I am working with a silicon PIN-type photodiode in photovoltaic mode, conducting measurements of open-circuit voltage and short-circuit current for varying intensities of incident light. The light is being measured as a proxy for another quantity, and so I must apologise for I can give no quantitative values of optical power. The photodiode has a claimed max “Operating Current” of 10mA, and an active area of 20mm^2. The measuring apparatus is a fluke 187 digital multimeter, which can measure to precisions of tens of microvolts and tens of nanoamps, though in the setup there is some background noise (probably electrical) that renders the last digit unreliable. Attempts have also been made to measure current with a moving coil microammeter, which has a resistance of 150Ω. As I understand it, with increasing optical power incident on the detector, the short-circuit current should increase linearly and be equal to the photocurrent. The open-circuit voltage should increase logarithmically, but for low optical powers, the increase is approximately linear*. *The open circuit voltage is given by the equation V = KT/q ln[(I/Io +1)], which for values of I/Io<<1 the Mercator series can be used to yield V=KT/q I/Io Open-circuit voltages have been detected at both at ultra-low light levels and in ambient environmental lighting. The open-circuit voltage s in these scenarios has ranged from 40μV to 350mV. Linearity is confirmed up until about 20mV. However, when the short-circuit current is measured, nothing is seen until the open-circuit voltage for the same light intensity exceeds 250mV. At this light intensity, the current is 1.4μA. My question: is the current just that small? Other measurements (albeit for p-n junctions) I have seen reported in the literature record that light intensities that lead to 1mV open-circuit voltages lead to at least 1μA short-circuit currents. Certainly, one would have thought that a current much closer to 10mA would have been seen in ambient environmental lighting. There is a second question I wished to ask, and it relates more to the theory. I understand that when p-type semiconductor is connected with n-type semiconductor, the majority carrier in each type diffuses along the concentration gradient. This leave behind ions, trapped in the crystal lattice, which generate an electrical field that opposes further diffusion. Forgive my ignorance, but why can this “barrier potential” not be read by a voltmeter placed in parallel around the junction? Greatly obliged for your assistance.
The fluke multimeter has some internal resistance. I don't know how much (and it depends on the measurement range: smaller ranges correspond to larger internal resistances), but that could reduce the current significantly. Unrelated to that, at the level of millivolts you can have temperature effects that lead to additional voltage differences. As far as I know, 1.4µA is not at the lower end of the fluke precision - do you see a sudden jump in the current? Current flows through the voltmeter (and the timescale is way too short to see that) until the potential is equal at both sides. That is true for every setup without a power supply.
Yes, that actually sounds quite reasonable. The key to understanding this is to realize just how small [itex]I_0[/itex] is. It's value is very temperature dependent (due to its dependance on [itex]N_i^2[/itex]), but even at I=1.4uA the ratio [itex]I/I_0[/itex] is likely to be order of around magnitude [itex]10^5[/itex] or so. Put this value into the equation for voltage and you'll find that it does indeed correspond to somewhere around 250 mV.
Update: I tested that with a Fluke289, and at 1µA, the voltage is well below 1mV (measured: 0.15mV, but that is not reliable*). That is not an issue if the multimeters are not completely different. *Edit: A more careful measurement showed 0.10mV.