where p (0 → 1) = dt/τ$_{c}$ and p (1 → 1) = 1 − p (1 → 0) = 1 − dt/τ$_{e}$ and also P$_{00}$ ( t ) = 1 − P$_{01}$ ( t ). This leads to a simple differencial equation: dP$_{01}$ ( t ) /dt = τ − 1 c − ( τ − 1 e + τ − 1 c ) P$_{01}$ ( t ). If P$_{00}$ (0) = 1, its solution is P$_{01}$ ( t ) = τ$_{e}$ ( τ$_{e}$ + τ$_{c}$ ) − $^{1}$[1 − exp( − t/τ$_{eq}$ )], where 1 /τ$_{eq}$ = 1 /τ$_{e}$ +1 /τ$_{c}$ . Similar results can be performed leading to P$_{11}$ ( t ) = τ$_{e}$ τ$_{e}$ + τ$_{c}$ [ τ$_{e}$ + τ$_{c}$ exp( − t/τ$_{eq}$ )]. Unless fluctuations on the current amplitude, whose average Δ depends on other microscopic factors of the device, this probability corresponds to the autocorrelation of the system A ( t ) = 〈 σ (0) σ ( t ) 〉 , where σ ( t ) corresponds to the state of the trap (occupied σ = 0 or empty σ = 1). In frequency domain this exponential decay of autocorrelation for one trap is described by Lorentzians, since the power spectrum density, i.e., the fourier transform of the autocorrelation is The known 1/ f noise results from a sum of these Lorentzians (a contribution of the many traps in device). For more details about the origin of 1/ f noise, see for example [7],[4], [13], [14]. So, coming back to relaxation phenomena, and starting from n (0) = 0 (all traps empty), we can calculate the average density of occupied traps at time t where Pr( k | n (0) , t ) is the probability of just k traps are occupied at time t , with k = 0 ...N$_{tr}$ . But the traps have different constants τ$_{e}$ and τ$_{c}$ and from that, we write with C$_{k}$ denoting every subset { d$_{1}$, d$_{2}$, ..., d$_{k}$ } from { 1 , 2 , ..., , n } . But { τ ( d$_{i}$ ) c , τ ( d$_{i}$ ) e } N$_{tr}$ i =1 are statistically independent and identically distributed, and we have : 5 Pr( k | n (0) , t ) = ∑ C$_{k}$ P$_{01}$ ( τ$_{c}$, τ$_{e}$ ; t ) k · P$_{00}$ ( τ$_{c}$, τ$_{e}$ ; t ) N$_{tr}$ − k = ( n k ) P$_{01}$ ( τ$_{c}$, τ$_{e}$ ; t ) k · P$_{00}$ ( τ$_{c}$, τ$_{e}$ ; t ) N$_{tr}$ − k Pr( k | n (0) , t ) = ∑ C$_{k}$ k ∏ i =1 P$_{01}$ ( τ ( d$_{i}$ ) c , τ ( d$_{i}$ ) e ; t ) N$_{tr}$ ∏ i = k +1 P$_{00}$ ( τ ( d$_{i}$ ) c , τ ( d$_{i}$ ) e ; t ) 〈 n ( t ) 〉 = 〈 ∑ N$_{tr}$ k =0 σ$_{i}$ ( t ) 〉 = N$_{tr}$ k =0 k Pr( k | n (0) , t ) (1) S ( f ) = ∫ ∞ −∞ e − 2 $^{πfti}$A ( t ) dt = 4Δ 2 ( τ$_{e}$ + τ$_{c}$ )( τ 2 eq + 2 πf $^{2}$)