Do you need to measure the antenna current of your HF transmitter? Or, like me, the primary current of your Tesla Coil? Here are some design guidelines and a real example.

There are many cases in which a current measurement that does not include DC is useful. The most common is the case of CTs (Current Transformers) for AC mains. Here I will discuss design of CTs for medium-high frequency, which are really straightforward to build. The formulas presented are valid also for AC mains units.

Description of probe
The probe in photo is designed to measure up to 50 A peak, from 7 kHz to several tens of MHz.
The schematic is in the attached Figure 1 and is quite simple: the wire whose current must be measured shall be passed through the toroid, which in my case is a common Amidon core FT 82-43, that has good performance up to at least 50 MHz.
The secondary consists in ten turns of wire evenly distributed along the core. If available, use a medium thickness Litz wire, but this is not mandatory. Due to the 1:10 turns ratio, the maximum current in the secondary is 5 A peak.
The secondary is loaded by 0.2 ohm, realized with a parallel of 5x 1 ohm resistors. When subject to 5 A peak, voltage across them is 1 V peak, a value very practical for scope measurement. At full-scale current, assuming sinusoidal current, dissipation on resistors is R*I^2/2 = 2.5 W or 0.5 W each, so continuous sinusoidal 50 A peak can be measured only with 0.5 W or bigger resistors. But, if the waveforms are pulsed or short measurements are taken, 1/4 W parts are ok. This was my choice, because I preferred to keep the design compact in order to have better RF behavior. OK, I must also confess that I had those resistors at hand...

Figure 2 shows a typical usage with a scope probe, using a BNC adapter for scopes. The device can also be used with a direct coaxial cable connection to the scope input, because 1 V peak is ideal for scope 1x operation: in this case, it is recommended to use a short cable to avoid reflections in the band of interest, because the coax will be mismatched on both sides.
What must be avoided is attaching the scope probe directly to resistors using the clips and skipping the BNC connector, because when measuring high RF currents, even the minimum unshielded loop in the probes will add artifacts to the measurements.
If the probe is 1x, you'll get a reading of 50 A/V, so for example 5 A/div when the scope is set at 100 mV/div.

A CT is not complex to design, but some formulas from electromagnetism are needed.
The design starts from the load resistor RL, that should be as small as practically possible in order to minimize the power loss introduced, because the circuit under measurement will "see" at least R/n^2, where n is the turns ratio and R is the sum of RL and of the secondary wire resistance. As already said, it is very important that the secondary is evenly wound, because otherwise the circuit tested we'll see also some stray inductance in series. However, if we choose too low a value for RL, we'll also have a very small voltage to measure, causing noise on the traces. Finally, we'll want RL at least equal or greater than the secondary resistance (but this is not mandatory).
In my case, I chose 0.2 ohm, in order to have 1 V with 5 A (50 A on the primary) and add at most 2 mohm to the circuit under test.
n, the number of secondary turns, determines the current ratio. In the case of a high frequency transformer, this number must be kept low to watch out from self-resonances caused by stray capacitance together with high inductance. In the case of mains CTs, the frequency is quite low (50-60 Hz) and n=1000 is a common value. Powers of 10 are commonly used, so also the current conversion ratio is immediate, but other values are possible.
The highest usable frequency for a toroidal ferrite CT depends on:

  • performance of ferrite material
  • self-resonance of secondary winding
  • resistor inductance (resistor itself and connections)

A design like mine can easily work up to several tens of MHz if a suitable ferrite, like material 43 from Amidon, is used. High permeability cores used for EMI suppression can be used as well, but only up to much lower frequencies. Low permeability cores used for power chokes and high-Q inductors are not recommended, because their inductance per turn is too low (impacts following point).

The choice of ferrite core also affects the lower usable frequency, for two reasons:
  1. in order to get good measurements, we'll want that the reactance of the secondary is much greater than RL. Since reactance is X=2π*f*L (f frequency in Hz, L inductance in H), this affects the lower frequency end. In practice, we'll want X > 10*RL at the lower end.
  2. we must avoid saturation of the core, that happens at high current and low frequency. Saturation of ferrites happen for field B=0.25-0.3 T, but in order to avoid nonlinear phenomena, we must stay below 0.2 T at the maximum current and minimum frequency.

For both reasons above, low usable frequencies with a small number of turns and high current tend to require large cores. The key core parameters are AL (the inductance per turn, generally -but not always- expressed in nH/turn for RF cores) and the core cross section S, that can be easily computed from core dimensions.
In case of core  Amidon FT 82-43, AL=470 nH/turn and the cross section is 24.6 mm^2 (easy to compute from outer diameter, inner diameter and height).

Inductance of 10 turns is 470*10^2=47000 nH=47 µH (remember inductance depends on n^2). If we accept that the lower cutoff frequency is for X=10*RL=2 ohm, by reversing the reactance formula we get f>6.8 kHz. This may be acceptable or not according to the application and a larger core, having a greater AL, will have a lower cutoff. In my case, for measuring a musical Tesla Coil, I was interested on frequencies >500 kHz, so the core was fine.

Regarding saturation, we have that in sinusoidal regime the magnitude of field B (magnetic flux density) is abs(B)=V/(n*2π*f*S), where V is the peak voltage across the winding. This expression can be derived by equaling the two expressions of magnetic linked flux Φc = L*i = n*B*S and replacing i with the absolute value of the AC current abs(i)=V/(2π*f*L). If V and n are fixed in previous design phases, the only way of reaching a lower f is adopting a larger S, namely a bigger toroid again.
In my case V=1 V, n=10, S=24.6E-6 m^2 (pay attention to use correct units), so we have B<0.2 T for f>3.2 kHz, which is fine again, at least for my application.
It may be confusing that current doesn't appear in the calculation: actually, it is hidden behind V, because we know that 1 V corresponds to 50 A.

I have presented an example of RF current transformer and explained the key design issues.
I hope that the design criteria and formulas presented may be useful to some readers. In case of doubts or if you think there is some mistake, please contact me.