How can value of acid dissociation constant is vary with concentration of acid? First of all I have learned that $K_\mathrm{a} = \frac{[\ce{H+}] [\ce{A-}]} {[\ce{HA}]}$ . Also I have been informed that $K_\mathrm{a}$ value is independent of concentration. 

So let us take an example. Suppose 1 mole of acid $\ce{HA}$ dissociates to form 0.9 moles of $\ce{H+}$ ions and 0.9 mole of $\ce{A-}$ ions and 0.1 mole of undissociated $\ce{HA}$. So, the $K_\mathrm{a}$ value will be $K_\mathrm{a} = \frac{[0.9] [0.9]} {[0.1]}=8.1$.

But if we take 2 moles of same acid then, 1.8 moles of H+ ions and 1.8 moles of A- ions will be formed leaving 0.2 moles of undissociated HA. This time $K_\mathrm{a} = \frac{[1.8] [1.8]} {[0.2]}=16.2$. 

So we can see that $K_\mathrm{a}$ has increased with increase in concentration. Can you tell me where my assumption is wrong?
Before we begin, note that $K_a$ is defined in terms of concentration and not amount. Furthermore, $K_a$ is a constant so it must be the same provided the conditions, other than concentration, are the same.

The nature of equilibrium is such that it does not always give the same absolute reaction (i.e. a fixed percentage of reaction, or the reaction does not proceed to the same "position" each time). This is unlike a fixed forward reaction which always ends at 90% completion. In other words, although the 1 molar solution of HA gave 0.9M of $H^+$ and $A^-$, the 2 molar solution would only give out...

$$ K_a = 8.1 = x^2/(2-x) $$

where $x$ is the number of moles of $H^+$ produced. When you solve this, you would get $x=1.66M$