Every large cardinal property admits a formalization with the desired property. That is, every large cardinal property $\text{LC}$ admits a ZFC-provably equivalent formulation $A$ for which $\newcommand{\ZFC}{\text{ZFC}}\ZFC^-+A$ is equiconsistent with ZFC.

Let's begin with a lemma. (See my paper, Nonlinearity and illfoundedness in the large cardinal hierarchy of consistency strength for more along these lines.)

**Lemma.** There is an arithmetic sentence $\sigma$ such that $\ZFC^-+\sigma$ is equiconsistent with $\ZFC$.

**Proof.** Let $\sigma$ assert, "for any proof of a contradiction in $\ZFC$, there is a smaller proof of $\neg\sigma$ in $\ZFC^-$." Assume $\text{Con}(\ZFC)$, and hence $\text{Con}(\ZFC^-)$. If $\text{Con}(\ZFC^-+\sigma)$ fails, then there is a proof of $\neg\sigma$ from $\ZFC^-$. Since $\text{Con}(\ZFC)$ holds, there is no smaller proof of a contradiction in $\ZFC$, and so we can also easily prove $\sigma$ is true, contrary to $\text{Con}(\ZFC^-)$. So $\text{Con}(\ZFC)$ implies $\text{Con}(\ZFC^-+\sigma)$. Conversely, if $\text{Con}(\ZFC)$ fails yet $\text{Con}(\ZFC^-+\sigma)$ holds, then there is a proof of a contradiction in $\ZFC$, but no smaller proof of $\neg\sigma$ in $\ZFC^-$, and from this we can prove $\neg\sigma$, contrary to assumption. $\Box$

**Theorem.** For every large cardinal property $\newcommand{\LC}{\text{LC}}$, there is a $\ZFC$-provably equivalent formulation $A$ such that $\ZFC^-+\exists \kappa A(\kappa)$ is equiconsistent with $\ZFC$.

**Proof.** Let $A(\kappa)$ assert "if the power set axiom holds, then $\LC(\kappa)$, and if it does not, then $\sigma$." In $\ZFC$, the power set axiom holds, and so this property is equivalent to $\LC(\kappa)$ in $\ZFC$. But if it fails, we are asserting $\sigma$. If $\text{Con}(\ZFC^-+\exists\kappa A(\kappa))$, then either we get $\text{Con}(\ZFC+\exists\kappa\LC(\kappa))$ or we get $\text{Con}(\ZFC^-+\sigma)$, both of which imply $\text{Con}(\ZFC)$. Conversely, if $\text{Con}(\ZFC)$, then $\text{Con}(\ZFC^-+\sigma)$ and hence by going down to the hereditarily countable sets, we get $\text{Con}(\ZFC^-+\exists \kappa A(\kappa)$. $\Box$

I don't really work in CZF, but I expect that this kind of trick is fairly general.