# pulse generation problem

Discussion in 'Electronic Design' started by Fred, Feb 12, 2007.

1. ### FredGuest

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

2. ### FredGuest

I knew I forgot a important detail

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

Fred

3. ### D from BCGuest

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

5. ### leggGuest

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

6. ### GibboGuest

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

7. ### GibboGuest

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.

8. ### JamieGuest

Set and Reset Flip Flop ?
sounds simply to me.

9. ### ehsjrGuest

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

Ed

10. ### FredGuest

Yep!

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