Delta Id/Id, CMOS Mismatch Caculation

For the CMOS Mismatch,
ΔId/Id=(ΔW/L)/(W/L)-ΔVth/(Vgs-Vth)
Could somebody tell me how to derive this formula, thank you very much!

First, pls write Id=1/2.u.Cox.(W/L).(Vgs-Vth)^2. Then calculate differentiation for a=ΔId/ΔW and b=ΔId/ΔVth individually. Finally, you got: ΔId=a.ΔW+b.ΔVth

thank you leo_o2, I understand what you tell me.
but from one lecture:
Id1=1/2.u.Cox.(W1/L1).(Vgs-Vth1)^2
Id2=1/2.u.Cox.(W2/L2).(Vgs-Vth2)^2
ΔId=Id1-Id2, Id=(Id1+Id2)/2
Δ(W/L)=(W1/L1)-(W2/L2)
(W/L)=((W1/L1)+(W2/L2))/2
ΔVth=Vth1-Vth2
Vth=(Vth1+Vth2)/2
Substitute, then:
ΔId/Id=(ΔW/L)/(W/L)-2ΔVth/(Vgs-Vth)

I don't know how to get ΔId/Id from the above formulas???????

I think the book of Fundamentals of Microelectronics of Rabaey has this formulas.

lqy said:

thank you leo_o2, I understand what you tell me.
but from one lecture:
Id1=1/2.u.Cox.(W1/L1).(Vgs-Vth1)^2
Id2=1/2.u.Cox.(W2/L2).(Vgs-Vth2)^2
ΔId=Id1-Id2, Id=(Id1+Id2)/2
Δ(W/L)=(W1/L1)-(W2/L2)
(W/L)=((W1/L1)+(W2/L2))/2
ΔVth=Vth1-Vth2
Vth=(Vth1+Vth2)/2
Substitute, then:
ΔId/Id=(ΔW/L)/(W/L)-2ΔVth/(Vgs-Vth)

I don't know how to get ΔId/Id from the above formulas???????

Click to expand...

Hello
Id= 1/2*u*Cox* (W/L)*(Vgs-Vth)^2
let's take derivative
ΔId = 1/2*u*Cox*Δ(W/L)*(Vgs-Vth)^2+1/2*u*Cox*(W/L)*2(Vgs-Vth)*(-ΔVth)
than
ΔId/Id = Δ(W/L) / (W/L) - 2*ΔVth/(Vgs-Vth)
that's your formula:wink:

Not sure if these 2 Id dependencies can be extracted from each other. Even if both dependencies - area or "Pelgrom", and offset voltage (variation) - are somehow correlated because both depend on similar physical parameter variations over area effects (gate-oxide thickness variations, mobility variations, and depletion, surface-state and threshold-implant charge variations), they are probably - to a good part - just different representations of the same physical reasons which lead to local area mismatch, s. e.g. David M. Binkley "Tradeoffs and Optimization in Analog CMOS Design", chap. 3.11 MISMATCH, p. 233 ff.

The above given equation ΔId/Id = Δ(W/L) / (W/L) - 2*ΔVth/(Vgs-Vth) could only be valid in strong inversion (because of the Id dependency on (Vgs-Vth)²); in moderate and weak inversion (Veff = Vgs-Vth ≈ 0 or ≤ 0) the second term would cancel the area dependency and would lead to an unrealistic result.

Thank you all for your help. Merry Christmas.