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mrnasty
Fordmods - Getting Side Ways
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 Posted: Wed Dec 29, 2004 7:02 pm Post subject: adding a light to my shift kit, problem, need a solution. |
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I added in a little red LED to know when my shift kit is on, i wired in series to one of the cables, and the shift kit is on and so is the light but problem is it draws to much of the current out of the cable, and by the time it gets back to the tranny theirs not enough current to make the shift noticibly harder/quicker.
If i use a bigger gauge wire, will this eliminate the problem? i know in domestic electrical thats usually is the solution (im a domestic sparky) but im unsure on how much current is actually produced by the car. |
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Pulco
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| Posted: Wed Dec 29, 2004 7:38 pm Post subject: |
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| cant u work it out? using ohms law as to how much current is drawn? or just chuck an ammeter on it, just a multimeter would do cause i dont think it would be a huge current, well it an only be a max of 30 amps or somethin flowing cant it because of the size of the fuses?? |
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data_mine
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| Posted: Wed Dec 29, 2004 7:42 pm Post subject: |
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The shift kit is built to handle AMPs of currnet, while a LED will take 50mA at most (and that'd be overdriving most of 'em).
Me thinks maybe something else is wrong.
I've thought about adding a bi colour LED, so I get green/red display for off/on (on/off ?) But hadn't gone much past thinking about it, and putting it in my 'Projects TODO' list. |
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Disco Frank
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| Posted: Wed Dec 29, 2004 8:33 pm Post subject: |
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nothign to do with current the kit works on resitance the LED has resticance in it, you need to take that into account
the best way to add a on/off ligght is to get a new double pole on/on switch
ie a switch that has 2 sets of poles on each end i am sure u know what i mean
like this!
then wire the kit up to one set of poles
MAKE SURE U WIRE THE KIT THE SAME IE THE MIDDLE WIRE MUST GO TO THE MIDDLE POLE
then run a +ve wire to one of the unused poles,
then run a wire from the 2nd unused pole to the led and then to earth |
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brett351
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| Posted: Wed Dec 29, 2004 10:33 pm Post subject: |
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| yeh what he said, or put a two pole relay in and use one of the poles on that. Either way you arre going to have to get the LED out of the CCT. |
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Andrew J
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| Posted: Wed Dec 29, 2004 11:32 pm Post subject: |
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| Or you could replace the resistor in the shift kit it self. |
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South
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| Posted: Wed Dec 29, 2004 11:53 pm Post subject: |
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Ahhhh Frank, You need to learn a bit more about electronics and physics my friend 
"Ohm's law is the most important, basic law of electricity."
"Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed."
One can then equate that by increasing the resistance in the circuitry the current flow will then be decreased. As the LED itself would not have a great deal of resistance the shiftkit should not be affected as you describe.
I can only assume you've installed the LED incorrectly. The easiest way to install an LED to indicate the kit is on (like you need it anyway, its pretty obivious) would be to use the switch Frank displayed. Dont forget you need a resistor for the LED otherwise it will burn out nice and quickly... |
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dang
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| Posted: Fri Dec 31, 2004 8:23 am Post subject: |
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Ahhh, South, you need to learn a little more about electronics....
A LED does not have 'resistance' as such, it is a nonlinear device. Current drawn will depend upon voltage across it. Reverse bias, current will be bugger all until you exceed the reverse breakdown voltage. Forward bias, voltage doesnt change much, which is why we limit the current with a resistor. Can't remember the diode equation off hand, but it's not important right now.
Mr Nasty, wired in series WILL NOT WORK. Because solenoid 5 needs hundreds of milliamps, running this through a LED will make it go pop.
Can't remember the shift kit cct, but you would definitely want to wire it in parallel some how. But definitely the double pole switch is the most efficient way to do it.
Dang. |
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mrnasty
Fordmods - Getting Side Ways
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| Posted: Fri Dec 31, 2004 12:33 pm Post subject: |
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ill have a fiddle and see what i can come up with. im not a huge fan of small fiddly electronics, thats why i wire houses/factorys/shops instead.
thanks fellas |
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cougar30
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| Posted: Fri Dec 31, 2004 1:03 pm Post subject: |
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| Disco Frank wrote: | nothign to do with current the kit works on resitance the LED has resticance in it, you need to take that into account
the best way to add a on/off ligght is to get a new double pole on/on switch
ie a switch that has 2 sets of poles on each end i am sure u know what i mean
like this!
then wire the kit up to one set of poles
MAKE SURE U WIRE THE KIT THE SAME IE THE MIDDLE WIRE MUST GO TO THE MIDDLE POLE
then run a +ve wire to one of the unused poles,
then run a wire from the 2nd unused pole to the led and then to earth |
thie switch with double poles is the easy way to go ...and cheep too |
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South
Fordmods - Smokin em up
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| Posted: Fri Dec 31, 2004 7:51 pm Post subject: |
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Dang: Here is a quote from Introductory DC/AC Electronics
"When forward biased, a P-N junction will act as a conductor and have a low but finite resistance value that will cause a corresponding voltage drop across its terminals. The forward voltage drop is approximately equal to the P-N junctions barrier voltage.
Forward Voltage Drop for Silicon = 0.7V
Forward Volatge Drop for Germanium = 0.3V"
Now tell me again that an LED, a semiconductor diode, does not have resistance.
Now who is it that needs to learn more about electronics? If your having problems understanding some of this, I will quite happily offer a tutoring service, $35/hr shall cover it... |
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Tomek
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| Posted: Fri Dec 31, 2004 8:02 pm Post subject: |
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| South wrote: | Dang: Here is a quote from Introductory DC/AC Electronics
"When forward biased, a P-N junction will act as a conductor and have a low but finite resistance value that will cause a corresponding voltage drop across its terminals. The forward voltage drop is approximately equal to the P-N junctions barrier voltage.
Forward Voltage Drop for Silicon = 0.7V
Forward Volatge Drop for Germanium = 0.3V"
Now tell me again that an LED, a semiconductor diode, does not have resistance. |
some of that info is really more then just introductory (prob. for uni its introductory) but i didnt no some of the stuff you wrote there and i did physics. |
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South
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| Posted: Fri Dec 31, 2004 8:08 pm Post subject: |
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Yeh High School Physics is s**t! You learn Ohms Law in the electrical/electronics side of things, and thats about it. Then you go off to TAFE/UNI and they start going on about this n that and your like WTF!!
Its a good book that one, definately worth every cent if your looking at going down the electronics road. |
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dang
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| Posted: Sat Jan 01, 2005 10:32 am Post subject: |
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South, just a friendly ribbing. Don't mean to offend. After 4 years doing electronics engineering at uni, I'll pass on the offer of tutoring. (BTW, $35 an hour is quite cheap - would I get what I pay for!!!)
Depends on your definition of resistance. The 'effective' resistance of a diode is defined as the forward voltage drop divided by the junction current. As the voltage across the diode will be much the same regardless of the forward bias current, the effective resistance will depend on the forward current.
Not sure why, but a LED drop is closer to 1V. So a LED at 20mA has an effective resistance of 1V/20mA = 50 ohms. At 10mA, we are up to 100 ohms.
The general definition of resistance is the slope of the V/I line of a linear device. Because a LED does not have a linear curve, it does not have a fixed resistance.
Your introductory text uses the term resistance to assist readers to grasp the concept.
However, I'm sure we both agree that the DPDT switch is tha answer to MrNastys problem.
Let's leave it at that, or I'll be forced to drag some of my texts out of mothballs and quote them.
BTW, Horowitz and Hill is highly recommended.
Dang |
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chr!s
Fordmods Stock as a Rock
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| Posted: Mon Jan 03, 2005 9:21 am Post subject: |
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bloody hell, i hope arguments dont go on like this on a cruise, youll get beaten up...
"high school physics is better!"
"No, tafe physics is better!"
Bahahahah... |
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South
Fordmods - Smokin em up
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| Posted: Mon Jan 03, 2005 1:16 pm Post subject: |
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I guess Ch!is failed Physics and English, seeing as he hasnt comprehended the simple conversation...
d**k! |
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chr!s
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| Posted: Thu Jan 06, 2005 4:32 pm Post subject: |
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| actually i passed and now im a teacher and i got just the reaction i wanted, thanks. nerd |
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mrnasty
Fordmods - Getting Side Ways
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| Posted: Wed Jan 12, 2005 6:29 pm Post subject: |
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| what the buggery is +VE, im assuming the V is voltage and the plus is positive (active in my terms) but what is the E? |
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South
Fordmods - Smokin em up
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| Posted: Wed Jan 12, 2005 11:42 pm Post subject: |
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| chr!s wrote: | | actually i passed and now im a teacher and i got just the reaction i wanted, thanks. nerd |
Oh, your a teacher... That explains it 
Whats your definition of being a 'nerd'?
Isn't that one who spends long time with head in book... Wait a minute, thats you... OH MY GOLLY GOSH!
spanker!
| Quote: | | what the buggery is +VE, im assuming the V is voltage and the plus is positive (active in my terms) but what is the E? |
Do not use active as a term to refer to a positive voltage. A device may be active in a low state, ie when you apply ground to it the device activates. You will find people refer to these as an Active High or Active Low. +VE is obviously a positive source, whereas -VE is ground or negative voltages...
That is a very basic run down, im sure some smart a*** gimp is going to say im wrong or is going to out do me |
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mrnasty
Fordmods - Getting Side Ways
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| Posted: Sat Jan 15, 2005 11:33 am Post subject: |
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only reason I use the term active is cause im a domestic sparky and theres no such thing as positive/negative in what I do,
just active/neutral/earth.
And none of this 12V stuff, 240/415V shall do me just fine thank you.  |
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