Mid Term Exam
October 25, 2005
Figure 3 |
V_{2} - V_{1} = V_{T} ln[ n_{2} / n_{1}]where V_{2} is the EPI voltage and n_{2} is the EPI doping.
V_{EPI} - V_{bl} = 0.0259 ln[5x10^{15} / 10^{17} ] = - 77.6 mV
The burried layer is at the higher potential.