Something between the Chernoff and Hoeffding bounds Suppose I have $n$ independent 0-1 random variables $X_1, \cdots, X_n$ and I want to show a concentration of $X = \sum_i X_i$.
I can use either the Chernoff bound or the Hoeffding bound.
Suppose $E[X] = O(1)$. Then, I should use the Chernoff bound which will give me Poissonian tails. On the other hand, if $E[X] = O(n)$ then using the Hoeffding bound will give me a better result -- namely Gaussian tails. 
However, what if, say, $E[X] = \sqrt{n}$. It is hard to believe that Chernoff is the best I can do. Is there some better inequality for this case? And what if $E[X] = n/\log n$ or anything else between $O(1)$ and $O(n)$?

Edit in response to Kodlu's answer: Using Chernoff, one gets that there is subgasussian concentration at an interval around the mean that gets bigger with increasing $\mu$. However, outside of this interval ($\delta > 1$), this still gives us only Poissonian tails. Now, my conjecture is that one can actually get better bounds.
 A: $\newcommand\ep{\varepsilon}$ $\newcommand\si{\sigma}$ $\newcommand\Ga{\Gamma}$ $\newcommand\tPi{\tilde\Pi}$ 
It follows from Theorem 2.1 of this paper or of its better version  that for a large class, say $\mathcal F$, of nondecreasing functions $f$, containing the class of all increasing exponential functions, we have 
$$Ef(X-EX)\le Ef(Y),\tag{1}$$
where 
$$Y:=\Ga_{(1-\ep)\si^2}+y\tPi_{\ep\si^2/y^2},$$
$\Ga_{a^2}\sim N(0,a^2)$, $\tPi_\theta$ has the centered Poisson distribution with parameter $\theta$, $\Ga_{(1-\ep)\si^2}$ and $\tPi_{\ep\si^2}$ are independent, 
$$y:=\max_i q_i,\quad \si^2:=\sum_i p_i q_i,\quad\ep:=\sum_i p_i q_i^3/(\si^2 y)\in[0,1],$$
$p_i:=P(X_i=1)$, and $q_i:=1-p_i$. 
We see that $\ep\in[0,1]$ "interpolates" between the Gaussian and (re-scaled centered) Poisson random variables (r.v.'s) $\Ga_{\si^2}$ and $y\tPi_{\si^2/y^2}$.
From here, one can immediately get exponential bounds on the tails of $X$, or one can get better bounds such as 
$$P(X-EX\ge x)\le\frac{2e^3}9\,P^{LC}(Y\ge x)$$
for all real $x$, where $P^{LC}(Y\ge\cdot)$ denotes the least log-concave majorant of the tail function $P^{LC}(Y\ge\cdot)$; see Corollary 2.2 and Corollary 2.7 in the linked papers, respectively. 
To see better how this works, consider the iid case, with $p_i=p$ and hence $q_i=q=1-p$ for all $i$. Then $y=q=\ep$ and 
(i) if $p$ is small then $\ep=q$ is close to $1$ (and $y=q$ is also close to $1$) and hence $Y$ is close to the centered Poisson r.v. $\tPi_{\si^2}$;
(ii) if $q$ is small then $\ep=q$ is small and hence $Y$ is close to the Gaussian r.v. $\Ga_{\si^2}$;  
(iii) if neither $p$ nor $q$ is small but $n$ is large then $\ep=q$ is not small and $\si^2$ is large, and hence $\tPi_{\ep\si^2/y^2}$ is close to $\Ga_{\ep\si^2/y^2}$ in distribution, so that 
$Y$ is close in distribution to the Gaussian r.v. $\Ga_{\si^2}$, just as in Case (ii). 

For $f(x)\equiv e^{tx}$ with real $t\ge0$, (1) becomes 
$$E\exp\{t(X-EX)\}
\le\exp\Big\{\frac{t^2}2\si^2(1-\ep)+\frac{e^{ty}-1-ty}{y^2}\,\si^2\ep\Big\}\tag{2};$$
cf. e.g. formula (1.5) in the better version  of the linked paper, which implies 
$$P(X-EX\ge x)
\le\inf_{t\ge0}\exp\Big\{-tx+\frac{t^2}2\si^2(1-\ep)+\frac{e^{ty}-1-ty}{y^2}\,\si^2\ep\Big\}$$
for real $x\ge0$. 
The latter $\inf$ can be explicitly expressed in terms of Lambert’s product-log function -- see the expression in formula (3.2) in the same paper; another useful expression for the same $\inf$ is given by formula (A.3) in this other paper. 
A: I am not an expert, bu I know that Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have
$$
\mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta},
$$
and
$$
\mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad 
$$
Similarly for any positive 
$\delta,$ we have
$$
\mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu,
$$
and
$$
\mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu.
$$
For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have
$$
\mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3]
$$
and
$$
\mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2]
$$
As an example, if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail)
$$
\mathbb{P}[|X- \mu|\geq \delta \mu]\leq 2\exp[-\delta^2 \mu/3],
$$
or equivalently letting $x=\delta \mu,$
$$
\mathbb{P}[|X- \mu|\geq x]\leq 2\exp[-(x^2/\mu^2) \mu/3]=2\exp[-x^2/3\mu]=
2\exp\left[\frac{-x^2}{3\sqrt{n}}\right].
$$
So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$
