do all relays need to be diode isolated?

Hey just researching some information and happened to run across information about using a diode to protect the relay coil from a reverse spike. At least this is what I believe is what the purpose is. Do all relays need this protection? My application for relays on my alarm are as follows: Vehicle is a 1992 ford mustang. 1) Relay to reverse polarity of my parking lights. Ford's parking light wire is (+). I don't want my back up battery(dei 520t) to power my parking lights if the main 12v supply is cut. BTW I have not purchased the relay yet. So I will wire a relay as such: pin 85: (-) ground signal supplied from alarm pin 86: (+) 12v constant direct from vehicle's main battery pin 30: (+) 12v constant direct from vehicle's main battery pin 87: (+) parking light wire 2) Relay to pop trunk. Wired into the (+)12v wire of the 2 wire switch in the glove compartment. I will be tapping into the (+)12v wire. The relay will send a (+)12v supply to this wire to pop the trunk. pin 85: (+) 12v constant direct from vehicle's main battery pin 86: (-) ground signal supplied from alarm pin 30: (+) 12v constant direct from vehicle's main battery pin 87: (+) trunk release wire 3) Relay that came with alarm viper 350 to kill starter. Wired as a basic starter kill diagrammed here on this page from the12volt https://www.the12volt.com/relays/page2.asp 4) A DEI 451M to control my 5 wire door locks. This appears to be two relays in on small box. Thanks for any info! I just joined this site and everyone is very helpful and I'm lucky to have found this site. One day I hope to be knowledgeable enough to help others here. Right now I'm a newbie

Opps the other relay I will be using is for my Sirens. One siren is the siren which comes with the alarm system for under the hood. I have purchased a second piezo siren from radio shack that draws a max of 150amps. I will wire the relay for this as such pin 85: (+) 12v from alarm siren output pin 86: (-) ground pin 30: (+) 12v tapped into from DEI 520t grey wire 12v supply to alarm's (+) 12v supply pin 87: (+) 12v to both sirens

Hey all, Trying to figure out how much current the 520t "diode box" is designed to output/handle. I am already going to upgrade the battery to a 12v 5ah. The 0.7ah is to small in my opinion. The reason I ask how much current it is designed to supply is because I want to wire a relay to the grey (+)12v alarm power supply to power additional sirens.   If I wire the sirens to the main vehicles battery 12v supply and the main car battery is disconnected the sirens will not work when wired through the relay I plan to add. Here is what I am planning to do with the relay: pin 85: (+) 12v siren output from alarm pin 86: (-) ground pin 30: (+) 12v supply from grey wire coming from 520t "diode box" pin 87: (+) to siren's Thanks for your help!

Not all relays need diodes across the coil - but it never hurts to add them. All outputs of an alarm that are designed to power relays should have the diode built in to the alarm. The battery backup should be used to power only a siren (no lights or anything else). This will keep the noise going on longer with the smaller battery. I would highly suggest getting a battery back up siren instead of a battery back up system. With the battery back up the thief can just cut the siren wire(s). With the battery back up siren you can cut all the wires to the siren and it will continue to make noise.Kevin Pierson

No relays need protection diodes. Only the circuit that attach to them do. And don't confuse a reverse-biased diode across a relay's coil (#85 & #86) with a isolation diode - they are two completely separate and different functions. One quenches spikes. The other stops circuits feeding other circuits. (Though that can include the spike....) PS - your piezo siren draws a max of 150mA, not 150A.

KPierson wrote: Not all relays need diodes across the coil - but it never hurts to add them. All outputs of an alarm that are designed to power relays should have the diode built in to the alarm. The battery backup should be used to power only a siren (no lights or anything else). This will keep the noise going on longer with the smaller battery. I would highly suggest getting a battery back up siren instead of a battery back up system. With the battery back up the thief can just cut the siren wire(s). With the battery back up siren you can cut all the wires to the siren and it will continue to make noise.

Hey, thanks for the response. From my understanding the 520t backup battery is designed to power the entire alarm not the siren alone. So I was going to use a relay for the sirens because my research on the threads from this site indicate that the siren output from the alarm is designed only for a single siren and adding more will damage that output so it's was recommended to use a relay to power the sirens. So instead of the alarm powering both the sirens I was going to just get power from the same point the alarm receives its power. Also I am hiding all the wiring with a stealth install. They wont see any wires for the siren or the siren itself. I will definitely look into the backup battery siren. How do I wire the backup battery siren with the other sirens? Use the relay I'm talking about and use the alarm siren output to power the coil for this relay?

oldspark wrote: No relays need protection diodes. Only the circuit that attach to them do. And don't confuse a reverse-biased diode across a relay's coil (#85 & #86) with a isolation diode - they are two completely separate and different functions. One quenches spikes. The other stops circuits feeding other circuits. (Though that can include the spike....) PS - your piezo siren draws a max of 150mA, not 150A.

Opps! My mistake, I can't edit any posts, not sure why You are correct it draws a max of 150ma. So would any of my relays need this isolation diode? Here is the thread from this site I am referring to.    https://www.the12volt.com/installbay/forum_posts.asp~TID~86348~PN~2~TPN~2 If you scroll down to "dual sport's" diagram, this is what I am referring to.

Neither of those are isolation diodes. They are both suppression (quenching, protection, etc) diodes that are there to short any negative transient spike from the coil - especially when turned off. Do you need it? You tell me.... As KP wrote - "it never hurts to add them". If you are unsure, you can omit them. Then if whatever controls them blows - or if something else blows - you can let us/me know that you did need them. PS - WRT that link - yes, that imaged text is wrong. I think that's the same text I donged somewhere else. The diode's line/kathode end is towards the +ve side of the coil.

Generally agree with Kevin and Peter, normally, you won't need the diodes when using a switch but ALWAYS fit them across the terminals as a precaution. I don't agree with Kevin about alarms having internal diode protection, also the circuits in these alarms such as aux and locks are such low current that they WILL be burnt out by a reverse. ALWAYS diode divide alarm inputs/outputs such as hood and trunk using the same trigger wire, separate door triggers , aux outputs. Always use 86 as the POS and 85 as the NEG coil terminals, it's the international convention, yes if the relay doesn't have an internal diode it makes no difference, but if you stick to that, less chance of mistakes. At the rates I buy them, 1N4004 diodes cost me less than a cent each, it's worth it for thoroughness and piece of mind. Remember the Ps, Proper Preparation, Prevents *iss Poor Performance.

Here is the datasheet to a very common output driver used in alarm systems: http://focus.ti.com/lit/ds/symlink/ulq2004a.pdf Page 3 shows the diode between the output and COM pin. I know that DEI used a similar chip in their older alarms - not sure what they are using now but I would guarantee that the diode is built in. A good illustration of this is the fact that the 451M does not have any diodes in it. However, it plugs directly in to most DEI alarms. In order for this to work the alarm would HAVE to have the diodes built in to the door lock outputs.Kevin Pierson