Yes, it’s possible. The formula for the number of degrees of freedom is
$\frac{\left( {s_1^2 \over n_1} + {s_2^2 \over n_2} \right)^2 }
{ \frac{s_1^4}{n_1^2 (n_1-1)} + \frac{s_2^4}{n_2^2 (n_2-1) }}$
where $n_i$ is the number of observations in the $i$th sample and $s_i$ is the standard deviation in this sample. If $s_1$ happens to be equal to $s_2$ and $n_1=n_2=n$, this reduces to $2(n-1)=2n-2$, i.e. the same number of degrees of freedom you would have with an equal variance t-test.
For your example $n=11$, so you would get 20 degrees of freedom, similar to your 19 degrees. So I would guess that your two standard deviations are very similar.
Note that for the equal sample size case, $2n-2$ is the largest number of degrees of freedom you can get. And it’s also easy to show that the lower bound on the number of degrees freedom you can get is $n-1$, which is what you would get if the sample standard deviation in one of the samples is very much larger than the sample standard deviation in the other sample ($n-1$ is the limit as $s_1/s_2$ tends to infinity).