How do I design a current transformer?
Definitions:

\$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 &gt;&gt; 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.