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The voltage across the photodiode is the same as the voltage of the power supply (they are in parallel). Put voltage of the power supply into the Shockley Diode equation to calculate the current through the ammeter if the current generated by exposure to light is $I_\mathrm{L}$, $$ I=I_\mathrm{S} \left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right) - ...



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