294 CHAPTER 6. BIPOLAR JUNCTION TRANSISTORS
AlGaAs
Eg= 1.7 eV
ΔEc= 0.2 eV
n-type: 10^17 cm-3
Graded region
n-type: 10^17 cm-3
Base
p+-GaAs
0.2 μm 0.1 μm
Emitter
n - GaAs
n-type: 10^17 cm-3
Figure 6.25: Figure for problem 6.21.
Problem 6.22A siliconpnptransistor at 300 K has a doping of
Nae=10^18 cm−^3 ,Ndb=5× 1016 cm−^3 ,Nac=10^15 cm−^3. The base width is 1.0μm.
The value ofDbis 10 cm^2 /s andτB=10−^7 s. The emitter base junction is forward biased
at 0.7 V. Using the approximation that the minority carrier distribution in the base can be
represented by a linear decay, calculate the hole diffusion current density in the base at (a)
VCB= 5 V (reverse bias), (b)VCB=15V.
Problem 6.23A uniformly dopednpnbipolar transistor is fabricated to within
Nde=10^19 cm−^3 andNdc=10^16 cm−^3. The base width is 0.5μm. Design the base
doping so that the punch through voltage is at least 25 V in the forward active mode.
Problem 6.24Annpnsilicon bipolar transistor has a base doping of 1016 cm−^3 and a
heavily doped collector region. The neutral base width is 1.0μm. What is the base
collector reverse bias when punch through occurs?
Problem 6.25The punch through voltage of a Gepnpbipolar transistor is 20 V. The base
doping is 1016 cm−^3 , and the emitter and collector doping are 1018 cm−^3. Calculate the
zero bias base width. IfτB=10−^6 s, what is theαof the transistor at a 10 V reverse bias
across the collector-base junction at 300 K? The hole diffusion coefficient in the base is 40
cm^2 s−^1.
Problem 6.26In a siliconnpntransistor, the doping concentrations in the emitter and
collector areNde=10^18 cm−^3 andNdc=5× 1015 cm−^3 , respectively. The neutral base
width is 0.6μmatVBE=0.7VandVCB=5V.WhenVCBis increased to 10 V, the
minority carrier diffusion current in the base increases by 5%. Calculate the base doping
and the Early voltage ifDb=20cm^2 /s andτB=5× 10 −^7 s.
