Bell states produce maximal entanglement between two qubits. On the other hand, two unentangled qubits provide no (i.e. minimal) entanglement at all.
However, I haven't seen any example of a weak entanglement and how it can be prepared.
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Sign up to join this communityThe state $|\psi\rangle = \dfrac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}} $ is not separable/untangled since a pure state $|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ is unentangled if and only if $ad - bc = 0$. But it is not a maximal entangled state either as @AdamZalcman pointed out in the comment as the Schmidt decomposition of this state do not have two equal Schmidt coefficient of $1/\sqrt{2}$. In fact, here the Schmidt decomposition of this state $|\psi\rangle = \dfrac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}} $ is:
$$|\psi\rangle = \sqrt{\dfrac{3 + \sqrt{5}}{6} }|u_1\rangle|v_1\rangle + \sqrt{\dfrac{3 - \sqrt{5}}{6} }|u_2\rangle|v_2\rangle $$
where $|u_1 \rangle , |u_2\rangle $ are the eigenvectors of the reduced density operator $\rho^A = \dfrac{1}{3}\begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix} $ correspond to the eigenvalues $\dfrac{3 + \sqrt{5}}{6} $ and $\dfrac{3 - \sqrt{5}}{6}$ respectively. Similarly, $|v_1 \rangle , |v_2\rangle $ are the eigenvectors of the reduced density operator $\rho^B = \dfrac{1}{3}\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix} $ correspond to the eigenvalues $\dfrac{3 + \sqrt{5}}{6} $ and $\dfrac{3 - \sqrt{5}}{6}$ respectively. Therefore, it is not a maximal entangled state.
You can prepare such state using this circuit in Quirk, which was also shown in Martin Vesely's answer here.
Another standard example are Werner states. These can be written as $$W_p\equiv p \frac{I}{4} + (1-p) \frac{|\Psi^-\rangle\!\langle\Psi^-|}{2},\qquad p\in[0,1],$$ and you can tune the entanglement by changing $p$. You can prove that the $W_p$ is entangled for $p<2/3$, and you go from a maximally entangled state at $p=0$ to a totally mixed state at $p=1$.