# Tests with a rogowski coil

1. Nov 28, 2012

### F.ono

I've been doing some test on a rogowski coil that I built. It is like an air-cored current transformer.

I am using it to measure high frequency pulses. The picture shows a discharging capacitor being captured by a current probe (above) and by the rogowski coil (below).
I don't know what is causing the inverted voltage peak. Would someone be able to explain this?
The picture below is the equivalent circuit of the rogowski coil. Vcoil is the voltage induced in the secondary winding and Vout is the actual voltage measured in its terminals.

2. Nov 28, 2012

### Carl Pugh

I would prefer for TEK #2 to be shown with the + and - reversed. (Upside down)

When di/dt is increasing, output is positive.

When di/dt is decreasing, output is negative.

The average voltage detected in the rogowski coil has to be zero, since the magnetic flux starts at zero and returns to zero after the pulse is over.

3. Nov 29, 2012

### Staff: Mentor

You'll find my answer very similar to that of Carl Pugh. You'd like the coil to act as a current sensor, but transformers don't work on current, their secondary voltage is determined by the primary current's rate of change. So when the current is at first rapidly changing, your coil probe returns a high voltage. When the capacitor current is momentarily steady at its peak value, di/dt = 0 and you can see the sensor's voltage passes through zero at that moment.

4. Nov 29, 2012

### F.ono

Thank you for your answers! It really helped me understanding what is going on.
For high frequencies, the rogowski coil is supposed to be self-integrating, that is why I didn't even think that what I was getting was the di/dt wave. But this self-integration characteristic depends on the 2pifL and Z relationship, which in this case does not occur.

5. Nov 29, 2012

### F.ono

In order to understand why the rogowski coil is not working as a self-integrating coil (as it is supposed to in high frequencies), I found the article attached.
Section D in that article explains the self-integrating behavior. But it is necessary to understand Equation (1), which I am not really getting.
The author says that he is not considering the capacitance. Okay, then according to the coil model, we have an equivalent impedance in series with a inductor. Shouldn't the equation be simply

(1/R)*(dphi/dt) = Ic

What is the (L/R)*(dIc/dt) in his equation?

#### Attached Files:

• ###### explicação.pdf
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6. Nov 29, 2012

### F.ono

Another question: What is the fundamental frequency of the pulse? Do I just have to consider the total time that it took from start to end and calculate 1/T?

7. Dec 1, 2012

### Staff: Mentor

Perhaps they are referring to the case of a repetitive waveform?

8. Dec 1, 2012

### Staff: Mentor

If an equivalent circuit involves R and L, you should expect that both R and L will appear in the analysis.
Something's seriously wrong if L goes missing, no reason given.

Draw a closed series circuit, just an L and an R. (In practice, I'd say R would include both the resistance of the coil plus that of your electronics input.) Across that 2 element circuit is applied the emf induced by the changing field, we usually know this voltage as N·dɸ/dt but this being the case of a single turn, here N=1.

Note: at the expense of a 600k download, I generally expect to find a document much more legible that what you've provided.

9. Dec 2, 2012

### Enthalpy

The Rogowski coil is self-integrating over a (high) frequency range where the self-inductance of the toroid winding creates a reactance that clearly exceeds the resistance of the circuit, which includes the winding's losses and the load resistance.

In this case, idealized, the induced current is created by a mutual inductance and limited by the self-inductance, which means that they're in phase and no differentiation (by the resistance) occurs. The load resistance creates a small maesurement voltage proportional to the current but doesn't significantly limit the current.

Second equivalent way to understand in: the measurement winding is nearly short-circuited, because the loop's resistance is clearly lower than the reactance (the imaginary part of the circuit's impedance). This (multiturn) loop in short circuit maintains a constant flux through itself, by circulating in itself a current that compensates the flux created there by the measured current. The induced current follows the measured current instantly and is a fraction of it, because the winding has several turns.

Note1: the "flux" for the self-inductance is greater than what flows through the toroid's section. It includes the flux that passes between the turns of the winding, what we would call a leakage inductance in a transformer. Hence the current ratio exceeds the number of turns.

Note2: the winding's losses are far greater than the DC resistance of the winding.

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Some ways to extend to lower frequencies this mode of operation:

- Reduce strongly the load resistance. Amplify if needed. But the following third option is better.

- Use a magnetic core. Nearly required at 50µs.

- Load the coil with a transimpedance amplifier, instead of a resistor followed by a voltage amplifier. The load resistance is then zero, and only the coil's own losses create a resistance. Extremely fast transimpedance amplifier exist commercially for photodiode receivers (>50Gb/s!), and at 50µs you can build yours easily.

- Have two windings at the coil: one that senses the flux, the other that zeroes it through an amplified feedback. This compensates fully the winding's losses. The current ratio nearly equals the number of compensating turns now. The magnetic core may not be required then. Mind the parasitic capacitance between the two windings; they can have different radius, and an electrostatic shield between them, slit so it doesn't short-circuit the flux.

Keep an eye at the DC behaviour of the circuit you design, and to its stability if using a feedback. This is now more a current transformer than a Rogowski coil, though they differ merely by the frequency range.

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