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Old 27 February 2021, 13:34   #1
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Resistor to Mimic Incandescent Bulb

I had a discussion on here as to whether removing the kill cord should cut power to the fuel shut off solenoid or provide power to the solenoid. In the end and following forum advice I went down the latter route.

It's a Yanmar diesel inboard.

In the event of going over the side, the solenoid is energised and cuts the fuel supply. The solenoid stays energised until either the ignition is turned off or the kill cord is replaced.

To try and prevent the solenoid from burning out, I added a relay that removes power from the solenoid once the oil pressure light comes on i.e. when the engine stops.

It doesn't work and I think it is because the oil pressure light is LED and it doesn't draw enough current for the relay to recognise it is on and therefore the solenoid remains energised.

Is there a resistor that I can fit in a blade fuse holder that will draw sufficient current to mimic an incandescent bulb?
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Old 27 February 2021, 13:54   #2
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Why not use a multi function relay that has multiple timers built in. This is just an example grabbed quickly https://www.amazon.co.uk/GRT8-A1-12V.../dp/B07W4TXKP8
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Old 27 February 2021, 14:00   #3
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Quote:
Originally Posted by GuyC View Post
I had a discussion on here as to whether removing the kill cord should cut power to the fuel shut off solenoid or provide power to the solenoid. In the end and following forum advice I went down the latter route.



It's a Yanmar diesel inboard.



In the event of going over the side, the solenoid is energised and cuts the fuel supply. The solenoid stays energised until either the ignition is turned off or the kill cord is replaced.



To try and prevent the solenoid from burning out, I added a relay that removes power from the solenoid once the oil pressure light comes on i.e. when the engine stops.



It doesn't work and I think it is because the oil pressure light is LED and it doesn't draw enough current for the relay to recognise it is on and therefore the solenoid remains energised.



Is there a resistor that I can fit in a blade fuse holder that will draw sufficient current to mimic an incandescent bulb?


Are you picking up from the right side of the oil light? One side will be high when the engine stops, the other will be to earth. Sounds like youíve picked up the wrong side. Unless Iíve mis-understood.
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Old 27 February 2021, 14:16   #4
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As long as the MAIN concern/priority is educating the Power Boating community to ACTUALLY USE their Kill Cords 100% of the time and they work!!....the rest is just samantics and or Bollox!
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Old 27 February 2021, 14:37   #5
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I would have gone the other way with the kill switch circuit. Firstly it is more simple and secondly its fail safe. if the kill circuit fails then the solenoid drops out and kills the engine. if the solenoid fails the engine stops but can be restarted by tying up the lever on the pump.you could use a smart charge relay that switches contacts when it detects that the engine is running 13.8 v and switches out when engine stopped 12.8v approx voltages.
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Old 27 February 2021, 14:43   #6
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https://www.ebay.co.uk/itm/160477-0-...MAAOSwFZdd~QeA
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Old 27 February 2021, 14:52   #7
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I have also seen a system that shuts off the air to the inlet manifold which kills the engine and has to be reset before use. This system also acts to prevent diesel engine run away which normally results in a destroyed engine. The military use this system as diesel engine run away can also result in fire
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Old 27 February 2021, 15:55   #8
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Quote:
Originally Posted by GuyC View Post
Is there a resistor that I can fit in a blade fuse holder that will draw sufficient current to mimic an incandescent bulb?
Tell me the wattage of the incandescent bulb and I'll tell you the resistance value and the power rating of the resistor.
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Old 27 February 2021, 19:13   #9
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Hi GuyC
I have designed safety circuits for production equipment. The main rule for safety circuits is, the circuit must be self interrogating.

This means that you must always know if your safety circuit is working properly and you will not be able to start equipment if it's not. Therefore a safety circuit is always a "closed circuit." In other words you must have current running through it all the time, so if there is a break in the circuit anywhere the equipment won't start.

Using the main rule of a safety circuit you have taken the wrong route. If you are going to supply power to the solenoid when the kill switch is operated but there is a break in the wiring to the solenoid, even though you have operated the kill switch, power won't get to the shut off solenoid.

Your first option of having the power to the fuel solenoid going through the kill switch is self interrogating as if there is a break in the wiring to or from the kill switch, you can't start the engine until you have fixed the problem.
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Old 28 February 2021, 06:52   #10
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Quote:
Originally Posted by Oldman2 View Post
Why not use a multi function relay that has multiple timers built in.
I do have a timer relay that provides power for a short period of time - about 10 seconds. I'll use this if I can't sort out another way.

Quote:
Originally Posted by Maximus View Post
As long as the MAIN concern/priority is educating the Power Boating community to ACTUALLY USE their Kill Cords 100% of the time and they work!!....the rest is just samantics and or Bollox!
Thnaks

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Originally Posted by mikehhogg View Post
I would have gone the other way with the kill switch circuit.
That was my initial thought but the consensus of the forum was to go the other way round. You want the engine to continue working if the power dies.

Quote:
Originally Posted by mikehhogg View Post
I have also seen a system that shuts off the air to the inlet manifold which kills the engine and has to be reset before use. This system also acts to prevent diesel engine run away which normally results in a destroyed engine. The military use this system as diesel engine run away can also result in fire
The Mermaid that was in my Pac 22 has this system. The air filter on the Yanmar makes fitting one more more difficult.

Quote:
Originally Posted by Salty Pete View Post
Hi GuyC
I have designed safety circuits for production equipment. The main rule for safety circuits is, the circuit must be self interrogating.

This means that you must always know if your safety circuit is working properly and you will not be able to start equipment if it's not. Therefore a safety circuit is always a "closed circuit." In other words you must have current running through it all the time, so if there is a break in the circuit anywhere the equipment won't start.

Using the main rule of a safety circuit you have taken the wrong route. If you are going to supply power to the solenoid when the kill switch is operated but there is a break in the wiring to the solenoid, even though you have operated the kill switch, power won't get to the shut off solenoid.

Your first option of having the power to the fuel solenoid going through the kill switch is self interrogating as if there is a break in the wiring to or from the kill switch, you can't start the engine until you have fixed the problem.
Fail safe in my case is for the engine to continue running.

Quote:
Originally Posted by Pikey Dave View Post
Are you picking up from the right side of the oil light? One side will be high when the engine stops, the other will be to earth. Sounds like youíve picked up the wrong side. Unless Iíve mis-understood.
Interesting and I'll go and have a look at this.

Thanks for all the responses.
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Old 28 February 2021, 14:33   #11
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If I'm understanding you correctly, you're picking up the high which illuminates the led but that circuit doesn't provide adequate power to energise your relay.
If I've got that right all you need is a wee transistor driver circuit. Picture below.

Connect the high from your led circuit to the 'signal in' of the driver circuit.
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Old 28 February 2021, 14:53   #12
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Originally Posted by jwalker View Post
If I'm understanding you correctly, you're picking up the high which illuminates the led but that circuit doesn't provide adequate power to energise your relay.
If I've got that right all you need is a wee transistor driver circuit. Picture below.

Connect the high from your led circuit to the 'signal in' of the driver circuit.
That does sound like the issue and the solution. I'll have to follow the lead PD gave me and check I've got the relay on the high side.

Thanks for info.
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