I want to make a Current transformer using toroidal ferrites to measure 220/50-60Hz AC current(40A max). Is it possible? And how can I calculate secondary turns count? Ferrite properties: H = 2000, Dimensions: 45*28*12mm.

I'm going to connect a current transformer to a mcu ADE7755


2 Answers 2



\$A_L\$: The \$A_L\$ factor of your core.
\$N_p\$: Primary turns. (Normally 1.)
\$N_s\$: Secondary turns. (Normally >50.)
\$L_p\$: Primary inductance.
\$I_p\$: Peak primary current.
\$\ell_c\$: Effective magnetic path length of the core.
\$A_c\$: Effective core cross sectional area.
\$\mu_c\$: Absolute permeability of the core (not relative!).
\$B\$: Magnetic field density.
\$R_b\$: Burden resistor.
\$R_r\$: Reflected burden resistor. (The burden resistor seen from the primary side.)
\$X_p\$: Primary side inductive reactance.
\$f\$: Working frequency.

The primary side inductance will be

$$ L_p = N_p^2A_L = \dfrac{N_p^2 \mu_c A_c}{\ell_c}. $$

The primary side reactance will be

$$ X_p = 2 \pi f L_p. $$

The reflected burden resistor will be

$$ R_r = \left( \dfrac{N_p}{N_s} \right)^2 R_b. $$

In order your current transformer to work, these two conditions must be met:

  • The reflected burden resistor must be much lower than the primary side inductive reactance; that is \$X_p >> R_r\$. The lower it, you will get more precision. If it is not low enough, you won't get signal on the burden resistor. Notice the trade off! If you make enough number of secondary turns, that will solve everything, but wire itself will behave as a burden and spoil the precision; also you will need a larger core. If you connect a very small burden resistor, again, you will lose precision because small resistor will have small voltage on it, and it is not easy to measure small voltages.
  • The core must not saturate. (\$B=\dfrac{L_pI_p}{N_pA_c}=\dfrac{N_p \mu_c I_p}{\ell_c}\$ must be smaller than the saturation level of your core. Usually 200mT. See the datasheet.)

Make your design according to these constrains.

  • \$\begingroup\$ I think that the core saturation criteria may need to take the burden into the account. The magnetic field density B is correct for an inductor (i.e. not loaded transformer). When, however, the secondary winding is loaded with (presumably low resistor in metering applications) I believe the actual primary saturation current is higher. Perhaps someone in the know could add the math into the model. \$\endgroup\$ Commented Dec 26, 2021 at 20:13

I had same problem few weeks ago. I did it experimentally for some reasons (cheap cores I used had unknown parameters). I don't recommend this method if you need good acurracy and/or if you know core parameters.

However for many applications "simplified" design process like this may be enough. Actually sometimes it may be better than calculating everything, because some core datasheets may be inacurrate.

So I just made CT with 200 turns, passed 1A current and measured current on secondary.

Then I created another transformer with more turns (added turns proportionally to get desired secondary current at 1A on primary).

After all this I had to make sure that core is not saturating - I was increasing current until secondary current was not rising proportionally to primary current. In my case - that saturation current was much higher than I needed. If it was too low - that would mean I need bigger core.


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