Q:

Can the Cramér-Rao Lower Bound be constant?

In many text books and research papers, I have seen that for the plot of Cramér-Rao Lower Bound vs Signal to Noise Ratio, the CRLB decreases with increasing SNR.

Is it possible that the CRLB remains constant for all SNR?
If so, what is the implication?

A:

Prologue
The CRLB of an unbiased estimator $\hat \theta$ for the paramater $\theta$ is 
$$
\text{var}(\hat \theta) \ge - \frac1 {\text E_x\left\{ \frac{\partial^2 (\ln f(x;\theta))}{\partial \theta^2} \right\}}
$$
with $f(x;\theta)$ being the likelihood function of $\theta$ under the observation $x$. It is the inverse of the so-called Fisher information.
Answer

Is it possible that the CRLB remains constant for all SNR?

Yes. For that, the expectation (over all $x$, hence the index)
$$\text E_x\left\{ \frac{\partial^2 (\ln f(x;\theta))}{\partial \theta^2} \right\} $$
just needs to be constant.
The "easiest" explanation would be if noise just didn't have anything to do with the likelihood – ie. noise is just not actually observed by the estimator.
But that's kind of least exciting result – "well, if I add a lot of noise to something and then observe something completely unrelated, the noise has no effect" isn't that surprising.
A lot more complicated, is the option that the Fisher information is actually constant, although $f(x;\theta)$ actually depends on the noise.
The expectation operator $E_x$ is definable as
$$
\text E_x \{g(x)\} = \frac1{\mu (\xi)} \int\limits_{x\in\xi} g(x)\,d\mu\text,
$$
i.e. as the integral over all possible $x$ divided by the measure $\mu$ of the set $\xi$ from which $x$ can be chosen.
Assuming smoothness of $x$ and compactness of $\xi\subset \mathbb R^1$ (just to make all the derivates easier to handle and to avoid cascading integrals), this specializes to:
$$\begin{align}
\text E_x \{g(x)\} &= \frac1{x_{max}-x_{min}} \int\limits_{x = x_{min}}^{x_{max}} g(x)\,dx\\
\text E_x\left\{ \frac{\partial^2}{\partial \theta^2}{ (\ln f(x;\theta))} \right\}  \cdot\left({x_{max}-x_{min}}\right) &= \int\limits_{x = x_{min}}^{x_{max}} \frac{\partial^2}{\partial \theta^2}{ (\ln f(x;\theta))} \,dx\\
&= \left[ \frac{\partial}{\partial \theta}{ (\ln f(x;\theta))}  \right]_{x = x_{min}}^{x_{max}}\\
&= \left[ \frac1{f(x;\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}  \right]_{x = x_{min}}^{x_{max}}\\
&= \frac1{f(x_{max};\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}_{x=x_{max}}
- \frac1{f(x_{min};\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}_{x=x_{min}}\\
&= c,&c\text{ const.}\\
\end{align}$$

