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pulse generation problem

I stumble a bit on the following design problem. Should be easy for you folks!!

I have a signal coming from one half of a 4538A monostable that generates

15uS pulses at 30uS interval. From this, I want to generate a second pulse which starts at the _end_ of a 15uS pulse and ends only at the _beginning_ of the next 15uS pulse.

However, sometimes the next 15uS pulse can be delayed by as much as 5mS, in that case the second pulse should also last that long, that is, until the beginning of the next 15uS pulse.

Note that the other half of the 4538 is free to use.

Any suggestions or ideas to put me on the right track are welcome.

Fred

Reply to
Fred
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I knew I forgot a important detail :(

The second pulse MUST be triggered by the transition at the end of a 15uS pulse.

Fred

Reply to
Fred

Translating... :) Source: Tp = 30uS Ton = 15uS

Function required: Neg edge trigger&reset on pos edge

Solution: Use inverter??

That's so easy it's gotta be wrong.....must be misreading this somewhere... :) D from BC

Reply to
D from BC 

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The 4538 can be triggered by a low-going edge, so use the low-going
edge of 4538A Q to trigger 4538B, set RC of 4538B to be longer than
5ms, and then when 4538A Q goes high again differentiate the output
of 4538A Qbar and use that spike to reset 4538B.
Reply to
John Fields

AC-couple the pulse signals into a schmitt inverter, with a 30uSec input-coupling time constant, discharging to the positive rail.

At start-up, you might get a false 'second' pulse, unless the rise-time of the supply is limited to below the 30uSec charging constant. Local RC supply decoupling can prevent this.

RL

Reply to
legg

Unless your description is wrong (or I misread it) I'm with D from BC on this one. Just invert what you already have.

--
Gibbo

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Reply to
Gibbo

I just looked at the datasheet. It's already got a Q bar output which gives you what you want. Or at least what you asked for.

--
Gibbo

This email address isn\'t real.
Reply to
Gibbo

Set and Reset Flip Flop ? sounds simply to me.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
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Reply to
Jamie

Me too. Inverter fits his needs - or we're both missing the point.

Ed

Reply to
ehsjr

Yep! That's the answer!

Even though an inverter may seem to be the obvious answer as other people said, it isn't. It's a bit tricky to explain but in the absence of first pulse, there must be no second pulse. That is, _both_ must be low. That's why the second pulse must start by the transition at the end of the first pulse (negative or positive, doens't matter as 4538's got Q and Qbar outputs).

Thanks guys! Fred

Reply to
Fred

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