[Brianj_92505]

I need a circuit to use for the stop solenoid on my tractor. It will have a 12 volt line coming in as well as the 12v from the ignition on switch. The Solenoid will be powered through a relay. What I need is when the ignition is switched off the relay gets turned on for about 3 or 4 seconds, long enough to stall the engine. I thought about something like a resistor and a large capacitor in series across the two power inputs and the junction connected to the base of a pnp transistor. When the ignition gets turned off the base should go low while the capacitor charges turning on the transistor and the turning off again when the cap is charged. The ignition side has a load on it so it provides a path to ground when turned off. Does this seem feasible?

 

[Harald Kapp]

Could the circuit used for turning on the lights in a car be used? These circuits turn on the lights when the door switch is activated and turn off the lights after some time. The circuit would be activated by the ignition switch instead of the door switch.

Google "car interior lights delay" for circuit diagrams or modules.

 

[duke37]

This seems strange logic.
You do not say what the solenoid does, it is unlikely that it stalls the engine. Does it push a stick against the flywheel?

The normal situation is to have a fuel valve which is turned on when the ignition is activated and sprung into the off position when the ignition is off. This is the fail safe method.

Early diesels had a stop knob which needed to be pulled out to stop the engine, these have been superceded for safety. If you have one of these, you could use a spring to pull the knob out and the solenoid to push it in.

My car has a fuel pump which only runs when the ignition is on.

 

[Brianj_92505]

The solenoid blocks fuel when energized which then stalls the engine. Why is that unlikely? I agree that a better design would have been a solenoid that allows fuel to flow only when energized but that’s not how Massey Ferguson chose to do it. The old relay is defective and it’s like $100 to replace it. Looking for alternatives.

 

[duke37]

I interpreted a stall as increasing the load to stop the engine.

I think the attached circuit will do what you want, it will need a FET of sufficient size to drive the solenoid and will need a flyback diode across the solenoid with sufficient current capability to pass the solenoid current for a fraction of a second.

 

[(*steve*)]

I think that circuit assumes that the ignition is pulled to Vcc/Ground.

What are the charge and discharge paths for the 100uF capacitor?

Is the 4093 a schmitt trigger input device? (Yes it is -- good)

edit: a 1M resistor across the 100u capacitor would be a minimum requirement. (100uF, really?)

 

[duke37]

1. The input goes from Vcc to ground as described in #1.
2. I can't write properly! It is not 100µF it is 100nF, charged and discharged from the input 10k.
3. Yes, the 4093 is a quad dual input Schmitt NAND. One gate not used.
I used a 4093 since I bought what I thought was one at a radio rally but was handed a tube with more than a dozen so almost everything I look at has to use one.
4. The edit - see1. The input resistor/capacitor is there to attenuate spikes from the dirty 12V supply.
5. A diode should perhaps be placed on the input of gate 3 to aid the internal diode, limiting the input to Vcc.
6. The supply to the 4093 should perhaps be fed with a resistor (220?) and bypassed with a capacitor (1000µF?).

 

[(*steve*)]

1. The input goes from Vcc to ground as described in #1.

Yep, OK.

2. I can't write properly! It is not 100µF it is 100nF, charged and discharged from the input 10k.

Yep, looking closely, it is different from the mu, and perhaps could be an n

3. Yes, the 4093 is a quad dual input Schmitt NAND. One gate not used.
I used a 4093 since I bought what I thought was one at a radio rally but was handed a tube with more than a dozen so almost everything I look at has to use one.

4. The edit - see1. The input resistor/capacitor is there to attenuate spikes from the dirty 12V supply.

There are many ways to protect the input. A good one is to split your 10k resistor into two 4k7 resistors and then place a pair of diodes at the junction (going to the supply rails) or a zener to ground. This is very similar to the protection in the device, but obviously significantly more rugged.

5. A diode should perhaps be placed on the input of gate 3 to aid the internal diode, limiting the input to Vcc.

I'll have to look at the recommended limits on current through these, but it should be limited by the output impedance of the preceeding gate.

6. The supply to the 4093 should perhaps be fed with a resistor (220?) and bypassed with a capacitor (1000µF?).

Yeah, some protection for the 4093 from power supply transients would be a good thing. I'd probably consider a 15V zener in parallel with the cap if you do this. It will protect you from reversed input voltage too (although it will get pretty warm in this case)

 

[KrisBlueNZ]

There are many ways to protect the input. A good one is to split your 10k resistor into two 4k7 resistors and then place a pair of diodes at the junction (going to the supply rails) or a zener to ground. This is very similar to the protection in the device, but obviously significantly more rugged.

I would be more concerned about the whole 4093 being splatted by load dump voltage coming in on the +12V supply. I wouldn't use CMOS in an automotive application without plenty of protection; this function could be easily performed by some simple discrete circuitry (a transistor and a MOSFET wired as a Schmitt trigger with an R-C circuit on the input) which wouldn't need protection.

 

[Brianj_92505]

Here is what I settled on.
An NE555 timer with a single resistor and capacitor. When the ignition is on there is an equal 12v at both ends so the trigger/threshold is at Vcc and the output is low. When the ignition is turned off the trigger goes low and the capacitor begins charging (the path to ground is through the fuel pump which is connected to the ignition circuit.) During charging the output goes high driving the relay which connects the Battery to the fuel cutoff Solenoid. After a brief delay (long enough for the engine to die) the relay closes and the solenoid is de-energized. I was able to fit the circuit inside the old relay housing so I can just plug it in and go. A couple of concerns: Is the environment too noisy for this circuit, and is the relay going to get fried from switching the large inductive load?

 

[duke37]

Two comments
1. The input is AC coupled so interference could give false triggering.
2. You could use a FET instead of the relay. A diode across the solenoid can eliminate high voltage high current transients.

 

[KrisBlueNZ]

Yes, I would be concerned about electrical noise on the supply rails. If you don't want to add proper surge suppression, at the very least, use an IC socket for the 555!

Noise on the ignition input might cause the relay to chatter but the 555 is not used as a monostable here, so noise won't trigger the circuit and cause the relay to close for a significant amount of time.

I'd use a lower capacitance for C1 and a correspondingly higher resistance for R1. At least ten times lower/higher. Also I would connect a diode (e.g. 1N4001) across R1 with its cathode to the 12V rail, and put a small resistor (e.g. 100 ohms) in series with C1; this is all just to protect the 555.

You should also add capacitors between VCC and GND on the 555 - say 1x 100 µF electrolytic and 1x 0.1 µF ceramic in parallel.

Re wear on the relay contact, yes a diode across the solenoid coil and/or across the relay contact will prevent damage to the contact when it opens, by suppressing the back EMF from the solenoid. And obviously make sure the relay contact is rated for at least twice the current drawn by the solenoid.

You could use a MOSFET instead of a relay. You would need an inversion stage, driving a high-current P-channel MOSFET. Make sure you put a diode across the solenoid.

If the circuit will be permanently connected across the battery, check its current consumption. It should only be 5~10 mA which is small potatoes for an automotive battery.

 

[Brianj_92505]

I installed my Delay-Relay© (trademarked) on my tractor today and it performed flawlessly. The delay is about five seconds longer than it needs to be but that just means the solenoid runs a little longer and it cuts out the instant the ignition is turned on again. A diesel engine isn’t easy to kill, mine even runs with the electric fuel pump unplugged so for the past two months I’ve had to put it in high gear, hold the break and release the clutch. It’s nice to be able to simply turn the key again. I’m going to add a diode at the actual solenoid and just leave the relay alone lest I mess up a good thing. Honestly if it fails it’ll probably be from too much vibration on my pitiful solder joints.

Kris, what is your obsession with MOSFETs...do you own stock in the company. J/K, your probably right . I've never played with them before and consider them new-fangled and frightening. I always think of SCR's as the alternative to relays for switching big loads but just like my tractor they don't like to shut off.

 

[KrisBlueNZ]

MOSFETs are very cool. They're not new-fangled nowadays, but there does seem to be a lack of accessible tutorials covering the basic practicalities of how to use them.

I'm not obsessed with them, but for many applications they are an excellent choice.

The best thing about MOSFETs is that they can switch significant current (dozens, even hundreds of amps), with ON-resistances in the milliohm range, with no moving parts, in compact packages, often for under a dollar (USD). They are fairly easy to use, and they don't even require any current to keep them in either state.