ABSTRACT
For a constant gate voltage applied at some time t = 0, N(t) assumes the form of a step input such that
N(t) = ((Vgs − VT0)Ci/q)U(t) (14.2) U(t) = 0 if t < 0
= 1 if t ≥ 0
Here, q is the electron charge, Ci is the TFT dielectric capacitance per unit area. Substituting for N(t) and taking the Laplace transform on both sides of Eq. 14.1,[
snf (s) sns(s)
] − [ nf (0) ns(0)
] =
[ −α− β −α −γ −γ
] [ nf (s) ns(s)
] +
[ α γ
] N(s) (14.3)
Assuming there are no trapped charges at t = 0, nf (0) = ns(0) = 0, and Eq. 14.3 [
nf (s) ns(s)
] =
[ s+ α+ β α
γ s+ γ
]−1 [ α γ
] N(s) (14.4)
for Non-Crystalline
We can define the state transition matrix Φ as
Φ(t) = L−1 ([
s+ α+ β α γ s+ γ
]−1) (14.5)
= L−1 (
(s+ λ1)(s+ λ2)
[ s+ γ −α −γ s+ α+ β
]) where 2λ1 = α + β + γ − ((α + β + γ)2 − 4βγ)1/2 and 2λ2 = α + β + γ + ((α + β + γ)2 − 4βγ)1/2. Note that λ2 > λ1 > 0. Taking the inverse Laplace transform, we find the state transition matrix to be
Φ(t) = 1
λ1 − λ1 × [−(λ1 − γ)e−λ1t + (λ2 − γ)e−λ2t −αe−λ1t + αe−λ2t
−γe−λ1t + γe−λ2t −(λ1 − α− β)e−λ1t + (λ2 − α− β)e−λ2t ]
(14.6)
The solution to Eq. 14.1 is then given by[ nf (t) ns(t)
] =
Φ(t− τ) [ α γ
] (Vgs − VT0)Ci
q dτ (14.7)
Solving this equation, we find
nf (t) = (Vgs − VT0)Ci
q
λ2 − λ1 [−α(1− e−λ1t) + α(1− e−λ2t)](14.8)
ns(t) = (Vgs − VT0)Ci
q
λ2 − λ1 × [( −γ + βγ
λ1
) (1− e−λ1t) +
( γ − βγ
λ2
) (1− e−λ2t)
]
The threshold voltage shift is related to the total trapped charge as
δVT (t) = q(nf (t) + ns(t))
Ci (14.9)
Noting that βγ = λ1λ2 we obtain from Eq. 14.8,
δVT (t) = (Vgs − VT0)(1− ϕe−λ1t − (1− ϕ)e−λ2t) (14.10) where ϕ = λ2−α−γλ2−λ1 . Since α > 0, β > 0 and γ > 0, it can be easily shown that 0 < ϕ < 1.
