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From: James Meyer
Subject: Re: Improving Rise/Fall times on Opto
References: <firstname.lastname@example.org> <email@example.com>
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Date: Fri, 06 Dec 2002 23:21:19 GMT
NNTP-Posting-Date: Fri, 06 Dec 2002 18:21:19 EST
On Fri, 06 Dec 2002 22:27:29 GMT, thorin wroth:
>On 06 Dec 2002, you wrote in sci.electronics.design:
>> "Sharpness of the pulse" was the only thing I read in the
>> message that sounded like what he wanted to improve. You're right, if
>> there is also a problem with the output pulse edge not lining up with
>> the input pulse edge quickly enough, then improving the "sharpness"
>> won't help the problem.
>The propagation delay is of no real concern. Im driving the input to the
>opto from an RC receiver via a switched current source to maintain the
>LED current at a fixed value.
>My real concern is the output swictching. Im using the emitter-follower
>constructed to drive the input of an Atmel AVR MCU. Timing is quite
>crictical because I am measuring the pulse-width of the input signal to
>get a speed demand value.
>The resolution I need across the pukse is about 8uS. With a rise/fall
>timein the opto of about 2uS at best Im concerned about the point at
>which the micro will switch at points when the input pulsewidth is
>borderlining between each 8uS timeslice..this is my reason for wishing
>the rise/fall to be as square as possible:) I also need minimal
>component coun and least $$ too:) anyone suggest a better way given the
OK. Since delay is no problem, I would look at putting an edge detector
before the opto and only pass a narrow pulse through the opto with each
transition, H-L and L-H, of the input signal. That effectively gives you a
pulse rate doubler with each H-L transition after the opto representing either a
H-L or a L-H transition of the input.
A classic edge detector is a 2 input XOR gate with the signal connected
to one input and the signal passed through an RC delay connected to the other
input. Then square up the pulse after the opto before it goes into the micro.
All the delays will be the same. H-L and L-H. So the result will be
just a shift in time while preserving what you really want to measure, the duty
The only drawback is the fact that you lose the original sense of the
input waveform. In other words, if the input is 50% then a change to 45% high
would look exactly like a change to 45% low. If the range of duty cycles is
always either less than or more than 50%, then the micro can figure out what's
going on quite nicely.
Jim "Do I win a prize?" Meyer
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