The device seems to have a capacity of some kind. It could be electronic or perhaps ionic. Try different scan speeds! You can calculate a capacity from such a forward and reverse curve by splitting the current into a "resistive" current and a "capacitive" current. Capacitor equation: I = C * dU/dt. If you perform two scans in opposite directions and look at a specific voltage U, the "resistive" current only depends on U and not on the scan rate. The "capacitive" current depends on scan rate and scan direction. At a given voltage U, the capacity would be (I1(U) - I2(U))/([dU1/dt](U) - [dU2/dt](U)): The current difference between two passes through the same voltage at different scan rate only comes from the capacitive contribution, and you can get the capacity by dividing the difference in current by the difference in scan rate, for example forward and backward scan. Such measurements are typical for electrochemistry and are known under the term "Cyclic voltammetry". Electrochemistry and semiconductor physics are virtually the same!
The ZnO is n-type? The curve looks like two antiparallel Schottky diodes. Au is maybe not the ideal contact metal in that case, try Ti or In. Or did you use an adhesion layer before depositing the gold? Are you measuring in four-point geometry? The contact resistance can be large compared to the resistance of the rod itself.
It is rectifying on both ends; you are putting low work function metal on an n-type oxide. If you want to see larger currents, put Al or Ag on both ends. If you want to see a diode behavior, put Au on one and Al or Ag on the other.
Did you reverse scan right after forword scan? Maybe you can reverse scan later,like a few minutes after forward scanning, and make sure scan under dark. And then to see if you still get this hysteretic thing.
The device seems to have a capacity of some kind. It could be electronic or perhaps ionic. Try different scan speeds! You can calculate a capacity from such a forward and reverse curve by splitting the current into a "resistive" current and a "capacitive" current. Capacitor equation: I = C * dU/dt. If you perform two scans in opposite directions and look at a specific voltage U, the "resistive" current only depends on U and not on the scan rate. The "capacitive" current depends on scan rate and scan direction. At a given voltage U, the capacity would be (I1(U) - I2(U))/([dU1/dt](U) - [dU2/dt](U)): The current difference between two passes through the same voltage at different scan rate only comes from the capacitive contribution, and you can get the capacity by dividing the difference in current by the difference in scan rate, for example forward and backward scan. Such measurements are typical for electrochemistry and are known under the term "Cyclic voltammetry". Electrochemistry and semiconductor physics are virtually the same!
never go to +/-4V before understanding the hysteresis and other irreversible changes that can happen when you scan it. Do the scan at lower range +- 1 or 2 V, and see whether the result is consistent and repeatable. As mentioned above, the contacts can be very important, and also make sure they dont short circuit from contamination. You are on good path, compare your results with literatures, but dont blindly trust them. Some were good, but I also saw people cycling V at +- 10-20V, which can cause Ohmic heating.
Hi, even though it is low voltage it could still be due to Joule heating as it would depend also on your Au contact with the nanorods, point contacts may lead to hysteresis. Just curious did you ground at least one end while recording these I-V curves? if not just try to check if there is any change.
Investigate memristors. We think that ZnO is memristive, as we've seen memristive-type responses with that ZnO nanoparticles in our own lab. The second curve especially looks memristive.
I thick there is capacitor effect in the figure. You can find that in many ReRRAM composed of high k dielectric materials. Few people care about it because it is uselees for RRAM.