11. Rise time and fall time can be equalized by
A. βn = βp
B. βn = 2βp
C. βp = 2βn
D. βn = 1/2βp
Answer: A
Rise time tr and fall time tf can be equalized by using βn = βp, which requires (Wp/Lp) = (µn/µp)(Wn/Ln).
12. Rise time and fall time can be also equalized by
A. Lp = Ln = λ
B. Lp = Ln = λ/2
C. Lp = Ln = 2λ
D. 2Lp = Ln = λ
Answer: C
Rise time and fall time can be equalized by taking Lp = Ln = 2λ which implies Wp/Wn = 2 and also µn/µp = 2.
13. Equalizing of rise time and the fall time is possible in
A. nMOS
B. pseudo nMOS
C. CMOS
D. pMOS
Answer: C
Equalizing of rise time and fall time is possible only in CMOS and not possible in nMOS and pseudo nMOS because of the ratio requirement.
14. High and low noise margins can be equalized by
A. βn = βp
B. βn greater than βp
C. βn lesser than βp
D. Lp = 2Ln
Answer: A
High and low noise margins can be equalized by choosing βn = βp, also Ln = Lp which implies Wp/Wn = 2.
15. Inverter pair delay D is given as equal to
A. tr
B. tf
C. tr-tf
D. tr+tf
Answer: D
Inverter pair delay D is given as the sum of rise time and fall time. This is proportional to (Rp+Rn)Cl where Rp and Rn are average resistances.
16. For minimum D consider
A. Ln = Lp = 2λ
B. Ln greater than Lp = 2λ
C. Lp greater than Ln
D. Lp = 2Ln
Answer: A
D increases with Ln and Lp so for minimum D we have to choose Ln=Lp=2λ. D does not vary significantly with (1) lesser than (Wn/Wp) lesser than (2).
17. Different parameter optimization is easily achievable in
A. nMOS
B. pMOS
C. pseudo nMOS
D. CMOS
Answer: D
Different parameter optimizations like noise margins equalization, rise time fall time equalization can be easily achievable in CMOS.
18. Minimizing A with respect to Wp.d. gives
A. Wp.d. = 2λ
B. Wp.d. = λ/2
C. Wp.d. = (k)1/2 × 2λ
D. Wp.d. = k × (λ)1/2 × 2
Answer: C
Minimizing A with respect to Wp.d yields a solution as Wp.d. = (k)1/2 × Wp.u. = (k)1/2 × 2λ.
19. Using Zp.u./Zp.d = k, Lp.u. can be obtained as
A. k × 2λ
B. k × λ
C. (k)1/2 × 2λ
D. k × 2 × (λ)1/2
Answer: C
Using this ratio Zp.u./Zp.d. = k, we obtain Lp.u. = (k)1/2 × Lp.d. = (k)1/2 × 2λ.
20. Minimum area can be given as
A. 4 × Ao × λ × (k)1/2
B. 4 × Ao × λ × k
C. 8 × Ao × λ2 × (k)1/2
D. 8 × Ao × λ × (k)1/2
Answer: C
Minimum area A can be given as 8 x Ao x λ2 x (k)1/2 which implies Zp.u. = (k)1/2 and Zp.d. = 1/(k)1/2.
