Solution7
(11.16)
(11.17)
(11.26)
Homework Solutions 8
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Waveform A is low for 60 + 40 + 60 + 40 + 60 = 260 ns
and high for 40 + 60 + 40 + 60 + 40 = 240 ns
Period is (260 + 240) = 500 ns
Frequency = 1/500 = 2 MHz
Percentage of cycle for which output 1s high = 240/500 = 48 %
Check: tp= (60 + 40)/2 = 50 ns
For eleven inverters, there are 2 x 11 = 22 transitions, whose average length 1s
tp = (1/(20 x 10°)) / 22 = 2,27 ns
The bit line capacitance is [n(2) +20] =2n +20 fF
For a 5V supply ad pre-charge to Vpp/2, there is a 2.5 V change on the 50 { F cell
capacitor to produce a 0.1V bit-line signal.
Thus 0.1 = 2.5 (50/(2n + 20 + 50))
or, 2n + 70 = 2.5 x10 x 50 => 2n = 1250 and n = 590 > 512
Thus we can use 512 rows for which log » 512 = 9
or, 9 bits addressing 1s needed.
For a sense amplifier of 5 x greater gain
0.1/5 = 125/(2n+70)
or, 2n + 70 = 125 x 50 = 6250
or, 2n = 6180 and n = 3090 > 2048
To address 2048 rows, we need log ; 2048 = 11 bits
( Page 13
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