Physics 127AL, Quiz 2 Solutions
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Physics 127AL, Quiz 2 Solutions
(a) To find VB(t), we look at both the DC component given by the voltage divider and the AC component as passed by the Vin (note that the capacitor blocks DC component of Vin while passing its AC component), thus, overall, we have:
VB(t) = × 10 + 0.1 sin(2几ft) = 1 + 0.1sin ( 2几ft) V
(b) In forward working mode, we have a base to emitter voltage ~0.6 to 0.8 Volt, VE(t) = VB(t) − 0.6 V = 0.4 + 0.1 sin(2几ft) V
(c) We utilize the fact that collector current is nearly identical to emitter current by IC(t) ≈ IE (t) = = (2几ft) = 0.04 + 0.01 sin(2几ft) mA
(d) VC(t) = VCC − IC(t)RC = 10 − 30k × [0.04 + 0.01 sin (21(几)k(ft))] = 8.8 − 0.3 sin(2几ft) V
(e) Vout(t) = −0.3 sin(2几ft) V (remember that the capacitor filters out the DC component, so it defers from VC(t) by the DC component, i.e., we do not get 8.8 Volt DC offset any more at Vout .
(f) At the frequency f=1MHz, the impedance of capacitor is small as Zcapacitor = −业(i)c ≈ −i Ω,
and we can approximate as
Zin = 10k ||90k||(F × 10k) ≈ 9 kΩ
(Refer to textbook page 103 for more information)
2023-06-15