Lecture 33 Part 2 Inverter Current &amp; Power Dissipation_1

# Lecture 33 Part 2 Inverter Current &amp; Power...

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EE 311 Friday April 2, 2010 Lecture 33 Part 2 5.10 in 5 th ed. (4.10 in 4 th ed.) CMOS Inverter Current & Power Dissipation Zubair Rehman

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2 CMOS Inverter Current Flow (No Load) At I peak For Symmetric CMOS V DSN = V SDP = V DD / 2 Q N (NMOS) & Q P (PMOS) are in Saturation I DN = 1 / 2 C ox μ n (W n / L n ) (V I - V tn ) 2 I DP = 1 / 2 C ox μ p (W p / L p ) (V DD - V I - |V tp |) 2 I DN = I DP for No Load Let 1 / 2 C ox μ n (W n / L n ) = 1 / 2 C ox μ p (W p / L p ) = K Let V tn = |V tp | = V t
3 I D = 1 / 2 [I DN + I DP ] I D = 1 / 2 K[(V I - V t ) 2 + (V DD - V I - V t ) 2 ] Where does I D v. V I have a maximum? (dI D ) / (dV I ) = 0 max (or min) I D (dI D ) / (dV I ) = 1 / 2 K[2(V I - V t ) - 2(V DD - V I - V t )] = 0 V I - V t = V DD - V I - V t 2V I = V DD V I = (V DD / 2) I peak = 1 / 2 C ox μ n (W n / L n ) (½V DD - V tn ) 2 = 1 / 2 C ox μ p (W p / L p ) (V DD - ½V DD - |V tp |) 2 = 1 / 2 C ox μ p (W p / L p ) (½V DD - |V tp |) 2 Non-Symmetric CMOS There will still be a peak current I peak but it will not occur at V I = (V DD / 2)

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4 For Q N & Q P in Saturation Region Define K n = C ox μ n (W n / L n ) & K p = C ox μ p (W p / L p ) I DN = 1 / 2 K n (V I - V tn ) 2 = I DP = 1 / 2 K p (V DD - V I - |V tp |) 2 Check: If K n = K p & |V tp | = V tn V I = V DD / 2 Sedra-Smith Definitions n = (W n / L n ) & p = (W p / L p ) Non-Symmetric CMOS There will still be a peak current I peak but it will not occur at V I = (V DD / 2) p n p n tn tp DD I p n tn tp

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