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Diode on Relay Coil?


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domiyork809 
Member - Posts: 29
Member spacespace
Joined: October 30, 2003
Location: United States
Posted: November 11, 2003 at 6:30 PM / IP Logged  

Hi,what is the use giving to a diode located in a relay, between the coil`s input inputs 85 and 86.

Please help new dealing with relays and diodes.
xetmes 
Silver - Posts: 586
Silver spacespace
Joined: May 18, 2003
Posted: November 11, 2003 at 7:09 PM / IP Logged  

it is used to disapate the large voltage created when the magnetic field collapses in the coil, the diode allows the volatge to become no greater than .7 V therefore not destroying electronics.

Or mathmatically:

v = L di/dt in an inductor

di/dt is the first derivative of current with relation to time, or the instantaneous rate of change

if current changes abrubtly, di/dt becomes extremly large, since it is the tangential slop of the current, multiplied by the inductance, the voltage is extreamly high.

Its there to protect the electronics that are connected to the relay from being destroyed from a very high reverse voltage.

Teken 
Gold - Posts: 1,492
Gold spacespace
Joined: August 04, 2002
Location: Aruba
Posted: November 11, 2003 at 7:20 PM / IP Logged  
The diode will prevent the CEMF ( counter electro magnetic force ) current from flowing back, and damaging components on the other side of the relay wiring.
The process by which CEMF is produced is called self-induction.
When installing the (blocking /quenching/ isolating) diode on the relay pack.
The negative striped side (cathode) will be facing pin 86 on the relay and the other side which is the (+) positive side (anode) goes to pin 85.
Regards
EVIL Teken . . .
NowYaKnow 
Gold - Posts: 1,217
Gold spacespace
Joined: December 18, 2002
Posted: November 11, 2003 at 7:27 PM / IP Logged  
Hey. That is known as a spike suppression diode or sometimes a quenching diode. To make a long story short what happens is that the coil in the relay creates a magnetic field that stores energy. When you pull power from the relay that stored energy has to go somewhere so it goes back through the relay. (Known as back EMF) Basically the diode takes the hit thus protecting the relay and other components. Hope that helps..Also once you add a diode to the relay you have to watch how you hook up 85 and 86 to make sure they are correct. Whereas normally without a diode they are interchangeable. Hope that helps,
Mike
domiyork809 
Member - Posts: 29
Member spacespace
Joined: October 30, 2003
Location: United States
Posted: November 12, 2003 at 6:22 PM / IP Logged  

 So to prevent components and circuits from being damaged all you need is a diode across 85 and 86 of the relay?? does not matter what is connected to the relay, lights, radio or any other electrical component.     

Thanking you for all this valuable help.

Teken 
Gold - Posts: 1,492
Gold spacespace
Joined: August 04, 2002
Location: Aruba
Posted: November 12, 2003 at 6:33 PM / IP Logged  
Technically no... Only a fuse or similar voltage / current limiting device will protect a component from failure.
Our reply was simply to answer your question as to what each device did, and how it interacts with the circuit.
In closing, please select a relay with the proper current rating you wish to sustain for prolonged periods of time.
Regards
EVIL Teken . . .
domiyork809 
Member - Posts: 29
Member spacespace
Joined: October 30, 2003
Location: United States
Posted: November 12, 2003 at 7:57 PM / IP Logged  
Thanks, Teken. Great help. 
Swim4food 
Member - Posts: 1
Member spacespace
Joined: August 15, 2003
Location: United States
Posted: November 12, 2003 at 8:18 PM / IP Logged  
Ok, here is my question.  I have a reverse polarity lock system on a 1995 Ford Taurus and on a remote start/remote entry system I had it required relays due to the high voltage that the locking devices require.  However, I needed an alarm so I bought the Clarion MS2007 and it supposedly has relays built in.  However, when I hooked it up the locks did not work so I proceeded to attempt to hook up the relays in a half attempt to see if it would work.  However, in my experiment I dont know that I hooked up the relays correctly and the locking module somehow got fried.  I am getting the whole thing replaced but my problem is what should I do next time.  Should I just hook up the relays again and make sure that they are hooked up correctly this time or should I attempt to use the supposed on board relays again laced into the locking mechanism.  You see I didnt realize that the mechanism was fried until after I tried the relays.  Of course after I fried this system it fried pretty much everything.  However, it was visibily clear that the locking mechanism was fried as the copper had come out of the silicon.  So what is your advice on this issue?  Thanks.
superchuckles 
Copper - Posts: 89
Copper spacespace
Joined: December 29, 2003
Location: United States
Posted: January 13, 2004 at 10:20 AM / IP Logged  
if it has diodes internally then opening the unit will likely void your warantee if any...... you could of course, use a set of external relays that don't have the supressor diode on them - of course, that will require more relays, but at least you'll just fry the relays if you somehow get it wrong with them..... be sure your wiring first, test everything before you bundle it all up and think you're done.
yardie0 
Member - Posts: 1
Member spacespace
Joined: February 17, 2004
Location: Canada
Posted: February 19, 2004 at 1:35 AM / IP Logged  
my question is, suppose 86 is connected to 12V and 85 connected ground. How should the diode be conneted? is the cathode connected to 6 and the anode connected to 85?
kirk
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