The correct answer is 64. Need procedure of calculation 258K Byte Flash EEPROM A

Question

The correct answer is 64. Need procedure of calculation

258K Byte Flash EEPROM ATDO VRL VDDA VSSA ATD1 VRL 12K VSSA 4K Byte EEPROM 2 Voltage Regulator PKO XADDR14 PK1' XADDR15' PK2 XADDR16 PPAGE Clock and Perlodlic Intern COP Watc PK4XADDR18 PK5i XADDR19 VSSPLLPLL Reset Clock Monitor ration MODB TEST DORA PTA J1850) TXB RXCAN CAND TXCAN CAN1 TXCAN CAN2 TXCAN CAN3 TXCAN Narrow Bus 8 Internal Logic 25V O Driver 5 PWMO PWM1 VSS1,2 VSSX A/D Converter 5V & Voltage Regulator Reference PWM4 VSSPLL PWM7 Voltage Regulator 5V & O SPI1

Explanation / Answer

Multiplexed Address/Data Lines on Port A & Port B provides address lines ADDR0 - ADDR15.

Hence address bus is 16 bits and total adressing capability will be 216 = 64 KB memory locaions

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