not a permanent design.
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Open the cover and take a look inside.
Make up a schematics on how things are wired.
If the transformer have not been used for a long time do a meggercheck. NB if the transformer is a centertap disconnect the centertap while doing the megger check. If you have 1Kohm/V or better you are good isolation wise.