
#1
Jan2614, 11:56 AM

P: 2

I am working with a silicon PINtype photodiode in photovoltaic mode, conducting measurements of opencircuit voltage and shortcircuit 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 shortcircuit current should increase linearly and be equal to the photocurrent. The opencircuit 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 Opencircuit voltages have been detected at both at ultralow light levels and in ambient environmental lighting. The opencircuit voltage s in these scenarios has ranged from 40μV to 350mV. Linearity is confirmed up until about 20mV. However, when the shortcircuit current is measured, nothing is seen until the opencircuit 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 pn junctions) I have seen reported in the literature record that light intensities that lead to 1mV opencircuit voltages lead to at least 1μA shortcircuit 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 ptype semiconductor is connected with ntype 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. 



#2
Jan2614, 12:24 PM

Mentor
P: 10,813

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? 



#3
Jan2714, 06:55 AM

Sci Advisor
P: 2,751

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. 



#4
Jan2914, 07:52 AM

Mentor
P: 10,813

PIN Photodiode, High Voc, Low Isc*Edit: A more careful measurement showed 0.10mV. 


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