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1 ECE 270 Learning Outcome Practice Exam B OUTCOME #1: an ability to analyze and design CMOS logic gates. Multiple Choice select the single most appropriate response for each question. Note that none of the above MAY be a VALID ANSWER. Place answers on the supplied BUBBLE SHEET only nothing written here will be graded. 1. One of your former best friends from another major (BFFAM) decided to insult you by telling his (other) friends that your IQ had assumed room temperature, i.e. (72)10. To put a better spin on this unfortunate reality, in your Faceplant posts you advertise your IQ to prospective employers as a base 8 number. Expressed as a base 8 number, (72)10 is: (A) (72)8 (B) (90)8 (C) (108)8 (D) (110)8 (E) none of these 2. The nominal (minimum) case for the outputs of logic family A to be able to successfully drive the inputs of logic family B is: (A) fanouta B 0 and DCNMA B > 1 (B) fanouta B 0 and DCNMA B < 1 (C) fanouta B 1 and DCNMA B > 0 (D) fanouta B 1 and DCNMA B < 0 3. The high impedance state of a tri-state buffer is created by: (A) turning on both the PMOS and the NMOS transistors at the output of the buffer (B) turning on the PMOS transistor and turning off the NMOS transistor at the output of the buffer (C) turning off the PMOS transistor and turning on the NMOS transistor at the output of the buffer (D) turning off both the PMOS and the NMOS transistors at the output of the buffer 4. A level of 2-input NAND gates followed by a 2-input OR gate is logically equivalent to: (A) a 4-input AND gate (B) a 4-input OR gate (C) a 4-input NAND gate (D) a 4-input NOR gate
2 ECE 270 Learning Outcome Practice Exam B Based on a growing consensus among your BFFAMs (see question 1), you get the bright idea that you might be able to earn back some respect among your Faceplant friends by working out the proof to the Consensus Theorem (specifically, its dual). Identify the theorems applied in each step of the proof, for questions 5 through 7, below. Dual of Consensus Theorem: (X + Y) (X + Z) (Y + Z) = (X + Y) (X + Z) BFFAM s Proof: LHS: (X + Y) (X + Z) (Y + Z) Step 1: = (X + Y) (X + Z) (Y + Z + 0) Step 2: = (X + Y) (X + Z) (Y + Z + X X ) Step 3: = (X + Y) (X + Z) (X + Y + Z) (X + Y + Z) Step 4: = (X + Y) (X + Y + Z) (X + Z) (X + Y + Z) Step 5: = (X + Y) (X + Z) Q.E.D. 5. The switching algebra theorem used to obtain the expression in Step 1 is: (A) identity (B) null elements (C) complements (D) involution 6. The switching algebra theorem used to obtain the expression in Step 3 is: (A) commutivity (B) complements (C) distributivity (D) combining 7. The switching algebra theorem used to obtain the expression in Step 5 is: (A) commutivity (B) combining (C) distributivity (D) covering The following table applies to questions 8 through 11:
3 ECE 270 Learning Outcome Practice Exam B Table 1. DC Characteristics of a Hypothetical Logic Family. VCC = 5 V VOH = 4.50 V VOL = 0.50 V VIH = 3.00 V VIL = 2.00 V VTH = (VOH VOL)/2 IOH = 5.0 ma IOL = 10 ma IIH = 50 A IIL = 0.1 ma 8. The DC noise margin for this logic family is: (A) 0.50 V (B) 1.00 V (C) 1.50 V (D) 2.00 V 9. The practical fanout for this logic family is: (A) 1 (B) 5 (C) 10 (D) When interfacing an LED that has a forward voltage of 2.0 V to this logic family in a current sinking configuration, maximum brightness will be achieved (within the rated specifications) using a current limiting resistor of the value: (A) 200 (B) 250 (C) 500 (D) When interfacing an LED that has a forward voltage of 2.0 V to this logic family in a current sourcing configuration, maximum brightness will be achieved (within the rated specifications) using a current limiting resistor of the value: (A) 200 (B) 250 (C) 500 (D) 600
4 ECE 270 Learning Outcome Practice Exam B The following circuit applies to questions 12 through 15: P-channel: on resistance = 30 off resistance = 3 M N-channel: on resistance = 20 off resistance = 2 M 12. When A = 0 V and B = 5 V, the potential at Vout will be: (A) 0.00 V (B) 2.00 V (C) 3.00 V (D) 5.00 V 13. When A = 0 V and B = 5 V, the amount of power dissipated by this circuit is: (A) 5 µw (B) 25 µw (C) 250 mw (D) 500 mw 14. When A = 5 V and B = 0 V, the potential at Vout will be: (A) 0.00 V (B) 2.00 V (C) 3.00 V (D) 5.00 V 15. When A = 5 V and B = 0 V, the amount of power dissipated by this circuit is: (A) 5 µw (B) 25 µw (C) 250 mw (D) 500 mw
5 ECE 270 Learning Outcome Practice Exam B The following figure applies to questions 16 and 17: Power Dissipation (mw) A B C D Power Supply Voltage 16. The curve that depicts power dissipation of a CMOS circuit as a function of power supply voltage (over the range of 1 to 5 volts) is: (A) A (B) B (C) C (D) D 17. If the power supply voltage of a CMOS circuit is reduced from 5.0 volts to 2.0 volts, its power dissipation will be reduced by: (A) 4 mw (B) 16 mw (C) 84 mw (D) 96 mw
6 ECE 270 Learning Outcome Practice Exam B The following circuit applies to questions 18 through 22: CMOS Inverter: IIH = +40 A IIL = - 40 A VIHmin = 3.5 V VILmax = 1.5 V Open-drain (O.D.) CMOS NAND gates: VOLmax = 0.5 IOLmax = +5 ma Off-state leakage current = +20 A 18. When A = 0 V, B = 0 V, and C = 0 V, the voltage at the inverter input will be: (A) 0.0 V (B) 0.1 V (C) 4.9 V (D) 5.0 V (E) none of these 19. Based on the open-drain NAND gate specifications provided, the value of pullup resistor shown in the circuit (1000 ) is: (A) < Rmin (B) Rmin (C) Rmin < R < Rmax (D) Rmax (E) > Rmax 20. When A = 5 V, B = 0 V, and C = 0 V, the voltage at the inverter input will be: (A) zero (B) greater than zero but less than the specified VOLmax (C) exactly the specified VOLmax (D) greater than the specified VOLmax 21. When A = 5 V, B = 0 V, and C = 0 V, the current sunk by the active open-drain NAND gate will be: (A) zero (B) greater than zero but less than the specified IOLmax (C) exactly the specified IOLmax (D) greater than the specified IOLmax 22. When A = 5 V, B = 5 V, and C = 5 V, the current sunk by each active open-drain NAND gate will be: (A) less than the case in which A = 5 V, B = 5 V, and C = 0 V (B) more than the case in which A = 5 V, B = 5 V, and C = 0 V (C) exactly the specified IOLmax (D) greater than the specified IOLmax
7 ECE 270 Learning Outcome Practice Exam B The following figure applies to questions 23 through 26 (each square is 10 ns): 5 V Vin V 5 V Vout 0 V 10 ns 23. The rise time (ttlh) for the open-drain NAND gate is approximately: (A) 10 ns (B) 20 ns (C) 30 ns (D) 40 ns (E) none of these 24. The fall propagation delay (tphl) for the open-drain NAND gate is approximately: (A) 10 ns (B) 20 ns (C) 30 ns (D) 40 ns (E) none of these 25. Estimate the value of the capacitor (C) based on the RC time constant: (A) 10 pf (B) 15 pf (C) 20 pf (D) 30 pf (E) none of these 26. Estimate the ON resistance of the open-drain NAND gate based on the RC time constant: (A) 100 Ω (B) 500 Ω (C) 667 Ω (D) 1000 Ω (E) none of these
8 ECE 270 Learning Outcome Practice Exam B For questions 27-30, assume the on resistance of each MOSFET is 10 and the off resistance is 1 M, and that the N- and P-channel MOSFETs in each circuit have been randomly wired together as shown. 27. The amount of current this circuit can source to a 1000 load (referenced to GND) is approximately: (A) 0 ma regardless of Vin (B) 2.5 ma if Vin is 0 V (C) 2.5 ma if Vin is 5 V (D) 5.0 ma regardless of Vin 28. The amount of current this circuit can sink from a 1000 load (referenced to 5 V) is approximately: (A) 0 ma regardless of Vin (B) 2.5 ma if Vin is 0 V (C) 2.5 ma if Vin is 5 V (D) 5.0 ma regardless of Vin 29. The amount of current this circuit can source to a 1000 load (referenced to GND) is approximately: (A) 0 ma regardless of Vin (B) 2.5 ma if Vin is 0 V (C) 2.5 ma if Vin is 5 V (D) 5.0 ma regardless of Vin 30. The amount of current this circuit can sink from a 1000 load (referenced to 5 V) is approximately: (A) 0 ma regardless of Vin (B) 2.5 ma if Vin is 0 V (C) 2.5 ma if Vin is 5 V (D) 5.0 ma regardless of Vin
Note that none of the above MAY be a VALID ANSWER.
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