Can Surge Protective Devices (or the MOVs) work without shunting to ground?

[tersh]

Does surge protective device really a ground to shunt the current or can it resist them by its own (when say connecting line to line (or directly put between the black and red wire in the 240v in US power system).


See technical paper at:
http://www.surgesuppression.com/ima...licationofSurgeSuppressionRevDate10202005.pdf

"Power was applied to Phase A and B and thermometer was used to measure the temperature of the MOV when surge was introduced (details in the paper above). Based on the much higher temperature rise in the discrete line-to-line MOV, the discrete line-to-line MOV absorbs a much larger portion of the surge current and energy than the combination of the two line-to-neutral paths.


See the following image:

Someone told me varistors don't really need ground to shunt the current. He said that there was no minimum input voltage anywhere. And the physics of an MOV do not show any minimum either.

He said that the Zinc Oxide grains were close enough to each other that they are in light electrical contact, with very little current able to flow between them; probably due to the oxide layer on their surfaces. Higher voltages are able to break down this oxide layer and allow more current to flow. That's how they suppress surges, by conducting the higher voltages, usually to ground, but blocking current when only the lower normal operating voltage is present. But without ground, then the varistors (MOVs) just heat up. That's how it can be connected line to line without ground.

So Surge Protective Devices can be connected directly between line to line right? and if there are surges, it just heats up (and degrades a bit). But compared to having access to ground. How good can it still dissipate the energy by heating up (without any access to ground)?

What do you think?

 

[gar]

181203-2329 EST

tersh:

You list yourself as an engineer. What kind, what is your background?

Your question is not something I would expect from an electrical engineer phrased as it is. We need more background to understand how to answer your question.

.

 

[GoldDigger]

A surge suppressor can do just as good a job at suppressing line to line transients when connected directly from line to line as it can with two devices each connected from one line to ground.
What it cannot do is provide any protection against common mode voltage (an abnormal transient voltage which affects both lines equally with the same polarity.)

To restate that, if the fault or transient condition applied +1000VDC to L1 with respect to ground and -1000VDC to L2, then a suppressor from L1 to L2 will see and excess voltage of 2000VDC and will conduct.
But if it applies +1000VDC to L1 and also +1000VDC to L2 a suppressor from L1 to L2 will not see any abnormal voltage and will provide no protection. In both cases two suppressors from L1 and L2 to ground (not neutral!!) would provide protection.

Another aspect to consider is whether a common mode transient will in fact do any damage to a connected load. The answer to that depends entirely on the details of the load and its insulation. 3

 

[tersh]

A surge suppressor can do just as good a job at suppressing line to line transients when connected directly from line to line as it can with two devices each connected from one line to ground.
What it cannot do is provide any protection against common mode voltage (an abnormal transient voltage which affects both lines equally with the same polarity.)

To restate that, if the fault or transient condition applied +1000VDC to L1 with respect to ground and -1000VDC to L2, then a suppressor from L1 to L2 will see and excess voltage of 2000VDC and will conduct.
But if it applies +1000VDC to L1 and also +1000VDC to L2 a suppressor from L1 to L2 will not see any abnormal voltage and will provide no protection. In both cases two suppressors from L1 and L2 to ground (not neutral!!) would provide protection.

But what if the L1 and L2 is not connected to any ground. What will happen to the AC surges that directly passes the two hot lines (without any path to ground)?


Gar. I am an electronic engineer with some background in electrical, but it was many decades ago. At that time. We were not taught with regards to Metal Oxide Varistors.

Another aspect to consider is whether a common mode transient will in fact do any damage to a connected load. The answer to that depends entirely on the details of the load and its insulation. 3

 

[GoldDigger]

But what if the L1 and L2 is not connected to any ground. What will happen to the AC surges that directly passes the two hot lines (without any path to ground)?


Gar. I am an electronic engineer with some background in electrical, but it was many decades ago. At that time. We were not taught with regards to Metal Oxide Varistors.

Whether or not L1 and L2 are directly or indirectly referenced to ground, any voltages applied to them can still be either differential voltages or common mode voltages with respect to ground.
If I raise L1 and L2 to +10000VDC by means of a lighting strike (capacitive current flowing), neither the load equipment nor the devices intended to protect that load will care one bit whether L1 and L2 were originally floating with respect to ground nor what their voltages to ground were before the event.

 

[tersh]

Whether or not L1 and L2 are directly or indirectly referenced to ground, any voltages applied to them can still be either differential voltages or common mode voltages with respect to ground.
If I raise L1 and L2 to +10000VDC by means of a lighting strike (capacitive current flowing), neither the load equipment nor the devices intended to protect that load will care one bit whether L1 and L2 were originally floating with respect to ground nor what their voltages to ground were before the event.

So an SPD (Surge Protective Device) with rating of for example Maximum Surge Current of 140,000A can take many surge hits (like 8kA each) from line to line without shunting them to ground/neutral (without any access to any ground/neutral).. but why do people usually believe that an SPD needs to be connected to ground so the current can escape to ground or neutral? Is this a common misconception? Many believe this and they are convinced they were right.

 

[gar]

181204-2536 EST

tersh:

An MOV is a two terminal non-linear resistance. It has bidirectional symmetry. Below a certain threshold voltage across the MOV it is a rather high resistance. Above the threshold it is a very much lower resistance, but still non-linear.

As GoldDigger pointed out whether either lead is grounded or not when the voltage across the MOV exceeds the threshold, then it conducts between the two leads. If one lead is earthed (grounded), then current will flow to earth when the threshold is exceeded. If the MOV is between two wires each floating off of ground, then the voltage between the wires will be clamped, but this alone has no effect relative to earth (ground).

The MOV does not clamp the voltage to a very low value like an SCR would do if the SCR was triggered. The MOV will look something like a constant voltage, the threshold voltage, with a low series non-linear internal impedance.

Relative to your last post. The MOV does just what it does. How it is applied in a circuit will determine what protective function it performs. Also MOVs are not very good clamps. Look at their characteristic curves. But they are better than nothing, and if the circuit they are combined with has enough peak voltage capability, then they do what you want. Further if you provide sufficient series impedance to the load shunted by an MOV, then the results can be good.

.

 

[tersh]

181204-2536 EST

tersh:

An MOV is a two terminal non-linear resistance. It has bidirectional symmetry. Below a certain threshold voltage across the MOV it is a rather high resistance. Above the threshold it is a very much lower resistance, but still non-linear.

As GoldDigger pointed out whether either lead is grounded or not when the voltage across the MOV exceeds the threshold, then it conducts between the two leads. If one lead is earthed (grounded), then current will flow to earth when the threshold is exceeded. If the MOV is between two wires each floating off of ground, then the voltage between the wires will be clamped, but this alone has no effect relative to earth (ground).

The MOV does not clamp the voltage to a very low value like an SCR would do if the SCR was triggered. The MOV will look something like a constant voltage, the threshold voltage, with a low series non-linear internal impedance.

Relative to your last post. The MOV does just what it does. How it is applied in a circuit will determine what protective function it performs. Also MOVs are not very good clamps. Look at their characteristic curves. But they are better than nothing, and if the circuit they are combined with has enough peak voltage capability, then they do what you want. Further if you provide sufficient series impedance to the load shunted by an MOV, then the results can be good.

.

UL has a requirement that there must be at least 10 meters (or 30 feet) to an SPD type 3 device. This 10 meters will act like the series impedance. Can you please share exactly what formulas are involved where if there is series impedance then the MOV can short (and conduct). And if there is no series impedance (no 10 meters wire). It won't short even if it's above the threshold (MCOV)? How? I'm also reviewing my ac lessons. Thanks.

 

[GoldDigger]

UL has a requirement that there must be at least 10 meters (or 30 feet) to an SPD type 3 device. This 10 meters will act like the series impedance. Can you please share exactly what formulas are involved where if there is series impedance then the MOV can short (and conduct). And if there is no series impedance (no 10 meters wire). It won't short even if it's above the threshold (MCOV)? How? I'm also reviewing my ac lessons. Thanks.

An MOV does not literally short. It takes on a much smaller impedance than it had just below the threshold voltage.
At any voltage above the threshold, as gar said, it will seriously conduct. Its current versus voltage curve is totally independent of what series resistance is present in the source circuit. What change is at which point along this curve the MOV is sitting at the peak of the event.
If you have a transient fevent4 which is capable of sourcing 24000A into a bolted short on a nominal 120V circuit, and the source of the fault current is at 4800V, the effective internal resistance of that source is .02 Ohm. A 120V design MOV might conduct 24000A at 480V (for example). That gives the MOV an effective internal resistance of .02 Ohm also at 480V. The MOV might be able to clamp the voltage at its terminals to about 480V. But if the series resistance between the source and the MOV (added wire length, inductance, etc.) is .2 Ohms the maximum current is now roiughly 2400A the MOV will be able to clamp the voltage to much less than 480V. More like, perhaps 300V, which could be below the damage threshold of the protected load. The MOV is now operating at a different point on its unchanged voltage/current curve.

 

[tersh]

A surge suppressor can do just as good a job at suppressing line to line transients when connected directly from line to line as it can with two devices each connected from one line to ground.
What it cannot do is provide any protection against common mode voltage (an abnormal transient voltage which affects both lines equally with the same polarity.)

To restate that, if the fault or transient condition applied +1000VDC to L1 with respect to ground and -1000VDC to L2, then a suppressor from L1 to L2 will see and excess voltage of 2000VDC and will conduct.
But if it applies +1000VDC to L1 and also +1000VDC to L2 a suppressor from L1 to L2 will not see any abnormal voltage and will provide no protection. In both cases two suppressors from L1 and L2 to ground (not neutral!!) would provide protection.

Another aspect to consider is whether a common mode transient will in fact do any damage to a connected load. The answer to that depends entirely on the details of the load and its insulation. 3

Let me use an illustration because something is confusing with some descriptions:

Let's say your circuit is purely 240v, without any neutral even connected to your panel and it's not even grounded (neither to local electrode or to centertap for sake of discussions). Does your description "

A surge suppressor can do just as good a job at suppressing line to line transients when connected directly from line to line" still valid??

But if it's still valid. Why did you state that "

But if it applies +1000VDC to L1 and also +1000VDC to L2 a suppressor from L1 to L2 will not see any abnormal voltage and will provide no protection".

In normal lightning induced surges. Is it +1000VDC to L1 and +1000VDC TO L2 or is it usually +1000VDC to L1 and -1000VDC to L2. Again, let's assume no neutral and no ground is connected.

It's confusing because you said line to line protection is as valid. But if it's both +, then the MOV won't conduct or work. And in follow up message when you said "

Whether or not L1 and L2 are directly or indirectly referenced to ground, any voltages applied to them can still be either differential voltages or common mode voltages with respect to ground. ".
But I was asking if L1 and L2 is totally floating without any ground present (in actual). Then how could it be exposed to +1000VDC and -1000VDC? Is it always +1000VDC and -1000VDC?

Thank you. ​

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[GoldDigger]

Even when a metal object (simple or complex) does not have a metallic connection to earth, you can still throw a charge onto it, making it one plate of a capacitor to ground, and producing an arbitrarily high voltage with respect to ground. That is a real, measurable, voltage.
You can also apply a large electric field near it and produce an AC voltage with respect to ground, again by capacitive coupling.

The specific numbers were just made up for the purposes of illlustration.
Yes, a line to line MOV will not help if a large voltage is induced onto both sides of the circuit at the same time.
Whether L1 and L2 move together or in opposite directions depends on the exact fault that causes the transient. A super voltage applied to the primary will cause L1 and L2 to move in opposite directions.
A large voltage induced on the entire secondary will cause L1 and L2 to move together.
A voltage induced into wiring by magnetic effects from nearby lightning currents could go either way.
If both L1 and L2 move together it is, by definition, not a line to line transient. So the MOV will not protect it, consistent with the earlier statement.

 

[tersh]

An MOV does not literally short. It takes on a much smaller impedance than it had just below the threshold voltage.
At any voltage above the threshold, as gar said, it will seriously conduct. Its current versus voltage curve is totally independent of what series resistance is present in the source circuit. What change is at which point along this curve the MOV is sitting at the peak of the event.
If you have a transient fevent4 which is capable of sourcing 24000A into a bolted short on a nominal 120V circuit, and the source of the fault current is at 4800V, the effective internal resistance of that source is .02 Ohm. A 120V design MOV might conduct 24000A at 480V (for example). That gives the MOV an effective internal resistance of .02 Ohm also at 480V. The MOV might be able to clamp the voltage at its terminals to about 480V. But if the series resistance between the source and the MOV (added wire length, inductance, etc.) is .2 Ohms the maximum current is now roiughly 2400A the MOV will be able to clamp the voltage to much less than 480V. More like, perhaps 300V, which could be below the damage threshold of the protected load. The MOV is now operating at a different point on its unchanged voltage/current curve.

In other words, the series impedance is only important if it's line to line surges (without any ground and also that is not common mode voltage where L1 and L2 move together). But if the MOV is connected across line to ground, then the surge current will just dump to ground, and here the series impedance is not important?

If you have a MOV with max surge rating of 10,000A. And there is a 3,000A surge current directly going into the MOV. So in the case of the MOV connected from line to ground, the MOV will dump the 3,000A to ground. While if the MOV is connected line to line, and the surge current of 3,000A is going for it (in opposite direction), then the MOV will not dump it elsewhere but just lower it by the series resistance or impedance thing, right?

In both cases, the damage to the MOV will be the same, regardless of whether it dumps the current or lower it by series impedance?

 

[GoldDigger]

In other words, the series impedance is only important if it's line to line surges (without any ground and also that is not common mode voltage where L1 and L2 move together). But if the MOV is connected across line to ground, then the surge current will just dump to ground, and here the series impedance is not important?

If you have a MOV with max surge rating of 10,000A. And there is a 3,000A surge current directly going into the MOV. So in the case of the MOV connected from line to ground, the MOV will dump the 3,000A to ground. While if the MOV is connected line to line, and the surge current of 3,000A is going for it (in opposite direction), then the MOV will not dump it elsewhere but just lower it by the series resistance or impedance thing, right?

In both cases, the damage to the MOV will be the same, regardless of whether it dumps the current or lower it by series impedance?

You are missing my point. In the case of a line to ground surge, the protective action of the MOV will still be greatest when there is a small but significant for this current level series impedance in both L1 and L2. If the series impedance is higher the MOV will end up having to drop less voltage by itself and so will have both a lower current and a lower voltage at its operating point than if it had to work with a lower source impedance. So the series impedance will allow the MOV to protect against higher open circuit voltage transients while dissipating less energy inside itself and therefore suffering less damage.

 

[tersh]

You are missing my point. In the case of a line to ground surge, the protective action of the MOV will still be greatest when there is a small but significant for this current level series impedance in both L1 and L2. If the series impedance is higher the MOV will end up having to drop less voltage by itself and so will have both a lower current and a lower voltage at its operating point than if it had to work with a lower source impedance. So the series impedance will allow the MOV to protect against higher open circuit voltage transients while dissipating less energy inside itself and therefore suffering less damage.

What if the series impedance is close to zero (just for sake of discussion even if it's hard to make it this way). Would the MOV connected line to ground still able to dump the current to ground.. and would the MOV connected line to line just heat up because it is dissipating it as heat already?

 

[GoldDigger]

What if the series impedance is close to zero (just for sake of discussion even if it's hard to make it this way). Would the MOV connected line to ground still able to dump the current to ground.. and would the MOV connected line to line just heat up because it is dissipating it as heat already?

Once again you seem to have a fundamental misconception of what is happening. The MOV is not "dumping" current anywhere. It is carrying current between the two points at which it is connected to the circuit (or in the ungrounded circuit case earth and the one connection point.
In all cases the power being dissipated as heat in the MOV is entirely a function of how much current is flowing through it, not where that current is coming from or going to.

 

[tersh]

Once again you seem to have a fundamental misconception of what is happening. The MOV is not "dumping" current anywhere. It is carrying current between the two points at which it is connected to the circuit (or in the ungrounded circuit case earth and the one connection point.
In all cases the power being dissipated as heat in the MOV is entirely a function of how much current is flowing through it, not where that current is coming from or going to.

gar earlier wrote: "As GoldDigger pointed out whether either lead is grounded or not when the voltage across the MOV exceeds the threshold, then it conducts between the two leads. If one lead is earthed (grounded), then current will flow to earth when the threshold is exceeded. If the MOV is between two wires each floating off of ground, then the voltage between the wires will be clamped, but this alone has no effect relative to earth (ground)."

He mentioned that if one lead was earthed (grounded), then current will flow to earth when the threshold is exceeded. Is this not dumping it to ground? Or let's not use the word "dumping". But "flow". So the current will indeed flow to ground? Or is this just tech slang that the current will flow (when it doesn't really)? Note majority of the population takes the picture so literally and site after site, they describe the conducting MOV shunts or dump current to earth.
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[junkhound]

Tersch: It may help you to visualize a MOV as back to back avalanche or zener diodes with high reverse losses and very soft breakdown transitions.

 

[kwired]

Put MOV in parallel to a load, now raise voltage to level that MOV will conduct a significant amount of current. Total volts applied is still same, but if MOV has lower resistance then more current flows through the MOV than the load.

Keep in mind these weren't designed to carry high current for long periods of time and is why they are effective for transient voltages. If you apply a high voltage for long time period you burn out the MOV and only the original load remains in the circuit, still subjected to the over voltage.

 

[gar]

181204-0947 EST

tersh:

As I said before in some fashion --- an MOV has a volt-current curve that defines what the MOV does. The external circuit that the MOV is connected to will determine where you are on this curve at some particular point in time.

A current source is an idealized concept. With electronic feedback circuits can be made that look like an idealized current source over some moderately limited range.

An approximate current source can be made from a large voltage source with a high internal impedance where the source voltage is large compared to the load voltage. So, if I have a 100 V battery with an internal impedance of 100 ohms, then the current is 1 A with the terminals shorted. If I change the load resistance from 0 ohms to 1 ohm, then the current changes from 1 A to 0.990 A. Fairly good current source for some applications, such as an LED supply, but not very power efficient.

Put an ideal 2 V MOV across this load. The MOV limits the voltage to 2 V across that load, and the current from the current generator is (100-2)/100 = 0.98 A if the MOV is limiting. As you adjust your load resistance from 2/0.98 = 2.04 ohms to infinity the MOV current goes from 0 to 0.98 A. If my load can withstand 2 V, then the load is protected. If the load can only tolerate 1.5 V, then the load is not protected and it blows-up.

.

 

[Besoeker]

181203-2329 EST

tersh:

You list yourself as an engineer. What kind, what is your background?

Your question is not something I would expect from an electrical engineer phrased as it is. We need more background to understand how to answer your question.

.

I think that's a tad harsh, old chap.

MOVs were not mentioned while I was an undergraduate. What one realises when you go at in real life is that your education, at best, just scratched the surface of a deep and wide ranging field. That's inevitably going to be the case with most degrees. They give you the basics, a starting point, the theory, the tools, but do not address very specific areas. Do you want to design distribution systems or motors or VSDs.......