Question for electronics gurus
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Fleet Owner
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From: Rio Rico, AZ.
Question for electronics gurus
So I installed a switch to the Exhaust Back Pressure Valve to utilize it as an exhaust brake. I didn't care about it coming on in the morning so I just cut the wire leading to the solinoid and wired up a switch that is connected to switched power. Without going into a long explantion why, I need to limit the voltage to the solinoid to 5 or 6 volts. The solinoid has only about 12 ohms to it so at 12 volts (if I remember Ohms law correctly) it's pulling down a full amp. IMO that's going to burn out the coil in the solinoid in short order.
So my question is: I want to go and get a 12 ohm resister at Radio shack, but they look pretty small. However looks can be decieving. Can they handle the current or should I just go find a big ballast resister somewhere?
So my question is: I want to go and get a 12 ohm resister at Radio shack, but they look pretty small. However looks can be decieving. Can they handle the current or should I just go find a big ballast resister somewhere?
Elder User
Joined: Jul 2002
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From: Kingsland, GA
As long as that sensor is designed to operate at 12 volts, then feeding it 12 volts isn't going to hurt a thing. That solenoid will only take as much amperage as it is designed to take. Remember, for the most part, amperage kills, not voltage. I realize that extremely high voltages will mess up things like sensors and such hat are designed to step 12 volts down to a certain voltage for a certain temperature/pressure. If you were to pipe 6 volts to that sensor, it would still need the same amount of total power (Remember, voltsXamps=power). So, if it's pulling 1 amp at 12 volts, it's going to try to pull 2 amps at 6 volts, and that's the surest way I can think of to burn it up.
If you really want to find out what the voltage to the solenoid is from the factory electronics, you'll have to put it in a situation where the EBPV would engage, then measure the voltage at the factory wires with a multimeter. My GUESS is that it runs on 12 volts, and that's where you should leave it until you are SURE...
If you really want to find out what the voltage to the solenoid is from the factory electronics, you'll have to put it in a situation where the EBPV would engage, then measure the voltage at the factory wires with a multimeter. My GUESS is that it runs on 12 volts, and that's where you should leave it until you are SURE...
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Fleet Owner
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From: Rio Rico, AZ.
Originally Posted by Rancha9
As long as that sensor is designed to operate at 12 volts, then feeding it 12 volts isn't going to hurt a thing. That solenoid will only take as much amperage as it is designed to take. Remember, for the most part, amperage kills, not voltage. I realize that extremely high voltages will mess up things like sensors and such hat are designed to step 12 volts down to a certain voltage for a certain temperature/pressure. If you were to pipe 6 volts to that sensor, it would still need the same amount of total power (Remember, voltsXamps=power). So, if it's pulling 1 amp at 12 volts, it's going to try to pull 2 amps at 6 volts, and that's the surest way I can think of to burn it up.
If you really want to find out what the voltage to the solenoid is from the factory electronics, you'll have to put it in a situation where the EBPV would engage, then measure the voltage at the factory wires with a multimeter. My GUESS is that it runs on 12 volts, and that's where you should leave it until you are SURE...
If you really want to find out what the voltage to the solenoid is from the factory electronics, you'll have to put it in a situation where the EBPV would engage, then measure the voltage at the factory wires with a multimeter. My GUESS is that it runs on 12 volts, and that's where you should leave it until you are SURE...
The reason that I wanted to step down the voltage at the solinoid is I sampled the signal at the wire for the solinoid. There is a constant 12 volts at the solenoid whether it's activated or not. The ground was constant. If you were to put a test light to the power side it is pulsing dimly, like a pulse width modulated signal. When the valve is activated the signal stays the same, blinking, just brighter. So I was wondering if the designers wanted to limit the draw at the solinoid for whatever reason. I just wanted to not fry the coil.
Anyone out there with a EBPV mod that has used it for a number of years, with a straight 12 v, have you had any trouble with the solinoid burning out?
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Joined: Feb 2003
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From: Utah
If it is 12 ohms, it will not draw 2 amps unless you put 24 volts on it. If you put 6 volts, it will draw .5 amps. amps = volts/ohms. If it has a pulse width modulated signal, then they are using that signal to set the amount of closure on the valve. If it gets really hot with 12 volts on it, then you might want to worry about it, but I doubt 12V will hurt it.
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From: Rio Rico, AZ.
Originally Posted by yellow73bb
If it is 12 ohms, it will not draw 2 amps unless you put 24 volts on it. If you put 6 volts, it will draw .5 amps. amps = volts/ohms. If it has a pulse width modulated signal, then they are using that signal to set the amount of closure on the valve. If it gets really hot with 12 volts on it, then you might want to worry about it, but I doubt 12V will hurt it.
I am curious as to the ability of the Radio Shack resisters to carry the amps involved. I get easily confused by watts and amps. What defines a watt and an amp? The highest rating is 1/2 watt. Now I could get some 100 ohm resisters and hook them up in parallel to make the total 10 ohm but is that necessary? I just want a resister that will carry the current and not overheat. If I need to get a great big ballast resister for an ignition coil or something that's fine, but if a couple of $.99 resister packs hooked up in parallel will do it than I am all for it.
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Joined: Feb 2003
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From: Utah
If you have a 12 ohm resistor, you should have half of your battery voltage across it. Since your battery voltage is about 13.8V (call it 14 for simplicity), half of that is 7 volts. The watts that will be dissipated is equal to V*V/R or I*I*R. In this case, since you will have about 7 volts across the resistor, the watts dissipated will be 49/12, or a little over 4 watts. Standard practice is to double the dissipated watts. You will need a 12 ohm 10 watt resistor to drop the voltage in half. Radio Shack has a 10 ohm 10 watt resistor (Cat # 271-132) that would work just fine. Your voltage at the valve would be about 7.6V with 14V in. You could also run 16 200 ohm 1/2W resistors in parallel and come up with 12.5 ohms (200/16). Your voltage across the valve will be 12*Vin/(12+r) where r is the value of your resistor, and Vin is the voltage at the battery.
Hope this helps.
Hope this helps.
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Fleet Owner
Joined: Jun 2004
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From: Rio Rico, AZ.
Originally Posted by yellow73bb
If you have a 12 ohm resistor, you should have half of your battery voltage across it. Since your battery voltage is about 13.8V (call it 14 for simplicity), half of that is 7 volts. The watts that will be dissipated is equal to V*V/R or I*I*R. In this case, since you will have about 7 volts across the resistor, the watts dissipated will be 49/12, or a little over 4 watts. Standard practice is to double the dissipated watts. You will need a 12 ohm 10 watt resistor to drop the voltage in half. Radio Shack has a 10 ohm 10 watt resistor (Cat # 271-132) that would work just fine. Your voltage at the valve would be about 7.6V with 14V in. You could also run 16 200 ohm 1/2W resistors in parallel and come up with 12.5 ohms (200/16). Your voltage across the valve will be 12*Vin/(12+r) where r is the value of your resistor, and Vin is the voltage at the battery.
Hope this helps.
Hope this helps.
