250+ TOP MCQs on Programmable Read Only Memory – 2 and Answers

Digital Electronic/Circuits Problems on “Programmable Read Only Memory-2”.

1. Silicon links are made up of _____________
A. Polycrystalline silicon
B. Polycrystalline magnesium
C. Nichrome
D. Silicon dioxide
Answer: A
Clarification: Metal links are made up of Nichrome materials. Silicon links are made up of polycrystalline silicon.

2. During programming p-n junction is _____________
A. Avalanche reverse biased
B. Avalanche forward biased
C. Zener reverse biased
D. Zener reverse biased
Answer: A
Clarification: The sudden heavy flow of electrons in the reverse direction and heat cause aluminium ions to migrate. So, during programming p-n junction is avalanche reversed biased.

3. The full form of FAMOS is _____________
A. Floating Gate Avalanche Injection MOS
B. Float Gate Avalanche Injection MOS
C. Floating Gate Avalanche Induction MOS
D. Float Gate Avalanche Induction MOS
Answer: A
Clarification: The full form of FAMOS is Floating Gate Avalanche Injection MOS. It is a floating gate transistor in which the trapped electrons is responsible for the dropping of the voltage.

4. PROM is programmed by _____________
A. EPROM programmer
B. EEPROM programmer
C. PROM programmer
D. ROM programmer
Answer: C
Clarification: PROM is programmed by plugging it into a special device called PROM programmer. The ROM cannot be clear and hence PROM is a one-time programmable device.

5. The PROM starts out with _____________
A. 1s
B. 0s
C. Null
D. Both 1s and 0s
Answer: B
Clarification: PROM is a one-time programmable device, which is programmed by the user. The PROM starts out with all 0s. These current pulses blow the fuse links, thus creating the desire pattern.

6. For the implementation of PROM, which IC is used?
A. IC 74187
B. IC 74186
C. IC 74185
D. IC 74184
Answer: B
Clarification: For implementation of PROM, IC 74186 is used. IC 74186 is of 512 bits (62 * 8 = 512). Thus, it has 62 rows and 8 columns.

7. IC 74186 is of ______________
A. 1024 bits
B. 32 bits
C. 512 bits
D. 64 bits
Answer: C
Clarification: IC 74186 is of 512 bits (62 * 8 = 512). Thus, it has 62 rows and 8 columns.

8. How many memory locations are addressed using 18 address bits?
A. 165,667
B. 245,784
C. 262,144
D. 212,342
Answer: C
Clarification: For n address bits, the memory location will consist of 2ⁿ bits. Using 18 address bits, 2¹⁸ = 262,144 (= 256 K) words are addressed.

9. How many address bits are needed to operate a 2K * 8-bit memory?
A. 10
B. 11
C. 12
D. 13
Answer: B
Clarification: For n address bits, the memory location will consist of 2ⁿ bits. Thus, for 2K, only 11 address bits are required, because 2¹¹ = 2K.

10. What is the bit storage capacity of a ROM with a 1024 × 8 organization?
A. 1024
B. 4096
C. 2048
D. 8192
Answer: D
Clarification: For n address bits, the memory location will consist of 2ⁿ bits. 1024 = 2¹⁰. So, 2¹⁰ * 2³ = 1024 * 8 = 8192 bit.

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