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For a MOD-12 counter, the Flip-Flop has a tpd = 60 ns. The NAND gate has a tpd of 25 ns. The maximum clock frequency is given by—
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- 37·74 MHz
- 377·4 MHz
- 3·774 MHz
- None of these
Correct Option: C
We know that Maximum frequency is given by
fmax ≤ | |
ts + n tPd |
where, n = no. of flip-flop
ts = propagation delay of gate
t Pd = propagation delay of flip-flop.
For Mod-12 counter we require 4 flip-flop
fmax ≤ | |
(25 × 10−9 + 4 × 60 × 10−9) |
or fmax ≤ 3·774 MHz
So, maximum frequency = 3·774 MHz
Hence alternative (C) is the correct answer