求助,光电探测器例子里的Pin是如何确定的?

fdtd
device

#1

就是photodetector里面的getting started中求响应度的代码:
Pin=0.9e-3; # Watts

I=-2*getdata(“CHARGE”, “anode.I”);
I_norm=I/max(I);
V=linspace(-0.25,2,26);
Resp=I/Pin;
plot(V,I_norm,“photo detector bias voltage(v)”,"I_(photo,norm)=I_(photo)/I_(norm) a.u ");
plot(V,Resp,“photo detector bias voltage(v)”,"Responsivity (A/W) ");
里的Pin=0.9e-3是如何确定的,我在例子中的fdtd文件vpd_optical.fsp (1.3 MB)
中找不到啊


#2

你说的是CHARGE仿真,FDTD仿真的结果一般是归化的,也可能是实际的入射光的光学强度。在这个例子中用的是指定的光学强度Watts/m^2,因此输出的吸收是针对这个强度的。这个强度对应于功率1瓦 The source intensity is set in such a way so that the generation rate is calculated for unity input power (1 W).

网页上还有介绍(数字我修改了):For this plot, for consistency purposes with the reference paper, an input power of 0.9 mW which corresponds to 1 mA of short circuit current in this case has been used. This is done by setting the value of “scale factor” of the gen object to 0.9e-3.
网上的数字原来有误,说是 0.009 mW 和 0.3e-9,很快会修改过来。文件里面的数字没错。

这个例子应该是单色光的结果,所以可以直接改变功率。Pin=0.9e-3; # Watts是我们根据文献中的电流倒推出来的。