Im hooking up a relay to drive an ignition switched fuse additional fuse box. I was wondering how much current the actual coil could handle, because right now i have the coil being energized by the switched radio power wire, which i know has a pretty good amount of current running through it. i just cut the red wire and am running the coil inline. will this destroy the relay? should i use the amp turn on lead instead?
The relay coil is self limiting with regards to current draw.. it won't draw any more current than it requires.
Jim
the coil of the relay is not the part of the relay that carry's the current. the switch state of the relay(legs) is what will handle the load. and the relay you have should have it's max current ratng on it. most are 20 - 30 amps. but if it does not ,get one that your sure of the ratings and that will handle what you intend to use it for.
the coil will only pull what it needs to close the legs witch ain't much. it will not pull anymore nor can you give it more than it needs.it makes that dission.
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JUST ONE MORE AMP!!!
hu,alpine cva 1005/dva 5205
sound processor,symmetry(first one).
sub amp,power 1000 the terminator.(1992).
subs,spl comp dual 1 ohms.
punch 150hd on a 10" ev.
alotofhighs
tibby01 wrote:
...right now i have the coil being energized by the switched radio power wire, which i know has a pretty good amount of current running through it. i just cut the red wire and am running the coil inline.
Umm.. are you running the relay coil in series on this circuit?.. or parallel?
Jim
good qeustion i over looked that. good eye.
because if its inline with a device that carry's a current more then the coil can handle, then it will fry. the coil will carry the load of what it's inline with.
-------------
JUST ONE MORE AMP!!!
hu,alpine cva 1005/dva 5205
sound processor,symmetry(first one).
sub amp,power 1000 the terminator.(1992).
subs,spl comp dual 1 ohms.
punch 150hd on a 10" ev.
alotofhighs
yeah, i had it in series with it, which is why i was asking. i have since wired it in parallel though, which should be fine, correct?
i knew the same current would be present in the coil and was wondering around what current is the max that the actual coil can handle.
fingaz22 wrote:
because if its inline with a device that carry's a current more then the coil can handle, then it will fry. the coil will carry the load of what it's inline with.
hum... this just don't sound quite right... something might have a problem with UNDER-voltage, but putting something in series with something else is never going to increase the current, unless one or both of the "somethings" is a power source (i.e. a battery) For some voltage "V" (say 12 volts) being fed to some resisitive load "R" (say 48 ohms) will allow a given quantity of current "I" to flow (about .25 Amps), all based on the math equation of V = I * R.
So lets say you have a relay with the above mentioned 48 ohms, and a radio with something like 12 ohms. The Relay alone will draw, or allow, .25 Amps. The radio, if wired alone, will draw 1 Amp (I * 12 = 12, or I = 12 / 12, or I = 1).
If these two devices are wired in series, the resistive load adds, so you now have 48 Ohms + 12 Ohms = 60 Ohms. Using the same equation, I * 60 Ohms = 12 Volts, I = 12 / 60, I = .2 Amps.
And as long as I'm on the soap box, both devices will also see less than 12 volts, with a ratio of their respective resistive loads. So for the relay, V = 48 Ohms * .2 Amps = 9.6 Volts, and for the radio, V = 12 Ohms * .2 Amps = 2.4 Volts. This means that while the relay will probably engage, with a pull-in voltage around 8 volts, and the radio will probably not do anything.
Now, I'm not at all saying that it won't damage something, because it could. Electric motors are particularly sensitive to under voltage problems. I am saying that this won't damage something because of too much current.
To answer the original question, it would be far better to run the relay parallel to the radio, not in series. In parallel, both devices see the full voltage, and neither will significantly effect the other, assuming that your source will handle the combined current of the devices.
Hope this helps.