292 CHAPTER 6. BIPOLAR JUNCTION TRANSISTORS
Problem 6.10The mobility of holes in silicon is 100 cm^2 /V·s. It is required that a BJT be
made with a base width of 0.5μm and base resistivity of no more than 1.0Ω-cm. It is also
desired that the emitter injection efficiency be at least 0.999. Calculate the emitter doping
required. The various device parameters are
Lb =10μm
Le =10μm
De =10cm^2 s−^1
Db =20cm^2 s−^1
What is the current gainβof the device? AssumeWbn=Wb.
Problem 6.11Consider anpnSi-BJT at 300 K with the following parameters:
Nde =10^18 cm−^3
Nab =10^17 cm−^3
Ndc =5× 1016 cm−^3
Db =30.0cm^2 s−^1
Lb =15. 0 μm
De =10.0cm^2 s−^1
Le =5. 0 μm
(i) Calculate the maximum base width,Wb, that will allow a current gainβof 100 when
the EBJ is forward biased at 1.0 V and the BCJ is reverse biased at 5.0 V.
(ii) Describe two advantages and two disadvantages of making the base smaller.
Problem 6.12TheVCE(sat)of annpntransistor decreases as the base current increases
for a fixed collector current. In the Ebers-Moll model, assumeαF= 0.995,αR=0.1,and
IC= 1.0 mA. At 300 K, at what base current is theVCE(sat)value equal to (a) 0.2 V, (b)
0.1 V?
Problem 6.13Consider annpnbipolar device in the active mode. Express the base
current in terms ofαF,αR,IES,ICS,andVBE, using the Ebers-Moll model.
Problem 6.14Derive the expressions for the emitter and collector current for apnp
transistor in analogy with the equations derived in the text for thenpntransistor.
Problem 6.15Annpnsilicon bipolar device has the following parameters in the
Ebers-Moll model at 300 K:
αF=0.99;αR=0. 2
Calculate the saturation voltageVCEforIC=5mAandIB= 0.2 mA. Why isIC/IBnot
equal toαF/(1−αF)?
