This set of Digital Electronic/Circuits Problems focuses on “Programmable Read Only Memory-2”.
1. Silicon links are made up of _____________
a) Polycrystalline silicon
b) Polycrystalline magnesium
d) Silicon dioxide
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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 _____________
d) Both 1s and 0s
Explanation: 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 implementation of PROM, which IC is used?
a) IC 74187
b) IC 74186
c) IC 74185
d) IC 74184
Explanation: 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
Explanation: 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?
Explanation: For n address bits, the memory location will consist of 2n bits. Using 18 address bits, 2^18 = 262,144 (= 256 K) words are addressed.
9. How many address bits are needed to operate a 2K * 8-bit memory?
Explanation: 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?
Explanation: For n address bits, the memory location will consist of 2n bits. 1024 = 210. So, 210 * 23 = 1024 * 8 = 8192 bit.
Sanfoundry Global Education & Learning Series – Digital Circuits.
To practice problems on all areas of Digital Electronic Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.