1000 volts per meter in surge protection device

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tersh

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I read that "The standard unit of electric field (E-field) strength is the volt per meter (V/m). An E field of 1 V/m is represented by a potential difference of 1 V existing between two points that are 1 m apart.".

Lightning strike has 1000 volts per meter as units of measurement.

But why do surge protection people interpret it as the length of wire making voltage rise by 1000 volts meter? So if you have 5 meters wire.. then they conclude voltage is 5000 volts, hence they advise to shorten the lead as much as possible. Did they misinterpret it or is it still accurate, how?​
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181130-0704 EST

On first reading I had no idea what was being described. A clear description would have been helpful.

Searching the Internet for a value for the voltage drop per meter along the arc of a lighting strike I did not find the 1000 V value. I did find some in the several thousand to 5000 V range. So 1000 V is not a bad figure to use for the purposes of the discussion.

A lighting arc as a voltage source will have a very low internal impedance. And a lighting strike represents a very fast rate of rise of voltage, and of short duration. Thus, high frequency components.

A piece of copper wire 1 meter long in and parallel to the arc will see the 1000 V across it. Both resistance and inductance come into play.

What is being described is a transient voltage device with a 500 V drop when conducting with leads on each end that are 1 meter long. This series combination is then connected across the load. This load itself has to be 2 meters long in order to be across a 2 meter length of the lighting arc. The voltage drop across each 1 meter wire would be 750 V. 2000 V gets applied across the load. This is an unrealistic analysis. But it does point out the value of short leads in an indirect fashion.

In lighting protection of equipment you want the transient limiter directly at and across the load. The less current to the limiter the lower is the clamping voltage. To lower the current you want long high impedance leads between the lighting strike point and the load. Here long leads are good.

At the entry point to a building you want high series impedance and a transient limiter. This lowers the voltage going into the building.

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gar said:
181130-0704 EST

On first reading I had no idea what was being described. A clear description would have been helpful.

Searching the Internet for a value for the voltage drop per meter along the arc of a lighting strike I did not find the 1000 V value. I did find some in the several thousand to 5000 V range. So 1000 V is not a bad figure to use for the purposes of the discussion.

A lighting arc as a voltage source will have a very low internal impedance. And a lighting strike represents a very fast rate of rise of voltage, and of short duration. Thus, high frequency components.

A piece of copper wire 1 meter long in and parallel to the arc will see the 1000 V across it. Both resistance and inductance come into play.

What is being described is a transient voltage device with a 500 V drop when conducting with leads on each end that are 1 meter long. This series combination is then connected across the load. This load itself has to be 2 meters long in order to be across a 2 meter length of the lighting arc. The voltage drop across each 1 meter wire would be 750 V. 2000 V gets applied across the load. This is an unrealistic analysis. But it does point out the value of short leads in an indirect fashion.

In lighting protection of equipment you want the transient limiter directly at and across the load. The less current to the limiter the lower is the clamping voltage. To lower the current you want long high impedance leads between the lighting strike point and the load. Here long leads are good.

At the entry point to a building you want high series impedance and a transient limiter. This lowers the voltage going into the building.

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This is the exact reference:

http://www2.schneider-electric.com/...277471/en_US/Surge protection devices SPD.pdf
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Is it true that voltage can rise per meter of wire with surge current. Some EE believe it's just an urban legend. What about you guys? Let's try to settle it once and for all. Does it have solid basis or not?
 
gar said:
181130-0933 EST

Rate of change of current and inductance were not mentioned in post #1.

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Click to expand...

Sorry. So based on the added information, is it true the voltage can rise the longer it is. But why do some electrical engineers don't believe in it? Does it have solid theoretical or physics backing?
 
gar said:
181130-0937 EST

Theory yes, but there has to be voltage to drive that current.

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Click to expand...

What do you mean? how about lightning? When it hits the lines.. does it mean the longer is the lines, the bigger is the voltage rise that even exceeds that of the lightning?
 
181130-1605 EST

tersh:

I have to leave shortly. You are listed as an engineer. If you are a recent graduate and still know some of the teachers, then I suggest you try to talk with one or more of them. Direct interaction is usually more useful.

Go back to differential equations and study transient circuits with an inductor and resistor, and also with KRC circuit.

Apply a voltage to an LR circuit where initial current is zero and observe the voltages and current of the components. You can do this on paper or experimentally.

Possibly more later after more qiestions.
 
181201-0831 EST

I am randomly looking for some information relating to this question. I might phrase the question as --- is a lighting strike a constant current source? Probably is with some qualifications.

The following link has no direct connection but is useful background ---
https://www.nssl.noaa.gov/education/svrwx101/lightning/faq/

More
https://www.researchgate.net/post/peak_value_of_lightning_strike_current
https://news.nationalgeographic.com/news/2004/06/flash-facts-about-lightning/

The following mentions a model based on a constant current source.
https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2004JD005202

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edit- ans to basic OP question- electric field strength and voltage drop on a wire are not directly related as implied in the post. Thus, there is no discrepancy in the numbers presented in the OP post.

Simplistically a meter of wire is 1 uH as a rule of thumb.
Much higher if there are any loops or the wire not straight. Magneitic forces on the wire due to high current try to straighten a wire, and will literally break a loop of wire. I have broken 250 MCM in some NWE lightning tests at megamps.

Typical lightning is 20kA (but has been measured as high as 3 MILLION amps)

Typical rise time is 6 usec.

Basic differential equation is V(t) = L* di/dt Volts/meter field strength in the vicinity do not factor into the voltage drop except for a small coupling effects due to fields of other nearby wires with lightning currents.

Can simplify to di = 20,000, dt = 6 us

So typically, L*di/dt = 1e-6*2E4/6E-6 = roughly 3,300 volts per meter.
The 1000 V rule of thumb probably assumes the 20kA of a typical lightning strike gets diverted to multiple paths before reaching a typical arrestor.

Correspondingly much high for the mega lightning bolts. Lower for less ;than 20kA lightning strikes.
 
Last edited:
The 1000V/m only applies for the short distances (i.e., SPD connection distances) under discussion. The high voltage is the result of the high impedance of the high frequency components. These high frequencies are "impeded" (converted to heat) and diminish in amplitude as the pulse travels along the wire. As the pulse is stripped of its higher frequencies, the impedance (due to inductance) will be less...and the resultant voltage drop will be less. To explicitly answer the OPs question, if you continued to measure the voltage drop over 1 meter intervals, it would continually diminish. It will not remain 1000 V/m. In the limit (if all frequency components are stripped), the voltage drop would be whatever DC current remained multiplied by the wire's resistance (i.e.., V = IR).

The main point of the example is to illustrate that the choice of an SPD must consider the voltage impressed across the connections (a function of connection length) as well as the voltage across the SPD device itself. For example, if equipment needs to be protected at 700V, one might acquire an SPD that triggers at 500V and think that's enough. But if the connection wires...due simply to the inductance in the wire...contribute an impedance that results in 500V of drop, then the NET delta is 1000V. That is, the SPD that triggers at 500V (across the SPD), won't trigger until the device under protection sees 1000V. And if that device fries at 700V....it's gone.

Big point: Wires have inductance. In daily work dealing with household current, this inductance is negligible or factored into the real resistance. Rapid pulse spikes (e.g., from lightning) create very high frequency components. As frequency goes up, inductance that is otherwise negligible begins to be felt. Now that wire that is a "short circuit" in daily life becomes a very high impedance when that lightning surge runs through it. Low impedance path in daily life becomes high impedance path (read also, high voltage delta) during lightning.
 
junkhound said:
edit- ans to basic OP question- electric field strength and voltage drop on a wire are not directly related as implied in the post. Thus, there is no discrepancy in the numbers presented in the OP post.

Simplistically a meter of wire is 1 uH as a rule of thumb.
Much higher if there are any loops or the wire not straight. Magneitic forces on the wire due to high current try to straighten a wire, and will literally break a loop of wire. I have broken 250 MCM in some NWE lightning tests at megamps.

Typical lightning is 20kA (but has been measured as high as 3 MILLION amps)

Typical rise time is 6 usec.

Basic differential equation is V(t) = L* di/dt Volts/meter field strength in the vicinity do not factor into the voltage drop except for a small coupling effects due to fields of other nearby wires with lightning currents.

Can simplify to di = 20,000, dt = 6 us

So typically, L*di/dt = 1e-6*2E4/6E-6 = roughly 3,300 volts per meter.
The 1000 V rule of thumb probably assumes the 20kA of a typical lightning strike gets diverted to multiple paths before reaching a typical arrestor.

Correspondingly much high for the mega lightning bolts. Lower for less ;than 20kA lightning strikes.
Click to expand...

Just to verify. When you described "roughly 3,300 volts per meter". Is it measured at the end of the line away from the current injection point or from point of current injection. Look at this schematic (I read this elsewhere):


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"
See schematic. L1 represents the inductance of the long wire. I1 is the 8kA 80/20 waveform. The longer the wire, the larger L1 will be, and the larger the voltage across I1 will be when the pulse waveform is applied. That is all the article is trying to say. Yes, a longer wire has more inductance, and consequently, it will have more voltage develop AT THE POINT OF CURRENT INJECTION. Not down at the end of the line away from the current injection point."

Do you agree with the above that the voltage develop at the point of current injection. Or do you think it's really down
at the end of the line away from the current injection point in the context of SPD "1000v per meter rise"?​
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It's easy to extend our models outside the physical reality. For example, we can "model" a current source...or a voltage source...or both. When we model a current source (without qualification), then voltage becomes unconstrained. Even with lightning, voltage isn't unconstrained. But if you look at our mathematical model of an unqualified current source, the equations are mathematically unconstrained.

Take a current source of 1A. Throw in a 1 ohm resistor. How many volts? One. Throw in a 5 ohm resistor. How many volts? Five. Throw in a 1 Mega Ohm resistor. How many volts? 1 MV!!!! Oh, and look at that power....1W to 1MW Wow, we can save the energy crisis with a 1A current source and really big resistors!!!!!

The current source is only good as a model of real life as long as we operate within the voltage capability of the source. If the source can't adjust the voltage as the resistance is increased, then current will drop. The 1A is unsustainable beyond the physical attributes of the source. But the MATH won't show it. The math will let you do whatever you want. Models are only sustainable (as models of reality) within the constraints of the assumptions on which that model is founded.

The example provided by the manufactures has a CURRENT source of 8KA, a rise time (8 usec) a sustain time (20 usec). Note there is no voltage in the assumption. We know, based on how inductance works, that the voltage is proportional to the change in rate of current, with the proportionality constant measured in what we call Henrys. Given enough Henrys (or impedance in general), the current source of 8KA becomes unsustainable. Yes, unsustainable...even for lightning. I'm not an SPD expert. My limit is pretty much at the understanding of Ldi/dt. I understand that. And I understand our models fall apart beyond the assumptions. The SPD experts will tell you that over short distances, the models and examples provided by the manufacture are operating within the bounds of the assumptions of what lightning can do. As one stretches that wire...although the Henrys per meter may remain the same....the current waveform will not...the resulting impedance will change...and the voltage will change.

You can "mathematically" create a thousand volts every time you add 1 meter of wire. You can crunch the numbers on paper. Try that with Mother Nature...and it's a fail. In the real world...no matter how hard we try to make our model do our bidding over physics....Mother Nature always wins.
 
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