Anyone know about LEDs???
#17
11-09-2000, 10:11 PM
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GAH, GIJoeCam, FireZap & Fellow F150-ians,
The approximation of 1.3K ohms will work fine but you probably won't find a resistor of that value. The most accurate way to calculate the proper series resistor value for any LED and DC voltage combo is thus... the source or DC voltage (13.8 for automobiles when running and this example) minus the forward voltage of the LED (Typically 1.7v for generic LEDs), that equals 12.1v... now to determine the resistor value, divide that voltage (12.1) by the current rating of the LED (typically 10 milliamps or .010 amps)... that equals 1210 ohms or rounded to 1.2K ohms. That is the ohm's law GAH spoke of. Fixed value resistors are only available in a certain values (don't remember them all now), one of which is 1.2K ohms. A few ohms either way is not usually a problem but if in doubt go with a higher value. Now (GAH) in your drawing (your first posted?) you should not drive the relay coil through the LED (i.e. in series). The relay is rated for 12v and does not need (or want) a resistor in series. Most of those relays have a coil resistance near 100 ohms. Using our previous formulas, the relay "desires" about 120 milliamps (or .120 amps) for proper operation. That's 10 TIMES what the LED is rated for and is not correct. It shouldn't even work (as drawn). To wire in parallel as you desire- Run a wire from the switch directly to the relay and also one to the LED and a 1.2k ohm resistor. (your drawing has the LED in SERIES with the relay) Of course you need to have the LED properly polarized. Usually one lead of the LED is shorter than the other and near a small flat section of the base of the LED. That is the Cathode or negative side of the LED and needs to be connected to the ground. The resistor can go on either lead of the LED since it is a series circuit. To be thorough one must also consider the wattage rating of the resistor. In this case a .250 watt (or 1/4 watt) rating is perfect. Wattage is Current times Voltage (or 12.1 X .010 in our example). That equals .120 watts or 1/8 watt. Half of the rating of the 1/4 watt resistor. This 50% margin is a good design rule and will assure the resistor won't overheat and fail. As FireZap said, Radio Shack sells LED's with the series resistor integrated and that is why they are rated for 12 volts. For your needs those will work fine if wired correctly.
Good Luck!
The approximation of 1.3K ohms will work fine but you probably won't find a resistor of that value. The most accurate way to calculate the proper series resistor value for any LED and DC voltage combo is thus... the source or DC voltage (13.8 for automobiles when running and this example) minus the forward voltage of the LED (Typically 1.7v for generic LEDs), that equals 12.1v... now to determine the resistor value, divide that voltage (12.1) by the current rating of the LED (typically 10 milliamps or .010 amps)... that equals 1210 ohms or rounded to 1.2K ohms. That is the ohm's law GAH spoke of. Fixed value resistors are only available in a certain values (don't remember them all now), one of which is 1.2K ohms. A few ohms either way is not usually a problem but if in doubt go with a higher value. Now (GAH) in your drawing (your first posted?) you should not drive the relay coil through the LED (i.e. in series). The relay is rated for 12v and does not need (or want) a resistor in series. Most of those relays have a coil resistance near 100 ohms. Using our previous formulas, the relay "desires" about 120 milliamps (or .120 amps) for proper operation. That's 10 TIMES what the LED is rated for and is not correct. It shouldn't even work (as drawn). To wire in parallel as you desire- Run a wire from the switch directly to the relay and also one to the LED and a 1.2k ohm resistor. (your drawing has the LED in SERIES with the relay) Of course you need to have the LED properly polarized. Usually one lead of the LED is shorter than the other and near a small flat section of the base of the LED. That is the Cathode or negative side of the LED and needs to be connected to the ground. The resistor can go on either lead of the LED since it is a series circuit. To be thorough one must also consider the wattage rating of the resistor. In this case a .250 watt (or 1/4 watt) rating is perfect. Wattage is Current times Voltage (or 12.1 X .010 in our example). That equals .120 watts or 1/8 watt. Half of the rating of the 1/4 watt resistor. This 50% margin is a good design rule and will assure the resistor won't overheat and fail. As FireZap said, Radio Shack sells LED's with the series resistor integrated and that is why they are rated for 12 volts. For your needs those will work fine if wired correctly.
Good Luck!
#18
11-10-2000, 01:49 PM
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XceLenT Driver,
Thanks for the corrections. I've updated the schematic in my previous post to reflect your comments. I'm a little confused as to why the previous circuit wouldn't work. The LED shown was rated at 0.1A. If you add the resistances (132 + 100) and divide into the voltage you'd have a current of about 57 mA. I'm not disputing your comments, I'm here to learn.
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'99 Ranger 4.0L 4X4 XLT
Thanks for the corrections. I've updated the schematic in my previous post to reflect your comments. I'm a little confused as to why the previous circuit wouldn't work. The LED shown was rated at 0.1A. If you add the resistances (132 + 100) and divide into the voltage you'd have a current of about 57 mA. I'm not disputing your comments, I'm here to learn.
------------------
'00 5.4L 4X4 Lariat (Black/Silver)
'99 Ranger 4.0L 4X4 XLT
#19
11-10-2000, 05:56 PM
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Yes, I believe that you want to limit the
current to around 20 mils for max life of the
led. When they get too bright the junction
inside that emits light overheats and then
acts just like a bulb on overvoltage: It
stops working. The principle is the same
as used to have to be done to the neon NE-2
type lights in that you had to add a resistor
in series to use them as indicator lights on 115 volts AC.
Just rambling. sorry.
------------------
2000 Lariat, 5.4, 3:55 L.S., 4wd, ext cab, Tow pkg, Chestnut/Gold, Century mid-hi cap, Ford short bug deflector, contoured Ford running boards and bedmat.
current to around 20 mils for max life of the
led. When they get too bright the junction
inside that emits light overheats and then
acts just like a bulb on overvoltage: It
stops working. The principle is the same
as used to have to be done to the neon NE-2
type lights in that you had to add a resistor
in series to use them as indicator lights on 115 volts AC.
Just rambling. sorry.
------------------
2000 Lariat, 5.4, 3:55 L.S., 4wd, ext cab, Tow pkg, Chestnut/Gold, Century mid-hi cap, Ford short bug deflector, contoured Ford running boards and bedmat.
#20
11-10-2000, 06:07 PM
I just got the e-mail from the 2 PHD EE's I e-mailed. They both said that all I need to do is step down the input voltage to 3 volts using a simple resistor, and that the intensity will change with the voltage (i.e. dimmable by increasing the resistance). The current does not need to be regulated, per se, but the input voltage does. Now, if I could just get to the bottom of the 12 volt/3volt mystery... The book lists them as 12 volt, but the tech says they are 3 volt.... hmmmmmm... I'l keep y'all posted.
-Joe-
-Joe-
#21
11-10-2000, 08:41 PM
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#22
11-10-2000, 10:03 PM
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GAH,
No problem helping... I owe this site more than I can ever repay.
Perfect schematic this time. Assuming you have properly polarized the LED it will light when the switch is turned on (the relay will work regardless of the LED). The reason it didn't work the first time is because the LED "blew" (or would have) as soon as you turned on the switch. Then the circuit will be broken and the relay is no longer supplied any voltage. Generic LEDs can only handle about 30 milliamps or so before they are damaged. If you wired the LED in backwards nothing would have happened since current will only flow in one direction through a diode (unless the voltage is sufficient to breakdown the PN junction and would again "blow" it, not the problem here). The 57 milliamps you calculated is correct mathematically but that is too much for the LED and probably not enough for the relay. Using I=V/R it needs about 130 milliamps to operate reliably. (13.2/100=.132)
GIJoeCam,
What your friends told you is true but it is not clear that you understand by what you wrote... Current DOES need to be regulated or more accurately stated: "limited". Generally, more than 30-40 milliamps or so and the LED is toast. That is what you are doing by adding resistance, limiting the current. The formula commonly known as "Ohm's Law" (V=IxR) expresses this. It can also be expressed R=V/I and I=V/R. Depending on if you are solving for Voltage-V, Current-I, or Resistance-R. By adding that resistance you are "dropping" some of the voltage across that resistor (when current flows) until the voltage the LED is left with is 1.7 or so. This forward voltage (1.7) varies depending on the type of LED. That is what I used for my earlier example since most are rated as such but your LED may be 3.0v (you mentioned a dual color LED earlier?). To solve for that LED start with the DC voltage (I think the automotive DC voltage when the car is running is about 13.8 volts). That's 13.8v minus the 3.0v the LED needs, that leaves 10.8v. That 10.8v is the amount of voltage the resistor has to "drop". Now we figure in the current using R=V/I. V is 10.8, I is .010 (or 10 milliamps since this is the lowest value you mentioned, for the red section right?) that means R=1080 ohms. But that value is not available in a fixed resistor so you will have to go with 1K or 1.2K ohm resistor. Either will probably be OK but if you want to be safe go with the 1.2K. This will extend the life of the diode beyond that of our trucks (unfortunately). The same holds true again for the size or wattage of the resistor you need. Watt's law can be expressed as: W=VxI, or in this case, 10.8 x.01=.108 or about a 10th of a watt. You could use an 1/8 watt resistor but go ahead and use the 1/4 watt (or bigger) if you have them already. This will insure a trouble-free design. As far as the 12 volt vs. 3 volt dilemma, who knows what the RS guy is talking about, go by what's printed on the package. If the package says it's for 12 volt usage it already has the proper resistor internally so treat it like a light bulb (except it must be properly polarized for it to illuminate). If it says 3 volts then use a 1.2k ohm resistor in series as we calculated. If you hook it up backwards it should not be damaged but it just won't light!
Hope this helps, I'm tired of typing for tonite
. Later fellow F150-ians.
------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
No problem helping... I owe this site more than I can ever repay.
Perfect schematic this time. Assuming you have properly polarized the LED it will light when the switch is turned on (the relay will work regardless of the LED). The reason it didn't work the first time is because the LED "blew" (or would have) as soon as you turned on the switch. Then the circuit will be broken and the relay is no longer supplied any voltage. Generic LEDs can only handle about 30 milliamps or so before they are damaged. If you wired the LED in backwards nothing would have happened since current will only flow in one direction through a diode (unless the voltage is sufficient to breakdown the PN junction and would again "blow" it, not the problem here). The 57 milliamps you calculated is correct mathematically but that is too much for the LED and probably not enough for the relay. Using I=V/R it needs about 130 milliamps to operate reliably. (13.2/100=.132)
GIJoeCam,
What your friends told you is true but it is not clear that you understand by what you wrote... Current DOES need to be regulated or more accurately stated: "limited". Generally, more than 30-40 milliamps or so and the LED is toast. That is what you are doing by adding resistance, limiting the current. The formula commonly known as "Ohm's Law" (V=IxR) expresses this. It can also be expressed R=V/I and I=V/R. Depending on if you are solving for Voltage-V, Current-I, or Resistance-R. By adding that resistance you are "dropping" some of the voltage across that resistor (when current flows) until the voltage the LED is left with is 1.7 or so. This forward voltage (1.7) varies depending on the type of LED. That is what I used for my earlier example since most are rated as such but your LED may be 3.0v (you mentioned a dual color LED earlier?). To solve for that LED start with the DC voltage (I think the automotive DC voltage when the car is running is about 13.8 volts). That's 13.8v minus the 3.0v the LED needs, that leaves 10.8v. That 10.8v is the amount of voltage the resistor has to "drop". Now we figure in the current using R=V/I. V is 10.8, I is .010 (or 10 milliamps since this is the lowest value you mentioned, for the red section right?) that means R=1080 ohms. But that value is not available in a fixed resistor so you will have to go with 1K or 1.2K ohm resistor. Either will probably be OK but if you want to be safe go with the 1.2K. This will extend the life of the diode beyond that of our trucks (unfortunately). The same holds true again for the size or wattage of the resistor you need. Watt's law can be expressed as: W=VxI, or in this case, 10.8 x.01=.108 or about a 10th of a watt. You could use an 1/8 watt resistor but go ahead and use the 1/4 watt (or bigger) if you have them already. This will insure a trouble-free design. As far as the 12 volt vs. 3 volt dilemma, who knows what the RS guy is talking about, go by what's printed on the package. If the package says it's for 12 volt usage it already has the proper resistor internally so treat it like a light bulb (except it must be properly polarized for it to illuminate). If it says 3 volts then use a 1.2k ohm resistor in series as we calculated. If you hook it up backwards it should not be damaged but it just won't light!
Hope this helps, I'm tired of typing for tonite
. Later fellow F150-ians.------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
#23
11-11-2000, 05:31 AM
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#24
11-11-2000, 10:33 AM
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GAH,
No problem, the more you help the more you learn. I was a bit confused on the dog reference for a while but I then realized what you had done. The HOME page is a bit bare these days but someday I'll get back to it and update it with some more stuff. Feel free to write anytime.
------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
No problem, the more you help the more you learn. I was a bit confused on the dog reference for a while but I then realized what you had done. The HOME page is a bit bare these days but someday I'll get back to it and update it with some more stuff. Feel free to write anytime.
------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
#25
11-12-2000, 04:55 AM
Xc,
Are you saying that resistors are available to regulate the current? I can calculate Ohm's law until I'm blue in the face, but I've never dealt with regulating the current before. Can I get resistors rated in watts at a particular voltage? I haven't searched yet...
-Joe-
------------------
98 Explorer (Lemon Law Case in Progress)
'00 F-150, X-cab, 4x4, Lariat, Off-road, Trailer Tow, Holandia Sunroof, built 11/99 picked up 10/00!
List of planned mods: Clear corners and tails, CB wiring, Stainless brush guard, at least 1 set of KCs (probably more!), custom radar detector mount, color matched trailer hitch with stainless steel ball, torsion bar tweak, bug deflector, spray in liner, and a tonneau of some sort...
Are you saying that resistors are available to regulate the current? I can calculate Ohm's law until I'm blue in the face, but I've never dealt with regulating the current before. Can I get resistors rated in watts at a particular voltage? I haven't searched yet...
-Joe-
------------------
98 Explorer (Lemon Law Case in Progress)
'00 F-150, X-cab, 4x4, Lariat, Off-road, Trailer Tow, Holandia Sunroof, built 11/99 picked up 10/00!
List of planned mods: Clear corners and tails, CB wiring, Stainless brush guard, at least 1 set of KCs (probably more!), custom radar detector mount, color matched trailer hitch with stainless steel ball, torsion bar tweak, bug deflector, spray in liner, and a tonneau of some sort...
#26
11-12-2000, 11:40 AM
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GIJoeCam,
NO PROBLEM on the clarifications. I'll be here to help as much as possible and as long as it takes. It can be confusing especially since we have to have this discussion one side at a time sort to speak.
Resistors basically have only 3 specifications... Resistance value in Ohms, Power capability in Watts, and their Tolerance in % (percentage). Without getting too far off topic, the percentage is how far + or - ohms they actually are from their intended value. For example a 1000 ohm, 10% resistor, would read between 900 and 1100 ohms on a Ohm meter and be considered "in spec.".
My point to this is there is no "Voltage" rating to resistors.
Resistance does have a regulating or limiting effect on current. Hence their name.
Now when current flows through a resistor there is some voltage "lost" or dropped depending on the value of the resistor and the amount of current. This is the V=IxR stated in Ohm's law. What this says is that all three components are related.
One way that was helpful for me to "visualize" Volts, Ohms and Amps when first learning was to use the water pipe or garden hose analagy... Think of the diameter of the hose as the amount of current or amps possible in a circuit, think of the water pressure as the voltage and a valve as a resistor. Without being specific in terms of numbers, do you see how a small hose say 1/2" diameter with high pressure could fill up a bucket at the same rate as a 1" hose with half the pressure? How about if each of the hoses had the same pressure... you could use a valve (resistor) to adjust the flow of the larger diameter hose to fill the bucket at the same time or rate as the smaller hose despite being bigger diameter?
The valve regulates or limits the flow of water. But the pressure on one side of the valve is greater than the other. That difference is, in a way, absorbed or dropped by the valve.
Now when reverting back to electricity, that drop in voltage that the resistor is doing must be released in the form of heat. Thats where the Wattage (can be thought of as heat dissipation) comes into play. You must make sure the wattage of the resistor exceeds that of the formula W=VxI. (This is the voltage dropped by the resistor times the current flowing through it) Usually a factor of two (2) is recommended. So the previous examples (posts) showed that the current was about 1/10th of a watt, pretty close to 1/8th. Resistors are usually available in 1/8, 1/4, 1/2, 1, 2, 5 watt (and up). So it would be best to choose the 1/4 watt to be safe and avoid possible overheating and failure of the resistor.
In my examples I am trying to explain the realation of these three components of simple series circuits using LEDs. Please let me know how I'm doing!
Standing by..... XceLenT Driver.
NO PROBLEM on the clarifications. I'll be here to help as much as possible and as long as it takes. It can be confusing especially since we have to have this discussion one side at a time sort to speak.
Resistors basically have only 3 specifications... Resistance value in Ohms, Power capability in Watts, and their Tolerance in % (percentage). Without getting too far off topic, the percentage is how far + or - ohms they actually are from their intended value. For example a 1000 ohm, 10% resistor, would read between 900 and 1100 ohms on a Ohm meter and be considered "in spec.".
My point to this is there is no "Voltage" rating to resistors.
Resistance does have a regulating or limiting effect on current. Hence their name.
Now when current flows through a resistor there is some voltage "lost" or dropped depending on the value of the resistor and the amount of current. This is the V=IxR stated in Ohm's law. What this says is that all three components are related.
One way that was helpful for me to "visualize" Volts, Ohms and Amps when first learning was to use the water pipe or garden hose analagy... Think of the diameter of the hose as the amount of current or amps possible in a circuit, think of the water pressure as the voltage and a valve as a resistor. Without being specific in terms of numbers, do you see how a small hose say 1/2" diameter with high pressure could fill up a bucket at the same rate as a 1" hose with half the pressure? How about if each of the hoses had the same pressure... you could use a valve (resistor) to adjust the flow of the larger diameter hose to fill the bucket at the same time or rate as the smaller hose despite being bigger diameter?
The valve regulates or limits the flow of water. But the pressure on one side of the valve is greater than the other. That difference is, in a way, absorbed or dropped by the valve.
Now when reverting back to electricity, that drop in voltage that the resistor is doing must be released in the form of heat. Thats where the Wattage (can be thought of as heat dissipation) comes into play. You must make sure the wattage of the resistor exceeds that of the formula W=VxI. (This is the voltage dropped by the resistor times the current flowing through it) Usually a factor of two (2) is recommended. So the previous examples (posts) showed that the current was about 1/10th of a watt, pretty close to 1/8th. Resistors are usually available in 1/8, 1/4, 1/2, 1, 2, 5 watt (and up). So it would be best to choose the 1/4 watt to be safe and avoid possible overheating and failure of the resistor.
In my examples I am trying to explain the realation of these three components of simple series circuits using LEDs. Please let me know how I'm doing!
Standing by..... XceLenT Driver.
#27
11-12-2000, 03:16 PM
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RedLariat,
Yes, lamps are not polarized...but that may be there only advantage (although I don't consuder it one). LEDs come in a variety of colors, sizes, shapes and intensities. Now even white ones are available. Cheaper?... not really, depends on where and how you buy them. Radio Shack may be the worst place in terms of price but pretty good in selection and availability. LEDs use much less current than lamps and will outlast our trucks if biased properly.
NOT trying to argue with you, but they are easy to use once you learn how. As you said most you have seen have a 1k resistor in series, as I tried to explain, so the work is done. Just duplicate the circuits 'till you heart's content! I'm just trying to help this site's members in the only area I really know anything about... electronics. Now that you know how to bias them, you may be able to use LEDs other than the ones pre-packaged for 12v. If you would rather use lamps I would encourage you to do so, go with what you know.
As always... sincerely XceLenT Driver
------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
Yes, lamps are not polarized...but that may be there only advantage (although I don't consuder it one). LEDs come in a variety of colors, sizes, shapes and intensities. Now even white ones are available. Cheaper?... not really, depends on where and how you buy them. Radio Shack may be the worst place in terms of price but pretty good in selection and availability. LEDs use much less current than lamps and will outlast our trucks if biased properly.
NOT trying to argue with you, but they are easy to use once you learn how. As you said most you have seen have a 1k resistor in series, as I tried to explain, so the work is done. Just duplicate the circuits 'till you heart's content! I'm just trying to help this site's members in the only area I really know anything about... electronics. Now that you know how to bias them, you may be able to use LEDs other than the ones pre-packaged for 12v. If you would rather use lamps I would encourage you to do so, go with what you know.
As always... sincerely XceLenT Driver
------------------
Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner
#28
11-12-2000, 07:55 PM
Yes, you are doing an excellent job of explaining this in much simpler and clear terms than i could dream of. However, the only thin I learned from it was that they do make reisstors to reduce the wattage. I've had 2 intro to EE classes and an EE lab as part of my Mechanical engineering degree and I have a heat transfer class this semester, so believe me, all this makes complete sense. I'm just glad someone can verbalize all this much more clearly than I can.
Now, back to the topic at hand... I suspect that it is, in fact, a 12 volt resistor, and that all I will need to worry about is the current. (If not, dropping the voltage is just a matter of adding another resistor in series) So if I am doing my math correctly, the voltage is 13.8 (err on the high side) and the currint I need is 10 mA or .01 A, then the wattage I need is .138 watts, correct? (a little more than 1/8 watt) If I wire that one up, that should provide me with my .01 amps through the LED, right?
If so, I'll try it out and let everyone know. It'll be a couple of weeks, though... Heat Transfer exam this week, Design of Machine Elements next week... Dynamics is in there somehere, and a Mechanics of Materials lab the following week! Is it december yet??!!!???!!??!! I plan to have all the supplies by Thanksgiving, so if they instructors are feeling generous, I'll do it then. Wish me luck! And thanks, again, for all the help.
-Joe-
------------------
98 Explorer (Lemon Law Case in Progress)
'00 F-150, X-cab, 4x4, Lariat, Off-road, Trailer Tow, Holandia Sunroof, built 11/99 picked up 10/00!
List of planned mods: Clear corners and tails, CB wiring, Stainless brush guard, at least 1 set of KCs (probably more!), custom radar detector mount, color matched trailer hitch with stainless steel ball, torsion bar tweak, bug deflector, spray in liner, and a tonneau of some sort...
Now, back to the topic at hand... I suspect that it is, in fact, a 12 volt resistor, and that all I will need to worry about is the current. (If not, dropping the voltage is just a matter of adding another resistor in series) So if I am doing my math correctly, the voltage is 13.8 (err on the high side) and the currint I need is 10 mA or .01 A, then the wattage I need is .138 watts, correct? (a little more than 1/8 watt) If I wire that one up, that should provide me with my .01 amps through the LED, right?
If so, I'll try it out and let everyone know. It'll be a couple of weeks, though... Heat Transfer exam this week, Design of Machine Elements next week... Dynamics is in there somehere, and a Mechanics of Materials lab the following week! Is it december yet??!!!???!!??!! I plan to have all the supplies by Thanksgiving, so if they instructors are feeling generous, I'll do it then. Wish me luck! And thanks, again, for all the help.
-Joe-
------------------
98 Explorer (Lemon Law Case in Progress)
'00 F-150, X-cab, 4x4, Lariat, Off-road, Trailer Tow, Holandia Sunroof, built 11/99 picked up 10/00!
List of planned mods: Clear corners and tails, CB wiring, Stainless brush guard, at least 1 set of KCs (probably more!), custom radar detector mount, color matched trailer hitch with stainless steel ball, torsion bar tweak, bug deflector, spray in liner, and a tonneau of some sort...
#29
11-13-2000, 12:02 AM
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GIJoeCam, a resistor must be able to withstand the heat that is being generated within it. The Watt rating (Joules/sec) on a resistor tells you how much energy the resistor can dissipate without being destroyed. As was mentioned by XceLenT Driver, W=(I^2)*R or W=V*I. They make resistors to "handle" the wattage not to reduce it. XceLenT Driver can confirm or correct my comments. I've installed probably 50 alarm systems over the last 10 years (hobby and extra money), some of them with very elaborate options, and never had one come back yet. But recently my wife's relative, whose car I'd worked on four years ago, developed an electrical problem. The dealer claimed that my wiring had caused the problem. After four years of trouble free driving it seemed unlikely to me. But now I'm not so sure and that uncertainty comes from lack of knowledge. The important thing is that we get it right and that's why I really enjoy these discussions and the expertise of people like XceLenT Driver. We all learn from this. Sorry to go on.
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'00 5.4L 4X4 Lariat (Black/Silver)
'99 Ranger 4.0L 4X4 XLT
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'00 5.4L 4X4 Lariat (Black/Silver)
'99 Ranger 4.0L 4X4 XLT
#30
11-13-2000, 12:40 AM
Senior Member
Join Date: Mar 2000
Location: Pa
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GI
every LED I've run into has at least a 1K 1/4 watt resistor in series to it. Why not use little 12v lamps instead? You don't have to figure out which polarity to hook them up and are cheaper.
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97 Lariet-sc-4.6-4x off road-auto-3.55 LS-
superchip-K&N-3"
pvc air box mod-Granetelli MAF
every LED I've run into has at least a 1K 1/4 watt resistor in series to it. Why not use little 12v lamps instead? You don't have to figure out which polarity to hook them up and are cheaper.
------------------
97 Lariet-sc-4.6-4x off road-auto-3.55 LS-
superchip-K&N-3"
pvc air box mod-Granetelli MAF