Arithmetic circuits with just one threshold gate
The answer is “yes” if $t=n^{O(1)}$. More generally, a threshold $\{+,\cdot\}$-circuit of size $s$ with threshold $t$ can be simulated by a $\{\lor,\land\}$-circuit of size $O(t^2s)$.

First, observe that it is enough to evaluate the circuit in $\{0,\dots,t\}$ with truncated addition and multiplication: in particular, if $a,a'\ge t$, then $a+b,a'+b\ge t$, and either $ab,a'b\ge t$ as well, or $ab=a'b(=0)$.

With this in mind, we can simulate the circuit with a Boolean monotone circuit by replacing each node $c$ with nodes $c_0,\dots,c_t$, where $c_i$ is intended to compute the predicate $c\ge i$. (We need $c_0$ only for notational convenience, it computes the constant $1$ function.) If $c$ is a Boolean input variable $x$, we take $c_1=x$, $c_2=\dots=c_t=0$. If $c$ is an addition gate, say $c=a+b$, we implement it via
$$c_i=\bigvee_{\substack{j,k\le t\\j+k\ge i}}(a_j\land b_k).$$
Multiplication gates are handled in the same way.

This takes $O(t^3)$ gates per one gate of the original circuit. As a minor optimization, we can reduce it to $O(t^2)$ by putting
\begin{align*}
c_t&amp;=\bigvee_{j+k\ge t}(a_j\land b_k),\\
c_i&amp;=c_{i+1}\lor\bigvee_{j+k=i}(a_j\land b_k),\qquad i&lt;t,
\end{align*}
so that each $a_j\land b_k$ is used as an input of only one of the $c_i$ gates.