210 CHAPTER 4. JUNCTIONS IN SEMICONDUCTORS:P-NDIODES
Figure 4.38: Diode for problem 4.29.
growth is dropped after growth of thep-layer is completed. The growth is then completed
with ani-layer of thicknessWiand the subsequentn-layer, as shown in figure 4.19. When
the capacitance of the diode is measured, it is determined that the diode is actually ap-n
junction and not thep-i-nthat was designed. The reason is that while waiting for the
temperature to drop after growth of thep-layer, oxygen (a donor) incorporated with
densityqσcm−^2 at thep-iinterface, (see figure 4.19).
(a) Derive the relation between the doping densities,Wi,andqσso that the measurement
is explained. AssumeNA=NDfor simplicity.
(b) Next, calculate a numerical value forWi.AssumeNA=ND=10^17 cm−^3 and
qσ= 5 × 1011 cm−^2.
Problem 4.30The critical field for breakdown of silicon is 4× 105 V/cm. Calculate the
n-side doping of an abruptp+ndiode that allows one to have a breakdown voltage of 30 V.
Problem 4.31Consider an abruptp+nGaAs diode at 300 K with a doping of
Nd=10^16 cm−^3. Calculate the breakdown voltage. Repeat the calculation for a similarly
dopedp+ndiode made from diamond. Use Appendix B for the data you may need.
Problem 4.32What is the width of the potential barrier seen by electrons during
band-to-band tunneling in an applied field of 5 × 105 V/cm in GaAs, Si and
In 0. 53 Ga 0. 47 As (Eg=0.8V)?
Problem 4.33Consider an Sip-ndiode withNa=10^18 cm−^3 ;Nd=10^18 cm−^3.
Assume that the diode will break down by Zener tunneling if the peak field reaches
106 V/cm. Calculate the reverse bias at which the diode will break down.
Problem 4.34Punch through diode: For junction diodes that have to operate at high
reverse biases, one needs a very thick depletion region. However, in forward-bias
conditions this region is undepleted and leads to a series resistance. One uses ap+-n-n+
