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Old 09-17-2010, 04:28 AM   #1
theflyer
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Easy Costruction and Installation of a FUEL GAUGE

Hi,

After my previous post where I described my simple circuit to display alternator charge status (see Gadgets & Farkles section) I will now present another of my achievements. This is an easy installation (without modify the bike tank) FUEL GAUGE.

(All new alert indicators are placed on a small additional panel with three high-brightness 3mm red LEDs, see image).
















Any comment is welcome !!!



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Old 09-17-2010, 06:06 AM   #2
theflyer
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Re: Easy Installation FUEL RESERVE Indicator

The third LED is used for indication of oil pressure and will be described in the next post.
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Old 09-17-2010, 07:11 AM   #3
Viirin
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Re: Easy Costruction and Installation of a FUEL GAUGE

Thats deadly and really looks like it should be on the bike
Good job man



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Old 09-17-2010, 07:22 AM   #4
blaine
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Re: Easy Costruction and Installation of a FUEL GAUGE

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Originally Posted by Viirin
Thats deadly and really looks like it should be on the bike
Good job man

:plus1: Excellent.Awesome job. :lol:
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Old 09-17-2010, 03:50 PM   #5
Water Warrior 2
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Re: Easy Costruction and Installation of a FUEL GAUGE

Absolutely brilliant.



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Old 01-24-2014, 01:00 AM   #6
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For anyone (like me) who didn't know what NTC was, it apperas to be a type of thermistor (that is, thermal resistor) with an inverse relationship between resistance and temperature (temp up, resistance down, the NTC stands for Negative Temperature Coefficient).
http://www.cornerstonesensors.com/Ab...hat&Print=Page
http://en.wikipedia.org/wiki/Thermistor#NTC
Googling, I can see that this is a fairly common use of NTC Thermistors, but why does it work? Is the liquid gas that much cooler than the air above it?
Also, the Flyer says that it's a 3.5 kOhm NTC... is that the R25 or what?
What ever happened to the Flyer? It looks like he never got around to posting the oil sensor instructions....
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Old 01-24-2014, 03:14 AM   #7
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It has been over 3 years so he likely found something else to ride or moved away. Riders come and go.
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Old 01-24-2014, 09:35 AM   #8
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Oh, sure, (sort of) answer the easy question, but not the ones I'm really curious about. I guess I could do a more intensive googling for those though.

Last edited by ImaginativeFig; 01-24-2014 at 09:37 AM. Reason: I keep putting my sort of in the wrong place
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Old 02-14-2014, 06:14 PM   #9
5th_bike
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Quote:
Originally Posted by ImaginativeFig View Post
why does it work? Is the liquid gas that much cooler than the air above it? ....
No, but as long as the NTC is in the gas (fluid), it is liquid cooled. Once it is out of the gasoline, it is vapor cooled. The current going through it, warms it up. Since it is not cooled by liquid anymore, it becomes much warmer than when it would be submerged. Then, more current will flow through, it will become even warmer etc. and the LED lights up.

I wouldn't want some hot resistor in my gas tank, though...

And I have no idea what the resistors in series are for (parallel to the LED branch), and I wonder what the resistor parallel to the LED is for, that one is probably to keep the LED from gently shining when the NTC is still submerged.

Last edited by 5th_bike; 02-14-2014 at 06:17 PM. Reason: typo
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Old 02-25-2014, 09:24 AM   #10
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Oh, that is an interesting consideration! Do you think it would get hot enough to cause potential ignition? Geez, I hope that's not why the flyer stopped posting.

Whenever I'm not in a class that actively uses circuits, I get a bit fuzzy on electrical stuff...

I found a post on another forum http://www.diystompboxes.com/smfforu...prev_next=prev that explains R4
Quote:
I don't know what that specific circuit does, but I do know what mounting an LED on a voltage divider does.

Assume a circuit with two 1K resistors in series, and an LED parallel with the 1K to ground.

As you increase the voltage across the two resistors from 0V what happens?

Initially, the LED does not have enough voltage to conduct, so it's effectively not there. The voltage at the middle of the two resistors goes up linearly at 1/2 the voltage applied to the top resistor.

Then the voltage at the divider gets to the LED turn-on voltage. Now the LED starts sucking current. But how much?

The LED forward voltage is roughly constant. Not exactly, but close. So with a constant voltage across the lower 1k, its voltage remains constant by the same "roughly". Now as you increase voltage, all of the increased voltage appears across the top 1K resistor, and the current is ((Vsupply-VLED)/1K) + (VLED/1K) The first term being the variable current in the top 1K resistor, the second being the constant current in the bottom resistor.

It's a turn-on delayer. The LED turns on at twice the voltage it otherwise might.
I think that when the NTC is submerged (temp down, res up), the current will run primarily through the R1/R2 leg (and I figure there are two used just to get the right value?) so actually, that leg is also a turn on delayer (right?).
The combination of R3 and the unsubmerged NTC must be supposed to draw the appropriate amount of voltage for the LED, though I'm a bit confused why R3 wouldn't be before they LED, as that's where I understand voltage limiting resistors are supposed to go, but then then again, as I said, I'm pretty fuzzy on this. I did fail my circuits class the first time around.

Last edited by ImaginativeFig; 02-25-2014 at 09:31 AM.
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