Would it be bad to hook a relay to be controlled by my ignition switch, so that when I turn the key, it is always on? Not sure what the duty cycle is on these. I was going to use a new, old style horn relay to control my compressor for air ride. I tried using the ignition feed on the fuse box, but it goes through one of the fuses in the box (10 or 15 amp), and will blow it from the draw. There is a 30 amp inline fuse for the compressor. My thought was to use the ignition feed to control the relay, which will put power to the system. But, I'm thinking it may overheat from always being "on" when the key is on. I want the compressor to be off when the key is off, to prevent it from running when I'm not in the car. The alternative would be to run the relay off a toggle switch, but I'd rather have it just work with the ignition. Thoughts?
use a relay that is rated for continuous duty , that horn relay is not. i believe the common 5 pin Bosch style relay is for continuous duty
X2. The horn relay will cook in short order, as will the common Ford starter solenoid. You've also got inrush current on a compressor motor. If the 30 amp fuse on the compressor is correct, and a common fuse, not a "slo-blo" you'll probably be ok at a 30 amp continuous relay. I'd rather see a 40 or 50 amp continuous, especially if you're going to lay it in the weeds and have to bring it back up on a regular basis. Put the relay close to the compressor, and allow for voltage drop to it, i.e., if the battery is up front and the compressor in the trunk, run some #8 wire, not #10. Make sure the ground is good, too.
I was considering a 12v golf cart solenoid. It looks like the Bosch/Viair/whatever relay is cheaper & the way to go. I believe they are 40 amp rated. 30 amp fuse is correct for the Air Lift system. I have a tendency to overkill stuff anyway, but good advice guys. Thanks
GOOD quality a/c relay or better yet an electric golf cart relay. They look like a ford starter relay. http://www.littelfuse.com/products/...urrent-relays/continuous-duty-spst/24106.aspx
You might consider a latching relay. The coil draws no current after it's latched closed. Will need a close and trip input however. Looks like they make them in the small 40 amp plug-in versions.
http://www.ozautoelectrics.com/narv...pin-normal-open-relay-resistor-protected.html Use a 70A relay, the 30 A fuse, the IGN source to operate the relay. Ignition to one side of the coil, other side of the coil to ground. Battery supply to 30 A fuse to one contact. Other contact to compressor. Simple.
A relay isn´t rated for 30 amp and will die from 32...it´s so many thousand opening closing cycles at rated current. So no need to go wild on this one. It´s going to be fairly good on the relay since compressor will most likely not be working when the relay opens. And what is worst for a relay is opening at full load. With only one opening closing cycle each ride it will last a long time. Get a good brand relay rated for the load you need. Use right size cables, if there is voltage drop in the cables the compressor will take the same power and compensate loss of voltage by taking more amps.
good choice, and if you are worried about life of relay, put a capacitor (.1uf @ 200V) across relay contacts.
I think that most systems use a 5 pin these days. This is my first attempt at air suspension, so I'm no expert. My thought was something that had screw on terminals (thus, the horn relay) to make it easier to hook everything up. If I get the 5 pin, I'll need the socket as well. My system was older, but never installed Air Lift Easy Street. Didn't come with a relay, or show one in the directions, but I probably need one. I'll have to do some searching on the net & check out your suggestions.
i like to use a relay with a mounting tab on it , no need for a socket. then i just use 1/4 insulated female connectors if you want a socket for the 5 pin relay.....BWD part # PT193 at O'reillys http://www.oreillyauto.com/site/c/detail/BWD0/PT193.oap?evaritm=BWD+--+R3177&rpl=1
I have a 40 amp relay that came with an electric fan. Looks like most all normal relays, small, black square thing, except the wires from the relay as as thick as a cigarette, definitely overkill. You won't burn that baby out from keeping it on all day.
https://www.amazon.com/Absolute-Hol...1466868752&sr=8-2&keywords=bosch+relay+holder much better deal on relay package.
I have air ride on a o/t truck, And have had a few. The way I've done it was by using a fuse tap at the fuse box from something non essential but obvious, I used my blinker fuse. Blinkers don't work, must be a problem. Then run that wire, 16ga maybe, through a switch for safety, to the back through the pressure switch then to operate a 60 "stinger" relay. I used 4ga power. Then tie your compressor or compressors onto that. Added a pic from another project. Never mind the two wires coming in the top right. That's for an EDC (Hope the pic works. I've never added one here.)
You may want to rethink this last statement. It isn't correct unless there are electronics on the compressor that can vary the input current based on voltage. The compressors DC motor will not do this, and while I don't know much about air ride compressors other than the ones on old GM cars, I assume that most of them do not have electronic front ends.
Nope, that's true. An electric motor is performing work and is rated in horsepower, and if input voltage is low will draw more current in an attempt to compensate (1 hp = 750 watts, and watts is volts X amps). I'll also note that ANY plug-in relay is really only good for 30 amps because that's all the spade connector is rated for. If you want a 'true' higher amp rating, you need a relay with bolted connections. Bosch does make a '40 amp' plug-in relay that's rated for 30 amps but has heavy-duty contacts to help with the current inrush. As said by several, use a name-brand, quality relay.
In most circuits a reduction of voltage will result in a corresponding reduction of current. There are differences depending on if the voltage is AC or DC, and also differences depending on a resistive or inductive load. The blanket statement you made is not correct. Headlights are resistive, DC compressor motors have a resistive and inductive component.
Lets make it simple you need lets say 1hp to turn the compressor around...if the voltage goes down the motor still puts out the same power to work so it compensates by drawing more amps...this has of course a limitation where the motor burns up or stops. And it doesn´t work on a pure resistive load but this thread was about a motor
What do you think of this? Airtex/Wells 1R2099, 4 terminal, 40A continuous, 70A rush-in. Comes stock on a 2009 Ford E450 (among others). No mounting tab, which makes it harder to use, but larger terminals for the larger gauge power wire.
Good reading here: http://www.watsons-streetworks.com/tips Click on "A current topic" You can get the relays both sizes from Fry's Electronics with the matching plug harness. Pay particular attention to the section regarding the ignition switch ,I ran the 70 amp relay on mine it paid off.
And as a blanket statement, the above isn't true either. The 'basic' rules are simple; in a purely resistive circuit, a reduction in SOURCE voltage will result a proportional reduction in the current. But in practice, it's not that simple. Very few electrical components on a vehicle are purely resistive in the static sense, and I'm including the wire/connectors, incandescent or halogen lights, and a cigar lighter if you have one. Here's an example.... Your basic 1157 lamp has the brake/turn filament rated at 23 watts with a 12 volt input. A quick calculation will show that the filament resistance has to be about 6.2 ohms for that watt rating and it will draw about 2.2 amps. This makes sense, as many LED vendors sell this size resistor to wire in if using a LED replacement and you need the 'same' load to make an electronic turn signal module work for example. Now measure the actual filament resistance with just the lamp and your meter. You'll find it's about 2.2 ohms.... Whaaa? What's happening is as the filament heats up and produces light, it's resistance goes up too. At rated output, it's 6.2 ohms. So for arguments sake, let's assume that a reduction in voltage will proportionally reduce the temperature the filament reaches and the resistance goes down proportionally also (it won't be particularly linear in actual practice, but this will give you the basic idea). So if we reduce voltage by 20% down to 9.6V and resistance the same (now at about 5 ohms), what amps does it draw? It calculates out to about 1.9 amps, or a reduction of only 15%. We lost 5% somewhere... Let's go another 20%, all the way to 40%. Voltage is now 7.2 volts, resistance is down to 3.7 ohms, and current calculates to.... 1.9 amps. We've reduced voltage by 40% (and have a lamp that probably needs a dark room to tell if it's even on) but are still using 85% of rated current. And where did the 40% of the voltage go? If it was dropped because of too-small wire or poor connections, it's being converted to heat. In this case because it's a small, relatively low current load, the power consumed will be about 9 watts or about .75 amp. So we started with 2.2 amps, reduced it to 1.9 but adding the .75 amp puts you at 2.65 amps, or a 20% increase in current.... Everything else will have some inductive component and reducing voltage to that part will cause an increase in current... Period. So the majority of the electrical system on a vehicle will need more amps, not less if voltage is low. And second, voltage 'lost' in the circuit isn't really lost, it's converted to heat and consumes it's own amps. This why I keep hammering on the importance of doing voltage drop calculations....
The easiest cheapest way to do this is to use two relays in parallel, ie: two common 30a relays are capable of carrying a total of 60 amps... (30a each contact).
It is true they can carry the double current, but they can not break the double current. Double relays don't break the curcuit at the exact same time, one will be slightly quicker than the other. That means only one of them will be conducting the full current for the last fraction of a second, and following that it has to deal with the arcing caused when breaking high current, especially if the load is inductive. That arcing burns the contact points (just like in a points type ignition with a bad condenser), killing the relay prematurely.
Just my 9 years auto sparky experience, and 25 years experience in the automated Industrial Packaging sector. do your homework if you don't believe me. Sorry to sound so negative, but when I see electrical advice - and it's bad...... it shits me to no end.
Thanks Mudgy. That's good stuff. I was an electronics tech - way back when - so I understand the theory, etc. Just wasn't sure of the duty cycle on a horn relay. Most of the Bosch type just don't look like they could handle a load like that, with those small terminals. A large gauge wire with a small spade terminal doesn't make sense. That's why I asked about the Airtex/Wells 1R2099. The power terminals are much larger = larger gauge wire. The instructions that came with the system only call for a 12 gauge power wire, & it was provided. These systems were obviously meant to be mounted in the trunk/truck box, so the designers must have assumed that was big enough. Seemed a little small to me, but what do I know? I'm only running one compressor & won't be overworking it. No making my car dance, etc. This is an older system, & maybe the newer ones have been upgraded to larger wire & relays because of former problems? I don't know. I have already ordered something & will be using it. If it doesn't work, I'll try something else. Thanks everybody.
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