Subject:
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Re: Question?
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Newsgroups:
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lugnet.robotics
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Date:
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Mon, 4 Feb 2002 05:24:37 GMT
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Original-From:
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T. Alexander Popiel <popiel@wolfskeep.#Spamcake#com>
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Viewed:
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461 times
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In message: <200202031206_MC3-F079-6BD5@compuserve.com>
PeterBalch <PeterBalch@compuserve.com> writes:
> > When set to use "active" sensors, the
> > inputs actually output power most of the time, switching briefly to
> > input mode (no power output) to take a reading.
>
> What I don't understand is how this works if the sensor is connected
> backwards.
>
> Sure, a bridge rectifier will get the power sorted out but how does the
> sensor send back a voltage in the right range? If say "light" is 8V and
> "dark" is 0V then when the sensor is connected backwards, "light" will
> be 0V and "dark" will be 8V.
My guess (having never looked at the schematics, or taken one apart)
is that when connected backwards, "dark" would be 0V and "light" would
be -8V. Again, a simple bridge rectifier (this time in the RCX) would
get it sorted out.
It might be simpler to look at it in terms of current flows; "dark"
is no current, and "light" is full current, regardless of polarity.
- Alex
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Message has 1 Reply:
 | | Re: Question?
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| (...) No, there's no bridge in the RCX, just a pull-up to 5V (10kOhm IIRC). If you look at the schematics (say at (URL) see that there is another bridge rectifier in the sensor between ground and the raw sensor output. The ground half of the bridge (...) (23 years ago, 4-Feb-02, to lugnet.robotics)
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Message is in Reply To:
| | Re: Question?
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| (...) What I don't understand is how this works if the sensor is connected backwards. Sure, a bridge rectifier will get the power sorted out but how does the sensor send back a voltage in the right range? If say "light" is 8V and "dark" is 0V then (...) (23 years ago, 3-Feb-02, to lugnet.robotics)
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