Quality factor limit for 2nd order RC passive band-pass filter
It is possible to prove mathematically that the Q factor of this band-pass RC circuit is limited to: $$ 0&lt;Q&lt;0.5 $$ 



First, we need to remember that the standard form of the transfer function for a second-order band-pass filter is: 
$$H(s)=\frac{K(2ζω_n)s}{(s^2+2ζω_ns+ω_n^2 )}$$
Where:


\$K\$ is the maximum gain of the filter
\$ζ\$ is the damping ratio
\$ω_n\$ is the undamped natural frequency


And the transfer function based on component values is:
$$H(s)=\frac{sC_1 R_1}{C_1 C_2 R_1 R_2 \cdot s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1}$$



To prove that 0 &lt; Q &lt; 0.5, I will start by setting up an equation that, when solved, give the poles (roots of the denominator) of the above transfer function:
$$C_1 C_2 R_1 R_2 s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1=0$$

Note that this expression is a quadratic equation in the form \$ as^2+bs+c=0\$. Its discriminant is given by \$Δ=b^2-4ac\$. So, we have:

$$Δ=(C_1 R_1+C_2 R_1+C_2 R_2)^2-4⋅C_1 C_2 R_1 R_2$$
Assuming that \$ C_1&gt;0\$, \$ C_2&gt;0\$, \$R_1&gt;0\$ and \$R_2&gt;0\$, it follows that \$Δ&gt;0\$ for any choice of components.

Now, doing the same process but using the standard form this time, we have:
$$s^2+2ζω_n s+ω_n^2=0$$
$$Δ=(2ζω_n )^2-4ω_n^2=4ζ^2 ω_n^2-4ω_n^2=4ω_n^2⋅(ζ^2-1)$$
But we already know that \$Δ&gt;0\$. So, using the standard form:

$$4ω_n^2⋅(ζ^2-1)&gt;0$$

The term \$4ω_n^2\$ is surely positive. Then, to satisfy the above inequality, its necessary that 
$$(ζ^2-1)&gt;0$$
Before proceeding, it's important to remember one more thing: 
$$Q≜\frac{1}{2ζ} \implies ζ=\frac{1}{2Q}$$
Then, by making the substitution, we have:

$$\left(\frac{1}{2Q}\right)^2-1&gt;0 \implies \left|\frac{1}{2Q}\right|&gt;1$$ 
Analyzing the expression of Q factor for the filter, we have, doing the same assumptions made above with \$Δ\$:
$$Q=\frac{\sqrt{C_1C_2R_1R_2}}{C_1R_1+C_2R_1+C_2R_2}&gt;0$$
Thus, we can finally conclude that:
$$\frac{1}{2Q}&gt;1 \implies 0&lt;Q&lt;0.5$$