Assuming your samples are independent, then Welch's t-test does seem to be appropriate here, since it appears you have unequal variances (but you can formally test this too if you want through Levene's Test for Equality of Variances).
That being said, since you have quite large samples from both device 1 and device 2, then you can appeal to the central limit theorem and use:
\begin{eqnarray*}
Z & = & \frac{\bar{X}-\bar{Y}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\sim N(0,1)\\
\end{eqnarray*}
under the null hypothesis of equal means. Here, $\bar{X}$ and $\bar{Y}$ and sample means from device 1 and device 2, respectively and $s_i^2$ and $n_i$ are the sample variance and sample sizes from the ith device $i=1,2$. Note that in large sample inference, you don't need to concern yourself with unequal variances.
Then a 95% confidence interval for your estimate would be given by:
\begin{eqnarray*}
\bar{X}-\bar{Y} & \pm & Z_{\alpha/2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}
\end{eqnarray*}
where $Z_{\alpha/2}$ is the upper $\alpha/2$ point of the standard normal distribution.
All this being said, I wholeheartedly agree with the answer provided by Stefan. These sample sizes are really large and he's provided sound advice that you should follow. You should focus on what is an important practical difference. Is a 0.0001 mean difference between device 1 and device 2 important to you? Is it still important if device 1 costs three times as much as device 2?