Madeline Horn edited Table_ref_table_Johnson1_shows__.tex  over 8 years ago

Commit id: 537c2c0c14fbcacd1cb7dfdaf79dea3581979169

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$V_{\textrm{meter}} = (^2 + ^2) G_1 G_2 / (10 \textrm{Volts}) \textrm{ [mV]}$  where the factor of $10 \textrm{Volts}$ comes from the amplifier and the $^2 + ^2$ is the measured voltage from the multi-mete before the error has been subtracted.  --- \textbf{are these really Johnson noise voltages? I suspect from the rest of your paper that these might actually values for $^2 + ^2$ in units of Volts$^2$. Obviously, this is an important distinction, and not just a matter of being picky, b/c these are very different things physically and numerically would lead to very different results for your calculations!} ---- from a resistance value of $10 K \Omega$, a temperature of $295$ Kelvin, a G1 of $X600$, a G2 of $X1000$, and we varied the $f1$ and $f2$ in order to change the bandwidth.