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spike's energy level

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dark

Member
Hi,

Can someone help me out with how the Joule rating for the spike is calculated say the following standard:
EN60255 direct spike Level 5 KV waveform 1.2/50μs impedance 500Ohm .

Then How much Joules this waveform has? I am little confused.

Regards
-Adi
 

crutschow

Well-Known Member
Most Helpful Member
What does 1.2/50µs mean? What is the shape of the waveform?
 

dark

Member
What does 1.2/50µs mean? What is the shape of the waveform?
I guess 1.2us rise time and 50us is 50% the fall value . EN60255 standard was asked me to in corporate I couldnt find much on the net :(


Regards
 

crutschow

Well-Known Member
Most Helpful Member
You would integrate the power over the waveform period to get the energy in joules. The following calculator should give you a ballpark number Welwyn - Calculation Tools.
 

dark

Member
You would integrate the power over the waveform period to get the energy in joules. The following calculator should give you a ballpark number Welwyn - Calculation Tools.
You are simply great , thanks for the nice link . One help more regarding the calculator does it works for current waveform aswell ? . The drawing shows in terms of voltage.

Thanks
 

crutschow

Well-Known Member
Most Helpful Member
One help more regarding the calculator does it works for current waveform aswell ? . The drawing shows in terms of voltage.
To get power you need both voltage and current. If you have a current waveform then you need to know the circuit resistance to calculate the voltage and power.
 

dark

Member
To get power you need both voltage and current. If you have a current waveform then you need to know the circuit resistance to calculate the voltage and power.
Hi,
The link you suggested accepts only peak voltage (no current) to determine the Joule?
 

duffy

Well-Known Member
Crutschow is giving you good advice on how this problem is solved, not a magic bullet.

That program also only works with square or exponential waveforms. A gaussian pulse is also common, but if you try to use a square wave or exponential as a model for, say, something approaching a Dirac delta impulse you could be off by a thousand to one.

It also assumes you have a resistance, not a complex impedance - you said your load was an impedance. If you ignore the reactive effects of your load, again, you could be off by a ridiculous amount.

Energy = Power x Time. You integrate (or more typically do a Riemann sum approximation) of the power waveform over time. Unless your problem fits that calculator exactly, you will get a much closer number by doing the Riemann sum approximation with pencil and paper by plotting a lousy half-dozen points. Or do it right and use a spreadsheet. First, you need to understand what it is you are doing.

Power isn't simply voltage or current, it isn't simply force or distance. This is a very important concept that over-unity nutters are always getting wrong. Your voltage will have one function into a complex impedance, your current will have another. You need both.
 

dark

Member
Crutschow is giving you good advice on how this problem is solved, not a magic bullet.

That program also only works with square or exponential waveforms. A gaussian pulse is also common, but if you try to use a square wave or exponential as a model for, say, something approaching a Dirac delta impulse you could be off by a thousand to one.

It also assumes you have a resistance, not a complex impedance - you said your load was an impedance. If you ignore the reactive effects of your load, again, you could be off by a ridiculous amount.

Energy = Power x Time. You integrate (or more typically do a Riemann sum approximation) of the power waveform over time. Unless your problem fits that calculator exactly, you will get a much closer number by doing the Riemann sum approximation with pencil and paper by plotting a lousy half-dozen points. Or do it right and use a spreadsheet. First, you need to understand what it is you are doing.

Power isn't simply voltage or current, it isn't simply force or distance. This is a very important concept that over-unity nutters are always getting wrong. Your voltage will have one function into a complex impedance, your current will have another. You need both.

Duffy , I talk with regards to "EN60255 direct spike Level 5 KV waveform 1.2/50μs impedance 500Ohm " and not Dirac delta or square impulse . The standard description shows the voltage rising as an exponential function and rise time and fall time has been mentioned. Crutschow's suggested link is producing results rather on the higher side but arnt too off the path . What my concern is in the standard mentioned above the peak current looks like 10Amps is it correct to deduce like this.

Thank you
 
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