Re: A Quick Question
- From: Charlie Siegrist <none.active@xxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 14 Jun 2008 13:09:36 -0500
On Wed, 11 Jun 2008 15:48:13 +0000, TomO wrote:
Maybe some of you knowledgeable folks have run across this before.
Not I, but I am insatiably curious when it comes to electricity.
Interesting discussion so far. Snip.
The question is whether or not this sensor has anything to do with the
regulator, or is it only there for the idiot lamp on the panel?
Probably the latter.
Should I
try to drill a hole in my new battery to insert the sensor?
Certainly not an option I would pursue.
I did try grounding and also connecting the sensor tip to +12v, but the
idiot lamp glared on in both of those cases.
Connecting the sensor tip to those voltages will probably yield no
helpful information, as it is not likely a conductor. Here's a link that
describes what you probably have, at least by similarity:
<http://www.osti.gov/energycitations/product.biblio.jsp?osti_id=6277756>
So the existing circuit is likely:
<panel>--<existing resistor>--<sensor as a voltage or current source>
Your tests should be done as such, after removing the sensor:
<panel>--<existing resistor>--<experimental voltages>
You might find in this case that full positive or full negative will
work. The trick then is to make sure that the resultant current will not
overheat the existing resistor. If you find that you must use a voltage
between full positive and full negative, then I would construct the final
circuit thusly:
<panel>--<existing resistor>--<V1>--<new resistor>--<+V>
With the new resistor chosen to give the required voltage at point V1,
and the wattage chosen appropriately. Choose V1 as a midpoint on the
range of voltages that keep the lamp off.
Alternatively, you might want to place the sensor into a known good, old
style, charged battery, measure the voltage at the sensor, then do the
calculations. That's the simple way, if you have the battery.
Ciao!
.
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