(a) Explanation Calculation: The given diagram is shown in Figure 1 The width to length ratios of the load transistors are given by, ( W L ) M L 2 = 1 ( W L ) M L 3 = 1 The width to length ratios of the driver transistors are given by, ( W L ) M D 2 = 8 ( W L ) M D 3 = 8 ( W L ) M D 4 = 8 The expression to determine the power dissipation of the circuit is given by, P D = i D V D D Substitute 250 μW for P D and 5 V for i D in the above equation. 250 μW = i D 5 V i D = 50 μA The expression to determine the value of the drain current is given by, i D = ( k ′ n 2 ) ( W L ) M L 1 [ − ( V T N L ) ] 2 Substitute 50 μA for i D , 60 μA / V 2 for k ′ n and − 2 V for V T N L in the above equation. 50 μA = ( 60 μA / V 2 2 ) ( W L ) M L 1 [ − ( − 2 V ) ] 2 ( W L ) M L 1 = 0.417 The expression to determine the value of the output voltage is given by, K D K L [ 2 ( v I − V T N D ) v O − v O 2 ] = [ − ( V T N L ) ] 2 Substitute − 0 (b) To determine The value of the output voltage v O 2 .

Microelectronics: Circuit Analysis and Design

4th Edition

ISBN: 9780073380643

Author: Donald A. Neamen

Publisher: McGraw-Hill Companies, The

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Chapter 16, Problem 16.21P

(a)

To determine

The value of the width to length ratio (WL)ML1 and (WL)MD1 .

(b)

To determine

The value of the output voltage vO2 .

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Question 2 A transistor is used as a switch and the waveforms are shown in Figure 2. The parameters are Vcc = 225 V, VBE(sat) = 3 V, IB = 8 A, VCE(sat) = 2 V, Ics = 90 A, td = 0.5 µs, tr = 1 µs, ts = 3 µs, tƒ = 2 μs, and f 10 kHz. The duty cycle is k 50%. The collector- emitter leakage current is ICEO = 2 mA. Determine the power loss due to the collector current: = = = (a) during turn-on ton = td + tr VCE Vcc (b) during conduction period tn V CE(sat) 0 toff" ton Ics 0.9 Ics (c) during turn-off toff = ts + tf (d) during off-time tot (e) the total average power losses PT ICEO 0 IBS 0 Figure 2 V BE(sat) 0 主 * td tr In Is If to iB VBE T= 1/fs

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Chapter 16 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 16 - Consider the NMOS inverter with resistor load in...Ch. 16 - The enhancementload NMOS inverter in Figure...Ch. 16 - Prob. 16.3EP Ch. 16 - Prob. 16.4EP Ch. 16 - Consider the NMOS inverter with enhancement load,...Ch. 16 - Prob. 16.2TYU Ch. 16 - (a) Consider the results of Exercise Ex 16.1....Ch. 16 - Prob. 16.5EP Ch. 16 - Prob. 16.6EP Ch. 16 - (a) Design a threeinput NMOS NOR Logic gate with...

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The value of the width to length ratio ( W L ) M L 1 and ( W L ) M D 1 .

Solution Summary: The author calculates the width to length ratios of the load transistors. The expression to determine the power dissipation is given by, l250W

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The value of the width to length ratio (WL)ML1 and (WL)MD1 .

The value of the output voltage vO2 .

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Chapter 16 Solutions