# Rogowski coil response factor frequency dependence

I am creating a high current (kA) pulse (<1 ms) by discharging a supercapacitor into a low resistance (<1mOhm) circuit. The switching is done by paralleled power-MOSFETs.

To measure the current I use a rogowski coil (RCT16-50000 from Accuenergy, no data sheet provided). According to Wikipedia, the output from the coil is:

This output is integrated by the math function of my PicoScope's software to deliver a function proportional to the measured current. This function must be multiplied with a (coil dependent) response factor to get the current. According to Accuenergy, my coil's response factor is 2.0 mV/kA @ 50 Hz and 2.4 mV/kA @ 60 Hz. They don't provide the full specs (A, N and l) of the coil (companies support is not responding).

I don't understand why the response factor is frequency depended, even though frequency is not a factor in the equation above.

Also, I don't know the frequency of my current pulse. Any ideas how to calculate the proper response factor of my coil for my pulse?

• Are you sure the coil can be used (easily) for such transient signals at, i.e. non-steady-state sinusoidal signals, at all?
– Curd
Commented Oct 18, 2017 at 9:55
• "no data sheet provided" Outch! Commented Oct 18, 2017 at 11:14
• @winny Yep, never again Accuenergy. But at least it helped my understanding of Rogowski coils ;)
– Matt
Commented Oct 18, 2017 at 14:22
• Off topic but PEM never fails to deliver when it comes to Rogowskis. Commented Oct 18, 2017 at 16:23

$$v(t)={\frac {-AN\mu _{0}}{l}}{\frac {dI(t)}{dt}}$$

From Curd's answer you get the response of a sinewave:

$$v(t)={\frac {-AN\mu _{0}}{l}}{I_0\omega \cos(\omega t)}$$

Let we introduce a constant $K_a$ as from your example:

$$K_a={\frac {-AN\mu _{0}}{l}}{\omega}$$

Now this is the output voltage from Rogowski coil with your constant:

$$v(t)={K_a}{I_0\cos(\omega t)}$$

But we could go in different way:

$${\frac {-AN\mu _{0}}{l}}=\frac{K_{a_{50Hz}}}{\omega_{50Hz}}$$

At the output of the Rogowski coil, put the integrator as this is the normal practice.

$${V_{\text{out}}=\int v\,dt={\frac {-AN\mu _{0}}{l}}I(t)+C_{\text{integration}}}$$

$${V_{\text{out}}=\int v\,dt=\frac{K_{a_{50Hz}}}{\omega_{50Hz}}I(t)+C_{\text{integration}}}$$

For $C_{\text{integration}}$ you can put a PI controller, that is subtracting the output voltage drift. If the mean value of the current is zero over long period of time, you can use the PI controller to compensate the the drift of the integrator output.

simulate this circuit – Schematic created using CircuitLab

• Brilliant, thanks. The relationship between the "response factor", the frequency and the coil's parameters was exactly what I was looking for. As I mentioned, I'd like to integrate the coil's output signal with my oscilloscope software. Are there any reasons not to do this? Also, In my understanding, the coil output is simply proportional to the derivative of the change of current. So, regarding @Whit3rd 's post, do I have to worry about the Fourier components of my current he mentions?
– Matt
Commented Oct 18, 2017 at 13:27
• @Matt: I wouldn'd go so far as Whit3rd and expect the response factor to be a completely arbirtary function of frequency but I think it could be a complex constant multiplied by the frequency. (LTI system with transfer function being a complex constant whose imaginary part is negligible only for low frequencies like 60Hz; in your case it seems you have to consider much higher frequency components; that's what my question comment below your question was aiming at)
– Curd
Commented Oct 18, 2017 at 14:22
• @Matt: to clarify my (and Whit3rd's) concerns: image that the coild has (of course) some inductance and also some capacitance between both terminals, i.e. it surely forms a parallel LC circuit (--> resonance and phase distorion). The questions is not if the real response differs from the simple formula but at what frequencies will it more than negligible. If you are lucky the coils resonant frequency is much higher than any substantial component of your current pulse. If not, you probably will see some "ringing".
– Curd
Commented Oct 18, 2017 at 14:38

I guess the response factor $k$ is defined for amplitudes of sinusoidal steady state signals of the given frequencies (50Hz and 60Hz).

$V_0 = k I_0$ where
$I_0$ is amplitude of sinusoidal current signal $I(t) = I_0 sin(\omega t)$ and
$V_0$ is amplitude of sinusoidal voltage singal (output)

Then the quotient of the given response factors 2.4 / 2.0 = 1.2 is exactly what you expect from the formula 60Hz/50Hz = 1.2.

Your formula from Wikipedia is frequency depedent: the frequency dendency is given by the d/dt operator: the higher the frequency (of the I(t) signal), the higher is the rate of change even if current amplitudes are the same:

$I(t) = I_0 \sin(\omega t)$ → $\frac{d}{dt}I(t) = I_0\omega \cos(\omega t)$
i.e. amplitude of rate of change is $I_0\omega$, directly depedent of frequency.

I don't understand why the response factor is frequency depended, even though frequency is not a factor in the equation above.

The explicit time dependence in the 'd/dt' derivative is a frequency term.

Because your pulse is not a steady sinewave, it must be treated as a composite of multiple Fourier components, and the formula applied separately to each component. The usual approach, is to sample the pulse at a fraction (one sixth should be enough) of the shortest timing that your pulse contains, and employ a FFT algorithm on the sampled data. Such a transform will give best results if the sampled data includes a quiet period before the pulse, and the entire pulse duration, and its decay train, and some extra duration (a quiet period AFTER the pulse).

The formula can be applied separately to each of the Fourier components. The sampled data is v(t), a list of samples at different times, and its transform is v(f), a list of components at different frequency and phase.

After finding I(f) from the v(f) terms, you can reverse-FFT the I(f) to find I(t). It won't be just 'a factor', but a complete synthesis, from your voltage data, of a current pulse.