Charge transport boundary condition at metal-semi interface and ohmic contact

Hello! I am simulating MSM Al-Ge-Al photodetector. I used recombination parameters from the VPD example. I was checking the band diagram and carrier conc. and have some query regarding boundary conditions:

  1. Workfunction difference of metal and doped semiconductor is around 0.233eV. How good approximation it would be to assume the contact to be ohmic? I am attaching two figures at V=0V, at dark condition:

  2. Here, Ge uniform doping level is low (5e14 cm^-3). I found following lines in lumerical website regarding setting interface boundary condition:
    “On conductor surfaces surface recombination model should not be applied to the majority carriers. This assumption is appropriate at a conductor interface where the doping concentration is large: the relative change in the carrier densities due to recombination may be large in for the minority carriers, but will typically be negligible for the majority carriers”
    Since the doping level is low here, is this assumption of not applying the model to majority carrier at conductor interface still valid? From charge concentration plot, I can see when optical signal is present, majority carrier ( p ) concentration is affected (irrespective to ohmic/Schottky contact). I am attaching figures for V=0.1V, light on condition:
    To check this, I also tried steady state simulation with turning on the majority model at conductor interface and I-V curve gives extremely noisy value:
    I am confused whether I did something wrong. I would appreciate any kind of suggestion/help regarding the ohmic/Schottky contact and recombination model for the detector. I am attaching the .ldev file and the generation file for your considerations:
    MSM_photodetector.ldev (7.6 MB)
    generate_111_Al_SiO2enclosed_L5micron.mat (6.4 KB)

Let me know whether you need any additional info.


  1. If you compare the dark and photocurrent results, it does not seem that using Schottky is a good approximation of ohmic since with Shocttky, you’d have two diodes on both sides of the device compared to when its ohmic and equivalent to a resistor. This is also obvious from the dark currents (one linear, the other exponential). Also, photocarrier collection would be more efficient with ohmic contacts which facilitate charge transport bidirectionally. This is also obvious from photocurrents when compared to dark currents.


  1. In your device, you have both sides undoped while a middle area is doped. With an undoped material, I’m not sure how you would defined the minority and majority carriers since this is usually considering a pn junction where electrons on p side are minority and vice versa. However, having the surface recombination applied to majority carriers at contacts should not be used since it will cause numerically instability (the solver needs to fix the majority carrier fermi level at the contacts as reference for calculations.

Thanks for the reply.

I am not clear about the first query. MSM structure normally consists of two Schottky electrodes (so theoretically there should be bending at the two metal-semi junction at equilibrium). If I use ohmic condition for the contacts, it will forcefully make the band flat at equilibrium. So, will this be a good approximation to use “ohmic” instead of using “Schottky” condition at the contacts of this device?

** I wanted to use ohmic approximation because for Schottky condition the current level seems to be very high even at dark condition and with ‘light on’ condition current level does not change much (you also showed that graphically).

For second question, I could understand that surface recombination should not be used for majority carrier for numerical calculation instability.
The Ge region is uniformly lightly doped with p dopant (I uploaded the corrected device file where complete Ge area is doped. The previous file had a mistake in doping definition, sorry for that). I meant hole to be majority carrier since Ge is p doped.

I see. Ohmic only would be a good approximation if you Schottky contacts had ohmic behavior. This is usually the case when the semiconductor is highly doped or the metal has a deep (for n type semi) or shallow (for p-type semi) work function. This condition will ensure the Schottky barrier is narrow enough (due to huge band bending) so that carriers can tunnel through. For example, the following plots show the Schottky case where the semiconductor is doped at 1e19 and the metal work function is shallow (~2eV). You can see how narrow the barrier is and the IV is almost linear showing ohmic behavior. But for the doping value and work function that you currently have, I don’t think ohmic would be a good approximation.
image image

With uniform doping, if you plot the charge density, it seems like n and p densities are comparable to each other (with only two to three orders of magnitude difference) so not having surface recombination applied to majority carriers may not be a good approximation.

Thank you for clarifying.
One question to add, I was trying to figure out what could be the reason for that extremely high dark current at low doping + Schottky case. For example, I=0.02A at V=1V from your post on May 17th. It seems unrealistic too, in many photodetector fab based papers, I found nA-microA range dark current.

From the band diagram and charge distribution, it seems that the 120nm wide Ge region is fully depleted at low doping level (n concentration higher) and V=0 (dark). At high doping, the mid Ge region is not depleted (p concentration equals the doping level), and dark current is low. Moreover, if I use large separation between the metals (i.e W=5micron), the dark current is small. I added figures of carrier concentration for different p doping and PD width at dark condition and V=0.
Is there any model parameter that needs to be set properly or is there any valid explanation for this extremely high dark current value? Any suggestion would be highly appreciable.


Looking at the bandstructure for low doping case (left), you can see that there is not much bandbending on the sides (due to very low level doping), so it is very easy for the junction to turn on at very low bias voltage and not much of the charge will get trapped in the middle and current is high. For the high doping case however (right), large bandbending can cause charges get trapped and reduce the current (can also be observed from quasi fermi levels on the right side).

For the wide device with low doping, I think the reduction in current is due to the fact that most of the device is not depleted and charges have to travel through diffusion and are subject to bulk recombination.

Hope this helps,

Thanks for the reply. Yes that is helpful.