# VLSI Questions and Answers – Optimization of Inverters-2

This set of VLSI test focuses on “Optimization of Inverters-2”.

1. 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
View Answer

Answer: c
Explanation: Rise time and fall time can be equalized by taking Lp = Ln = 2λ which implies Wp/Wn = 2 and also µn/µp = 2.
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2. Equalizing of rise time and fall time is possible in
a) nMOS
b) pseudo nMOS
c) CMOS
d) pMOS
View Answer

Answer: c
Explanation: 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.

3. 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
View Answer

Answer: a
Explanation: High and low noise margins can be equalized by choosing βn = βp, also Ln = Lp which implies Wp/Wn = 2.

4. Inverter pair delay D is given as equal to
a) tr
b) tf
c) tr-tf
d) tr+tf
View Answer

Answer: d
Explanation: 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.

5. For minimum D consider
a) Ln = Lp = 2λ
b) Ln greater than Lp = 2λ
c) Lp greater than Ln
d) Lp = 2Ln
View Answer

Answer: a
Explanation: 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).
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6. Different parameter optimization is easily achievable in
a) nMOS
b) pMOS
c) pseudo nMOS
d) CMOS
View Answer

Answer: d
Explanation: Different parameter optimizations like noise margins equalization, rise time fall time equalization can be easily achievable in CMOS.

7. Minimizing A with respect to Wp.d. gives
a) Wp.d. = 2λ
b) Wp.d. = λ/2
c) Wp.d. = (k)1/2 x 2λ
d) Wp.d. = k x (λ)1/2 x 2
View Answer

Answer: c
Explanation: Minimizing A with respect to Wp.d yields a solution as Wp.d. = (k)1/2 x Wp.u. = (k)1/2 x 2λ.

8. Using Zp.u./Zp.d = k, Lp.u. can be obtained as
a) k x 2λ
b) k x λ
c) (k)1/2 x 2λ
d) k x 2 x (λ)1/2
View Answer

Answer: c
Explanation: Using this ratio Zp.u./Zp.d. = k, we obtain Lp.u. = (k)1/2 x Lp.d. = (k)1/2 x 2λ.

9. Minimum area can be given as
a) 4 x Ao x λ x (k)1/2
b) 4 x Ao x λ x k
c) 8 x Ao x λ2 x (k)1/2
d) 8 x Ao x λ x (k)1/2
View Answer

Answer: c
Explanation: 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.
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10. When Zp.d. or Zp.u. increases, delay
a) increases
b) decreases
c) remains the same
d) delay becomes zero
View Answer

Answer: a
Explanation: Pd is minimized by increasing Zp.d.. Large Zp.d. requires large Zp.u. which results in increase in delay D of the inverter pair.

11. For minimum D which relation is choosen?
a) Zp.u. = 1/2k
b) Zp.u. = k
c) Zp.d. = 1/k
d) Zp.d. = 1
View Answer

Answer: c
Explanation: For minimum D, Zp.u. is 1 and Zp.d. is equal to 1/k with Wp.u. = 2λ and Wp.d. = k x 2λ.

12. Noise margin measures the changing strength of
a) input voltage
b) output voltage
c) threshold voltage
d) supply voltage
View Answer

Answer: a
Explanation: Noise margin measures by how much the input voltage can change without disturbing the present logic output state.

13. Which has better noise margins?
a) nMOS
b) pMOS
c) CMOS
d) BiCMOS
View Answer

Answer: c
Explanation: CMOS has better noise margins than nMOS especially at low conditions because ratio adjustment is easier in CMOS.
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advertisement Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Instagram | Facebook | Twitter