Newbie question: EMDR project

Snit · 2011-05-21T15:34:26+00:00

No idea how to proceed with making what is likely a pretty simple device,but I have *no* background in electronics. OK, here is what I want:Two handheld cell-phone style vibrators that "take turns" vibrating, eachconnected to a central control device which allows changing the speed of theswitching back and forth.As a bonus, but not needed, it would be good to have a way to connectheadphones which would be able to play some sound that is timed to the sameback-and-forth behavior: when the left "vibrator" (is there a better word!)is active, a sound is played for the left ear; same thing for the right.I suspect for someone who has a clue this is not that hard to make... butfor me: completely out of my league. Any ideas? Wrong group?-- [INSERT .SIG HERE]

May 21, 2011 26 Replies

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Phil Hobbs 15 years ago

It doesn't work well except with CMOS 555s, and then only when the output is lightly loaded. The logic levels of the bipolar part don't track the power supply voltage, so the supply rejection goes south.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

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John Fields 15 years ago

The problem isn't that the logic levels of the bipolar part don't track the supply voltage, because they do. They _have_ to because the references for the threshold and trigger comparators are generated by a three-resistor voltage divider connected between Vcc and ground with one of the comparators' inputs connected to the two junctions. That way the references are ratiometric and are _always_ 1/3 Vcc and 2/3 Vcc. The problem is that if the bipolar part is operated at a low voltage, Vce(sat) of the upper totem-pole emitter follower can fall below 2/3Vcc, making it impossible to switch the front end.

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Phil Hobbs 15 years ago

_Logic_ levels, not comparator thresholds--the output high and low levels are not too close to the rails on the bipolar part, which makes the ratiometric property fail to work.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

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John Fields 15 years ago

But the output _does_ track the supply in that for any given current out, the output voltage will vary as the supply does. It's exactly the same with CMOS but, since the CMOS output is closer to the rails, it extends the working range of the ratiometric property.

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Phil Hobbs 15 years ago

You're trumping up a silly semantic issue by foisting your definition of 'track' where it doesn't fit. That winds up being nonsense, as you say, but it's your nonsense and not mine.

If the output high level is, say, two VBEs below the positive supply, the charging current and the upper comparator threshold won't track each other as the supply voltage changes. That makes the frequency and duty cycle supply voltage dependent.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

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Snit 15 years ago

John Fields stated in post snipped-for-privacy@4ax.com on

5/24/11 1:25 PM:

Thanks!

[INSERT .SIG HERE]

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John Fields 15 years ago

Really? If the supply voltage changes and the output voltage of the logic follows the change, how is that not "tracking"?

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