hw4_newsolns

hw4_newsolns - EECS 170D Homework #4 Due in class Thursday...

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EECS 170D Homework #4 Due in class Thursday November 15 th , 2007 Problem 1: Assume and inverter driving an identical inverter. The following technology parameters are given 1. λ= 0.5um 2. VDD= 3V 3. Cox= 2 fF/um 2 4. CJN= 0.3 fF/um 2 5. CJP= 0.5 fF/um 2 6. CJSWN= 0.4 fF/um 2 7. CJSWP= 0.7 fF/um 2 8. CGDO = 0.4 fF/um 2 9. KPn = 75uA/V 2 10. KPp = 25uA/V 2 11. Φ 0 = 0.6V 12.VTOn =0.7V 13.VTOp =-0.7V. As you can derive from a good layout, the minimum drain area is W*5λ (Assume a rectangular shape with at least one contact). With the above provided information derive the value of CL as a function of Wn and Wp for a) Low to High transition. b) High to Low transition. Hint: The Keq values for the High-Low transitions are given below so you only need to compute the values for Low-High.: Solution:Assuming that area = w*5λ Applying the equation (Keq) from the book (equation number 3.11) and using the voltage swing indicated below we can compute the values of Keq as follows: KeqB (VR from 0V to -1.5V) = 0.697 KeqSW (VR from 0V to -1.5V) = 0.785 KeqB (VR from -1.5V to -3V) = 0.463 KeqSW (VR from -1.5 to -3) = 0.60 CL as a function of Wn and Wp a) CL (Low to High) = 2.CGDO.(Wn+Wp)+ 0.697*(Wn*2.5*CJN)+0.785*(Wn+5)*CJSWN+ EECS170D Fall 07
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hw4_newsolns - EECS 170D Homework #4 Due in class Thursday...

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