Madeline Horn edited Once_we_had_taken_values__.tex  over 8 years ago

Commit id: 107b1ce32c60365022cfde23cff9c0383c8e63b8

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Once we had taken values across the photo-diode from 0 to 120 mV in steps of 10 mV, had our error in all the values, we were able to start calculating the current from both mulitmeters. In order to find the current from the voltage across the photo-diode, we only had to use Ohm's Law (V = IR) beause we knew the value of R, 10K ohms. Finding the current for Vsq was harder because we had to account for the change through the amplifiers and multiplier. The equation we used to find the filtered current was: $10*  \textrm{ Vsq} = 2 \textrm{ e} \textrm{ idc} \Delta f (\textrm{ G1} \textrm{ G2} \textrm{ Resistance})^2$, where e is the charge of the electron that we are looking for, R is the resistance of 10K ohms, and G2 varied