and 48-pin
TSOP packages. The word-wide data (x16) appears on
DQ15

The device requires only a single 3.0 volt power supply
for both read and write functions. Internally generated
and regulated voltages are provided for the
program and erase operations. ;) ;) :) :(

The device offers access times of 70, 80, 90, and 120 ns,
allowing high speed microprocessors to operate
without wait states. :D :D :cool:

The device requires only a single 3.0 volt power supply
for both read and write functions. Internally generated
and regulated voltages are provided for the
program and erase operations. :P

The device offers access times of 70, 80, 90, and 120 ns,
allowing high speed microprocessors to operate
without wait states. :o :cool: :mad: :o

The device requires only a single 3.0 volt power supply
for both read and write functions. Internally generated
and regulated voltages are provided for the
program and erase operations. ;) ;) :) :( :P :) :D

The device offers access times of 70, 80, 90, and 120 ns,
allowing high speed microprocessors to operate
without wait states. :D

and 48-pin
TSOP packages. The word-wide data (x16) appears on
DQ15

Write cycles also internally latch addresses and data
needed for the programming and erase operations.
Reading data out of the device is similar to reading
from other Flash or EPROM devices. :D :o :o :cool: :o

The device requires only a single 3.0 volt power supply
for both read and write functions. Internally generated
and regulated voltages are provided for the
program and erase operations. ;) ;)

Commands are written to the command register
using standard microprocessor write timings. Register
contents serve as input to an internal state-machine
that controls the erase and programming circuitry. :D ;) :P :mad: :mad: :o :cool:

Write cycles also internally latch addresses and data
needed for the programming and erase operations.
Reading data out of the device is similar to reading
from other Flash or EPROM devices. :D :cool:

int calcrc(char *ptr, int count)
   {
   int  crc = 0;
   char i;
  
   while (--count >= 0)
   {
       crc = crc ^ (int) *ptr++ << 8;
       i = 8;
       do
       {
           if (crc & 0x8000)
             crc = crc << 1 ^ 0x1021;
           else
             crc = crc << 1;
       } while(--i);
     }
     return (crc);
  }

int calcrc(char *ptr, int count)
   {
   int  crc = 0;
   char i;
  
   while (--count >= 0)
   {
       crc = crc ^ (int) *ptr++ << 8;
       i = 8;
       do
       {
           if (crc & 0x8000)
             crc = crc << 1 ^ 0x1021;
           else
             crc = crc << 1;
       } while(--i);
     }
     return (crc);
  } :D

int calcrc(char *ptr, int count)
   {
   int  crc = 0;
   char i;
  
   while (--count >= 0)
   {
       crc = crc ^ (int) *ptr++ << 8;
       i = 8;
       do
       {
           if (crc & 0x8000)
             crc = crc << 1 ^ 0x1021;
           else
             crc = crc << 1;
       } while(--i);
     }
     return (crc);
  } :D :D

int calcrc(char *ptr, int count)
   {
   int  crc = 0;
   char i;
  
   while (--count >= 0)
   {
       crc = crc ^ (int) *ptr++ << 8;
       i = 8;
       do
       {
           if (crc & 0x8000)
             crc = crc << 1 ^ 0x1021;
           else
             crc = crc << 1;
       } while(--i);
     }
     return (crc);
  } :D :D :D

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当有人回复这个主题时 :D通过email通知你 :) :D

当有人回复这个主题时 :D通过email通知你 :) :D :P