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The solutions to Homework 11 (HW11) of an Electronic Circuits course. The solutions involve calculating the values of unknown resistances (R2) and voltage at the base (VBE) using two equations. The document also includes the analysis of DC circuits and the determination of equivalent circuits and half-circuits.
Typology: Exercises
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Vth = 30 R^2 k×+^2 R.^52 , Rth = 3030 kk×+RR^22 VCE ≥ VBE (To be guaranteed in active mode, soft saturation is not allowed) VCE = VCC − (IC 2 k + IE 100) IC = βα αR(Vthth+−βRVBEE^ )
∵ IC = Vth−
( VBE + IαC RE ) Rth
So VCE = VCC −
[ (^) βα(Vth−VBE ) αRth+βRE 2 k^ +^
β(Vth−VBE ) αRth+βRE ×^100
VCE ≥ VBE means 2.5-
30 k+R 2 −VBE
)
100 ( (^) R 2 × 2. 5 30 k+R 2 −VBE
)
And VBE = VT ln
IC S
= VT ln
[ (^) βα(V I th−VBE^ ) S (αRth+βRE )
There are two unknowns (R 2 and VBE ) and two equations (1 and 2) Since 2 is a nonlinear equation, the problem can be solved by itera- tion maximum R 2 = 20. 343 k
(5.46)
(a)
Rup = R 1 + (^) gm^12 ||rΠ 2 Rdn = ∞ Rout = R 1 + (^) gm^12 ||rΠ 2 Rin = rΠ + (1 + β)RE |AV | = R^1 +^ gm^12 ||rπ^2 RE + (^) gm^11
(b)
(a) DC Analysis: IC = β
=> β 6 (1.24+.^2 −V 0 BE.α 4 β^ ) Guess VBE = 0.7 => IC = 1. 072 mA Verify VBE : VBE ' VT ln
IC S
= 0. 726 V not 0.7V reiterate VBE = 0. 726 V ; IC = 1. 0163 mA Verify VBE : VBE ' VT ln
IC S
= 0. 725 V converged! VBE = 0. 725 V, VCE = 2. 5 −
(1.0183) 1k + 0. 4
IC = 1. 0163 mA,IB = 10. 163 μA
(9.24)
Equivalent circuit:
β => ∞ Gm = (^) viino ≈ −gm 1 Rout = [1 + gm 2 (ro 1 ||rπ 2 )] ro + (ro 1 ||rπ 2 ) Av = −GmRout = gm 1 [{1 + gm 2 (ro 1 ||rπ 2 )} ro 2 + (ro 1 ||rπ 2 )]
(10.29)
