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Old 13th March 2008, 03:48 PM   (permalink)
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raviram87 is a jewel in the roughraviram87 is a jewel in the rough
Default analog input to flip flops

i had a doubt on d flip flops...

can i use an analog signal (say a sinusoidal input) as a clock input to a positive edge trigerred flip-flop? or should the clock input necessarily be a digital one?
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Old 13th March 2008, 04:30 PM   (permalink)
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Nigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to beholdNigel Goodwin is a splendid one to behold
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Quote:
Originally Posted by raviram87
i had a doubt on d flip flops...

can i use an analog signal (say a sinusoidal input) as a clock input to a positive edge trigerred flip-flop? or should the clock input necessarily be a digital one?
It needs to be a digital one, feed your input via a schmitt trigger to make it a squarewave.
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Old 13th March 2008, 05:06 PM   (permalink)
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on1aag is a jewel in the roughon1aag is a jewel in the rough
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Hi Raviram87,

Quote:
Originally Posted by raviram87
i had a doubt on d flip flops...
can i use an analog signal (say a sinusoidal input) as a clock input to a positive edge trigerred flip-flop? or should the clock input necessarily be
a digital one?
In theory Nigel Goodwin is right but there are always exceptions.
Compare the datasheets of the CD4013 and the HEF4013.
You will find that the HEF4013's clock input has a schmitt-trigger input
which means that it will accept also signals with a lower rise and fall
time without oscillating. Check also the datasheet of the CD4017 &
CD4022, same thing here. But your signal needs to stay at all times
within limits of the power supply range of the digital part.
Therefore if you want to apply a sinusoidal signal to a digital part
you will have to make shure that it stays within those limits !
Therefore you need to keep the voltage at the input of the digital part
at approx. 1/2 of it's supply voltage with a resistive divider and apply
the sinewave through a capacitor. This will shift the dc level of the
sinewave up to 1/2 of the supply voltage of the digital part.
If the peak-to-peak amplitude of the sinewave is larger than the
supply voltage of the digital part you'll have to attenuate it or you
could also add a resistor in series with the input capacitor to limit
the current through the input protection diodes to a safe value.

on1aag.
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Old 13th March 2008, 11:16 PM   (permalink)
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raviram87 is a jewel in the roughraviram87 is a jewel in the rough
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thanks for the replies.... :-)
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