= Photodiode operating mode
Photovoltaic mode uses no external reverse bias: a <transimpedance amplifier> can hold zero terminal <electric potential difference>, while an open <electrical circuit> develops a photovoltage. Photoconductive mode biases the <diode cathode> positive relative to the <diode anode>, generally reducing junction <capacitance> and increasing speed and <dark current>.<Si photodiodes by Hamamatsu Photonics K.K.>[Si photodiodes, pp. 2–3, §2-1; pp. 6–7, §§2-4 and 2-6; p. 9, §3-1: zero-voltage current, reverse operation and amplifier connection]{ref}
A simple calculation separates the readout from the conversion. If a linear <transimpedance amplifier> has feedback <electrical resistance> $R_f=1.00\,\mathrm{M\Omega}$, the $650\,\mathrm{nm}$ model above produces an output change of magnitude $R_f I_{\rm ph}=0.393195\,\mathrm V$. The output sign depends on the connection. Increasing $R_f$ increases that <electric potential difference>, but does not create more collected <charge carriers>.
For a reverse-biased <photodiode> feeding an <electrical load> $R_L$, <Ohm's law> gives a readout magnitude $V_L\simeq I_{\rm ph}R_L$ while the response remains linear. With negligible <dark current> and other series drops, the remaining reverse bias is $V_R\simeq V_b-I_{\rm ph}R_L$, where $V_b$ is the supply magnitude. Thus the chosen <electrical load> can consume the available bias: a larger readout does not automatically reduce <measurement uncertainty>. For the same constructed example, $V_b=5\,\mathrm V$ and $R_L=1\,\mathrm{M\Omega}$ leave about $4.60680\,\mathrm V$; this algebra alone does not establish a real device's linear range.
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