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  • Electrical Engineering Archive: Questions from 2023-03-23

    Design the Pseudo NMOS inverter in Figure Q3 with ID = 100 uA and VL = 0.2 V. State any assumptions made. Assume the following parameters: Supply voltage VDD = 2.5 V, Transconductance parameter (N) K'N = 100 uA/V2 Transconductance parameter (P) K

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    A pseudo-NMOS Inverter is utilizing FETs with the following parameters: Vtn = 1V, Vtp = -1V, kn

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    Consider a pseudo NMOS inverter shown below. Use the following parameters for your calculations. Neglect Early effect (VA is infinite) Transistor parameters: kp

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    For the following pseudo NMOS circuit, find the value of r, so that NMH is 50% of VDD. Also find the value of VoL and the current flowing through QN when output is at its lowest value. Use the following values for your calculation: VDD = 10 V, IVt| = 1.5 V, kn'(W/L) = 1 mA/V2 b. Sketch the following function in pseudo NMOS technology. F = (AB)_bar + AB

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    Consider a pseudo-NMOS inverter, L=1 um, unCox = 115 uA/V2, upCox= 30 uA/V2, Vtn = -Vtp = 0.6 V, and VDD = 4 V. Let the W/L ratio of Qp be (1 um/1 um) and r = 8.

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    A pseudo-NMOS inverter has VDD = 2.5 V, Vtn = 0.5V, Vtp = -0.5V, kn' = 115 uA/V2. kp' = 30 uA/V2 and (W/L)n = 1.5 and r = 4. Determine (W/L)p.

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    Find the current I and the voltage Vo in the network below if Kn = Kp = 1 mA/V2 for both transistors and VDD = 11 V, Vtp = -6V, Vtn = 2V, Kp = Kp'(W/L), Kn=Kn'(W/L).

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    The circuit shown in Figure 7 is supposed to work as an inverter where it has the following values for its parameters: VDD = 1.8V, lambda_1 = lambda_2 = 0, unCox = 100 uA/V2, upCox = 50 uA/V2, VTH,N=0.4V, VTH,P= -0.5V. (W/L)2 = 3/0.18 1. Assuming that vout,min = 0.1 V when vin = VDD, determine the mode of the transistors M1 and M2. 2. Using the previous results, determine the value of (W/L)1 that will guarantee that vout, min = 0.1 V, when vin = VDD.

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    Consider a pseudo-NMOS inverter fabricated in a 0.18-um CMOS technology for which unCox = 120 uA/V2, upCox = 40 uA/V2, Vtn = -Vtp = 0.5 V, and VDD = 2.5 V. If (W/L)n = 0.325um/0.18um and r = 4, a total capacitance at the inverter output of 21 fF.

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