Page 133 - Complementarity and Variational Inequalities in Electronics

124 Complementarity and Variational Inequalities in Electronics

Indeed, we have

∗ ∗ ∗
1
2
1
u − E 1 − Ri ∈ Ri − ∂ R + (−i )
u − E 1 1
∗
∗
∗
⇔ − i ∈ i + ∂ R − (i )
1
2
1
R R
u − E 1 −1 u − E 1
∗
∗
∗
∗
∗
⇔ − i ∈ i + ∂ R − (i ) ⇔ i = (I + ∂ R − ) ( − i )
2
1
1
2
1
R R
u − E 1 u − E 1
∗
∗
∗
∗
∗
⇔ i = min{0, − i }⇔ i + i = min{i , }.
2
1
2
1
2
R R
We also have
∗ ∗ ∗
2
2
1
u − E 2 − Ri ∈ Ri + ∂ R + (i )
u − E 2 1
∗
∗
∗
⇔ − i ∈ i + ∂ R + (i )
2
2
1
R R
u − E 2 −1 u − E 2
∗
∗
∗
∗
∗
⇔ − i ∈ i + ∂ R + (i ) ⇔ i = (I + ∂ R + ) ( − i )
1
2
1
2
2
R R
u − E 2 u − E 2
∗
∗
∗
∗
∗
⇔ i = max{0, − i }⇔ i + i = max{i , }.
2 1 1 2 1
R R
We obtain
⎧
u−E 1
R
⎪ if u<E 1
⎪
⎨
∗
i = 0 if E 1 ≤ u ≤ E 2
⎪
⎪
⎩
u−E 2 if u>E 2 .
R
Indeed, we know that i ≤ 0 and i ≥ 0. Therefore, if u>E 2 , then
∗
∗
1 2
∗ u−E 1
∗ u−E 2
max{i , }= u−E 2 , and if u<E 1 , then min{i , }= u−E 1 .If E 1 ≤
1 R R 2 R R
∗ u−E 1
∗ u−E 2
∗
∗
u ≤ E 2 , then min{i , }≥ 0, max{i , }≤ 0, and thus 0 ≤ i + i ≤ 0.
2 R 1 R 1 2
So, for a driven time-dependent input t à u(t), the time-dependent current
∗
t à i (t) through the resistor R is given by
⎧
⎪ u(t)−E 1 if
R
⎪ u(t) < E 1
⎪
⎨
∗
i (t) = 0 if E 1 ≤ u(t) ≤ E 2 (4.111)
⎪
⎪
⎪
⎩ u(t)−E 2 if u(t) > E 2 ,
R
and the output-signal t à V o (t) defined by
∗
V o (t) = V 2 (t) + E 2 = u(t) − Ri (t)
is then given by the expression