"For a
CMOS gate operating at a power supply voltage of 5 volts, the
acceptable input signal voltages range from0 volts to 1.5 volts for
a “low” logic state, and 3.5 volts to 5 volts for
a “high” logic state."
HIGH = 1 = ON
LOW = 0 =
OFF
Binary 1 = +3.5 - +5 VDC = ON = HIGH
Binary 0 = 0 VDC - + 1.5 VDC = OFF = LOW
128|064|032|016|008|004|002|001
000|000|000|000|000|000|000|000 = 0
000|000|000|000|000|000|000|001 = 1
("One" bit is turned on)
000|000|000|000|000|000|001|000 = 2
("Two" bit is turned on)
000|000|000|000|000|001|000|000 = 4
("Four" bit is turned on)
000|000|000|000|001|000|000|001 =
9 ("One" bit + "Eight" bit turned on) (1+8)
000|001|000|000|001|000|000|000 =
72 ("Sixty Four" bit + "Eight" bit turned on)(64+8)
If you were to read the voltages in a memory chip address,
there would be a series of 1's & 0's, with the 1's represented by a voltage
of +3.5 to +5 volts, and the zeroes represented by a voltage of 0
to + 1.5 Volts. You would need a REALLY tiny voltmeter probe, but that's
what would be in the chip.
Ones are ON, zeroes are OFF.
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