Lecture33 Part 1 Average Discharge Current and Transistion Times_1

# Lecture33 Part 1 Average Discharge Current and Transistion Times_1

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EE 311 Friday April , 2010 Lecture 33 Part 1 10.2.3 in 5 th ed. (13.2 in 4 th ed.) CMOS Inverter Transitions Times Average Current Method Zubair Rehman

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2 Determine Average Discharge Current I D,AV = 1 / 2 [I DN (0 + ) + I DN (t PHL )] I DN (0 + ) = 1 / 2 K n ’ (W n / L n )(V DD -V tn ) 2 Saturation Region (Note that V GS = V DD ) I DN (t PHL ) K n ’ (W n / L n )[(V DD -V tn )(V DD /2) - 1 / 2 (V DD /2) 2 ] Triode Region I D,AV = 1 / 2 K n (W n / L n )[(V DD - V tn ) 2 /2 + (V DD -V tn )V DD /2 - 1 / 2 (V DD /2) 2 ] I D,AV = 1 / 2 K n (W n / L n ){V DD 2 ([1-(V tn /V DD )] 2 / 2+ [1 - (V tn /V DD )]/ 2 - 1 / 2 ( 1 / 2 ) 2 )} I D,AV = ¼ K n ’ (W n / L n )(V DD 2 )[1 - (2V tn / V DD ) + (V tn / V DD ) 2 + 1 - (V tn / V DD ) - 1 / 4 ] I D,AV = 1 / 4 K n ’ (W n / L n )(V DD 2 )[1 + 3 / 4 - (3V tn /V DD ) + (V tn / V DD ) 2 ]
3 Q = C V C = C * (V DD / 2) Q = I AV t pHL = C * (V DD / 2) I D,AV = 1 / 4 K n ’ (W n / L n )(V DD 2 )[1 + 3 / 4 - (3V
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Unformatted text preview: tn /V DD ) + (V tn / V DD ) 2 ] t PHL = (C V DD ) / (2 I AV ) = C / (2 I AV / V DD ) t PHL = C / {K n (W n /L n )V DD [1 + 3 / 4- (3V tn / V DD ) + (V tn / V DD ) 2 ]} t PHL = 2C / {[K n (W n / L n )V DD ] * [1 + 3 / 4- (3V tn / V DD ) + (V tn / V DD ) 2 ]} t PHL = {C / {K n (W n / L n )V DD } * F 2 Where F 2 = {2 / [1 + 3 / 4- (3V tn / V DD ) + (V tn / V DD ) 2 ]} For V tn / V DD = 0.2 F 2 = 2 / [1.75 - 3*0.2 + (0.2) 2 ] = 1.68 = 1.7 Recall that the exact method with an integration lead to a factor 1.61...
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Lecture33 Part 1 Average Discharge Current and Transistion Times_1

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