isolating a 12 V relay

[maxg]

I need to connect a 12V relay to a circuit that provide the 12V under specific condition, but the coil of the relay change the charateristics of the circuit, how do I isolate the relay from the circuit? i have access to 12V dc power.
thank you

 

[KMoffett]

Can you post a schematic of the circuit that provides "12V under specific condition"?

Ken

 

[maxg]

it is the output of a car generator, there are 3 wires coming out, one have 12 V always on, the second have 12 V after you turn the key once, the last have 12 V when the engine is running, I want ot connect a relay to this wire, but if do that the faulty battery charge come on, so I assume that the relay is changing the impedance of the circuit

 

[KMoffett]

maxg,

Not being into automotive electrical systems...I'm going to pass...and hope there are others here that will jump in and help you.

Ken

 

[Rolf]

{snip} I assume that the relay is changing the impedance of the circuit

You are probably correct. Measure the resistance of your relay coil and then substitute larger resistance until the battery fault indicator dosent come ON. This is the lowest resistance you can load down the circuit. To overcome this problem it would probably be necessary to use a transistor to drive the relay.

 

[maxg]

relay isolator

I have measured the relay 80Ω, so I am going to try an NPN transistor 200mA current and 2.2 KΩ load resistor from the alternator side plus a diode across the coil for back EMF protection I'll let you know if it work, I still can not find the impedance between the base and emiter in the transistor spec unless they using some strange terminology, Thanks guy's

 

[maxg]

isolator relay

well it did work! I have used a NPN transistor 500mA IC current, a 2k7 load resistor on the base side and a 20V .5A diode across the relay for protection, the battery charge light stay put, and my circuit working.
thanks

 

[Rolf]

well it did work! I have used a NPN transistor 500mA IC current, a 2k7 load resistor on the base side and a 20V .5A diode across the relay for protection, the battery charge light stay put, and my circuit working.
thanks

Great! For circuit protection I would increase the diode to a 1000v (1N4007), can't hurt and they are dirt cheap.

Thanks for posting "The Rest of the Story"!

Rolf

 

[SMUGangsta]

have you tried disconnecting the alternator to ensure the light does come on when its supposed to? is something i have been thinking about doing.

 

[Willbe]

Since the inductive spike goes negative with respect to ground the PIV of the diode only needs to be about 15v plus a safety factor, say a 25 PIV diode. The diode has to momentarily withstand the coil current so the coil current would be the surge rating of the diode.

It should be a switching type, though. If you get away with using a 1N400x I'd say parasitic resistances and capacitances elsewhere in the circuit are absorbing just enough of the coil's stored energy so the transistor doesn't immediately blow.

 

[maxg]

12 V relay isolator

quite correct I used a 1n4001F and the transistor look ok so far, I haven't try to disconnect the alternator as there are 5 wires so you don't prove any thing but I have re connected the alternator directly and the same thing hapen, the transistor is not so fast so I don't think the back EMF will go more than 20V

 

[Willbe]

quite correct I used a 1n4001F and the transistor look ok so far, I haven't try to disconnect the alternator as there are 5 wires so you don't prove any thing but I have re connected the alternator directly and the same thing hapen, the transistor is not so fast so I don't think the back EMF will go more than 20V

-e/ΔI = L/ΔT, with I = 12/80 and e = -20v, then L/ΔT = 20(80)/12 = ~130.

If the transistor switches in 100 nS then L = 1.3 nH. I think a short piece of straight wire has more inductance than this.