Page 150 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 141

with

μ = μ 0 (1 − α I α N )

and
⎧
⎪ M 11 = R g (1 + L + R 1 K) + α N R g S(1 + L + R 1 K) − α N R g KT,
⎪
⎪
⎪
M 12 =−α I R g (1 + L + R 1 K) − R g S(1 + L + R 1 K) + R g KT,
⎨
⎪ M 21 = R g (L + R 1 K) + α N (R g (L + R 1 K)S − T(R g K + N)),
⎪
⎪
⎪
M 22 =−α I R g (L + R 1 K) − R g (L + R 1 K)S + T(R g K + N).
⎩
Suppose now that the emitter electrical potential (resp. the collector elec-
trical potential) of the (practical) transistor is given by ϕ E ∈ (R;R ∪{+∞})
(resp. ϕ C ∈ (R;R ∪{+∞}))asinSection 2.3.2 of Chapter 2 (practical diode
model). We set
2
∀z = (z 1 ,z 2 ) ∈ R : (z) = ϕ E (z 1 ) + ϕ C (z 2 ).
Our model now reads

2
2
x ∈ R : Mx + q,v − x + (v) − (x) ≥ 0, ∀v ∈ R . (4.130)
n
It is clear that ∈ D (R ;R ∪{+∞}). Moreover, the principal minors of the
matrix M are
1 1
1 (M) = M 11 , 2 (M) = M 22 , 12 (M) = det(M).
μ μ

We can easily check that M 11 > 0 and M 22 > 0. Indeed, replacing L, N, S,
and T in M 11 and M 22 , after some calculations, we obtain

R g R C
M 11 = R g + (1 − α N )( + R 1 R g K) > 0 (4.131)
R L
and

R g R C
M 22 = NT + + (1 − α I )(R g L + R 1 R g K) > 0. (4.132)
R 6
Moreover, since det(A) = R g K + N(1 + L + R K )> 0, det(B) = R g T> 0,
and det(C) > 0, we see that det(M) > 0. The matrix M is thus a P-matrix, and
n
we may use Theorem 7 to conclude that for each q ∈ R , problem VI(M,q, )
has a unique solution.