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Science Forum Index » Electronics - Basics Forum » Relay setup with two controls.
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| Mike H |
 Posted: Mon Apr 21, 2008 4:25 am |
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I am working on a design for a trunk mount battery on my car. Since I
visit some drag strips, I need to follow NHRA and IHRA rules for such
a mount. This requires a shut off switch that will cut power to the
engine such that it shuts off, and the chassis wiring will no longer
have power.
I haven't worked with such things for a while, so my mind is
questioning my design. In essence, I need to cut power from the
battery, as well as shut the engine off as the alternator will keep
the car running even if the battery is disconnected. In my design I
will be pulling the 12v the Ignition system needs to operate to kill
the engine, thus the alternator.
Now I can just stick an relay in series with the ignition wire that
will be energized if the trunk switch is Closed. But I prefer to not
have a relay energized when the ignition is off. Thus the idea is a
second relay driven by the ignition system 12v lead. This relay would
be in series with the wire powering the first relay. Thus with the
ignition switch off, the NO relay is open, killing power to the first
relay.
I have a diagram here:
http://i8.photobucket.com/albums/a23/tcw_fort/carparts/wiringcutoffswitch.jpg
I couldn't find a good way to diagram this with AACircuit.
The question is. Will this work? Will the line feeding pin 86, then
running through the second relay (K2) have 12v at the output at Pin
87 of K2? What impact, if any does having relay K1 driven by the
connection from K1 pin 86 to K1 pin 85 ground have on the voltage
available at Pin 87 of K2?
The goal here is that Relay K2 is only Actuated if both the DPDT
switch in the truck is closed and the ignition switch is closed (thus
power on the wire referred to as Circuit 16, pin 13). And that if the
DPDT switch is opened, K2 will open (K1 will stay closed until the
alternator stops but K2 will open immediately with the open of the
DPDT switch)
Thank you for any assistance. |
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| Mike H |
| Posted: Mon Apr 21, 2008 5:48 am |
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On Apr 21, 9:45 am, Ecnerwal <LawrenceSM...@SOuthernVERmont.NyET>
wrote:
The wire I am tapping into is kind of the ignition switch and it will
shut the car off if left open. It is an output from a Integrated
Relay Control Module that feeds 12v to the ignition system, as well as
various other sensors. (Distributorless Ignition System) It receives
power when the ignition switch is in Start or RUN which itself closes
a relay in the IRCM (also known as a CRCM or Constant Relay Control
Module)
The purpose of the relays I use is so that the length of the main
power wire for the ignition system is kept short and not tied into a
switch in the rear. Note that the IRCM is in the front left of the
car, and the battery is going to the rear left. About 15 feet. Relays
are Bosch automotive relays so they should be up to the task. Yet I
take your comment of fewer parts, less to go wrong to heart. I do
tend to over think problem and a proposed solution.
Looking at your proposal, if I have a 14gauge wire carrying up to 14.6
volts (alternator output when charging) and as low as 10.5 volts (low
battery power until alt spins up. but enough to fire the ignition
system), what would be the best gauge wire, and method of connecting
it to limit voltage drop over 30 feet (15feet one way) to a minimum.
If I'm going to switch a power source for sensors, I need to be
certain that voltage drop is as minimal as possible otherwise sensor
readings may be mis-interpreted by the EEC-IV processor which monitors
the sensors.
Thanks |
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| Ecnerwal |
| Posted: Mon Apr 21, 2008 9:45 am |
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In article
<1bdc7786-1ae1-46c9-ab59-edb89980d8e1@a9g2000prl.googlegroups.com>,
Mike H <mike8675309@gmail.com> wrote:
Quote: http://i8.photobucket.com/albums/a23/tcw_fort/carparts/wiringcutoffswitch.jpg
The ignition switch turns the car off, correct? So, feed the ignition
switch from the "low amp line" and it should work as desired without
either K1 or K2. Verify before starting that removing input power to the
ignition switch actually kills the engine, I guess.
Fewer parts, less to go wrong.
Otherwise, what you draw should work.
--
Cats, coffee, chocolate...vices to live by |
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| Ecnerwal |
| Posted: Mon Apr 21, 2008 1:09 pm |
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In article
<ab0508ee-4786-4904-8add-6263a524dc41@c19g2000prf.googlegroups.com>,
Mike H <mike8675309@gmail.com> wrote:
Quote: Looking at your proposal, if I have a 14gauge wire carrying up to 14.6
volts (alternator output when charging) and as low as 10.5 volts (low
battery power until alt spins up. but enough to fire the ignition
system), what would be the best gauge wire, and method of connecting
it to limit voltage drop over 30 feet (15feet one way) to a minimum.
If I'm going to switch a power source for sensors, I need to be
certain that voltage drop is as minimal as possible otherwise sensor
readings may be mis-interpreted by the EEC-IV processor which monitors
the sensors.
What matters for voltage drop is not the 10.5-14.6 voltage, but the
current. V=IR, which applies to this problem as "voltage drop" = current
in wire * resistance of wire.
As a specific example or seven (for 30 feet):
10 amps on 16 gage wire - 1.2 V
1 amp - 0.12V
10 amps on 14G wire - 0.75V
1 amp - 0.075V (see a pattern? we'll drop the 1 amp case from here out)
10 amps on 6G - 0.12V
10 amps on 2G - 0.047V
10 amps on 4/0G - 0.015V
You end up making a practical call on wire size .vs. current drawn .vs.
length .vs. acceptable voltage drop for the application. If the circuit
is drawing an amp, 14 ga is probably fine.
The computer presumably already deals with the (large) variability of
the overall voltage in the system, so this may be less critical that you
guess at first. Depends on the sensor type and how it's being
interpreted, which I don't know the specifics of.
--
Cats, coffee, chocolate...vices to live by |
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