HW2_Solved - EECS 170D Homework #2 Due in class Thursday...

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EECS 170D Homework #2 Due in class Thursday October 20 th , 2005 Problem 1: Diode Connected Load 1 VIN VDD M1 M2 VOUT 100k VDD! = 2.5V V to = 0.5V L M1, M2 = 0.25 μ m W M1 = 1.0 μ m W M2 = 2.0 μ m L S = 0.25 μ m k n = 100uA/V 2 λ = 0 V -1 γ = 0.2 V 1/2 φ F = -0.3V Consider the inverter circuit shown in Figure above with an ideal square-wave input. Assume that short-channel effects are negligible – meaning V DSAT >> V DS , V GS -V T. Find V OH and V OL . Hint: Both the load and driver transistors are NMOS, so answer cant be 2.5V and 0V! Solution: (a)Finding Voh: Set Vin=0 Only M1 is on and it is pulling up against the resistor. Vout (Voh) will be between Vdd – Vt and ground. However assuming that 100k is a large enough resistor that the current can be assumed negligible and hence output can be VDD-Vt which means your Voh has decreased and can have values starting from 2.5-0.5=2V To calculate the exact value you can start by equating the currents as shown below IM1=IR EECS170D Fall 05
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EECS 170D Homework #2 Due in class Thursday October 20 th , 2005 ( 29 V Vor Voh k Voh IR Voh e IM 7 . 1 24 . 2 100 5 . 0 5 . 2 25 . 0 1 2 100 1 2 06 = = - - = - The obvious solution in this case is Voh=1.7V. (b)Remember, Vt is not equal to Vto anymore for M1 since there is back body biasing with Vsb = Voh (the bulk is grounded) and hence the Vt of the transistor increases which means that the drain current ID decreases and hence Voh decreases further. (c) Finding Vol: Set Vin=2.5V
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HW2_Solved - EECS 170D Homework #2 Due in class Thursday...

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