332 CHAPTER 7. TEMPORAL RESPONSE OF DIODES AND BIPOLAR TRANSISTORS
Example 7.1Consider annpntransistor with the following properties at 300 K:Emitter current, IE =1.5mA
EBJ capacitance, Cje =2pF
Base width, Wb =0. 4 μm
Diffusion coefficient, Db =60cm^2 /s
Width of collector depletion region, Wdc =2. 0 μm
Collector resistance, rC = 30Ω
Total collector capacitance, (Cs+Cμ)=0. 4 pF
Saturated electron velocity, vs =5× 106 cm/sCalculate the cutoff frequency of this transistor. How will the cutoff frequency change (i)
if the emitter current level is doubled? (ii) if the base thickness is halved?
The emitter resistancer′
eis given by (see equation 7.5.11 for the resistance of a
forward-biased diode)r′
e=dIE
dVBE∼=kBT
eIE=
0. 026
1. 5 × 10 −^3
=17.3Ω
This gives
τe=r′
eCje=(17.3)(2×^10− (^12) )=34.6ps
The base transit time is
τt=
Wb^2
2 Db
=
(0. 4 × 10 −^4 )^2
2 × 60
=13.3psThe collector transit time isτC=Wdc
2 vs=
(2. 0 × 10 −^4 )
1 × 107
=10psThe collector charging time isτc=rc(Cμ+Cs) = 30(0. 4 × 10 −^12 )=12psThe total time is
τec=34.6+13.3 + 10 + 12 = 69.9ps
The cutoff frequency isfτ=1
2 πτec=
1
2 π(69. 9 × 10 −^12 s)=2.3GHzIf the emitter current is doubled (assuming no other change occurs), the timeτeis reduced
by half. This gives a cutoff frequency of 2.54 GHz. Similarly, if the base width is reduced
by half, the base transit time becomes 3.3 ps and the cutoff frequency becomes 2.65 GHz.
In this problem the dominant source of delay is the emitter junction.