What I mean is with an RDSon of less than an ohm and a current of, say, 1 amp, the voltage drop across the mosfet will be less than 1V. That means, the Vgs (if the gate is tied to Vcc) will also be less than a volt. At least that is the way it is explained in the IR literature.

I think we are saying the same thing, namely that Vgs need to be 10V for regular mosfets. If you notice the times on our posts, I was writing while you were posting. I don't think mine adds anything to your post, except introduction of the term Vds. John

Edits: I mean Vgs not Vds
Edit#2: BTW, I notice the OP wants something on the order of a 60 second delay. The left circuit works fine for a few hundred ms, but would keep the mosfet in the linear region for a relatively long time , if one had it set for a 60 second delay. For such long delays, I would consider something like a comparator between the switching transistor and the RC timer to give a rapid turn-on/turn-off period.

The problem is, enhancement mode MOSFETs have a Vgs threshold, below which they basically don't conduct. The threshold varies with part number, and even quite a bit among individual units of the same part number, but it is always on the order of volts - not millivolts. Below is the V-I curve of a diode-connected (Vgs=Vds) IRF530, which I picked at random from the LTspice model collection that I have. The IRF530 is spec'ed at Rds=0.11 ohms max with Vgs=10V. However, it's maximum threshold voltage (Vgs=Vds, Id=250uA) is 4 volts. The curve below is actually a plot of that worst-case part. The minimum threshold voltage is 2 volts.
In the plot below, you can see that, with Vds=5V, Id~2.5A (Pdiss=12.5W). With a 12V supply, this would occur with Rload~2.8 ohms. You only get 7V across that load.
However, if Vgs=10V (Rds=0.11 ohms), Vds would only be about 450mV, and Id~4.1A (Pdiss=1.9W).

I have included the LTspice .asc file, in case anyone wants to play with the sim.

 

I have got to learn a simulator! It took me a couple of years to go from sketching on paper to drawing in CAD to doing it the other way around. There is still hope for me yet. John

 

Wow. All I can say is that I am very grateful to you all for yoru discussions, and I am totally lost lol.

I was looking, quite simply, to have an adjustable delay on a 12v output triggered by turning off a 12v input. Even a cap across a relay will hold it open a bit....

Seems like it's not as simple as I thought. Would I actually be simpler going back to the 555 design I had originally?

One more thing - the delay when the 12v input is cut is fine but the circuit must not create an output when the 12v input is there.

I attach the 555 circuit I had help working out and wonder if anyone can suggest something simpler as it looks like my idea of a cap/tranny/pot was a bit over simplistic from what some of you are saying.

So long as it stands up to living in a car for a few years and I can trim the value to the seconds I want (roughly 20-40 sec but I gave it a bit wider margin) and once trimmed it stays there stably over the years, I'll be happy.

Greg.

 

You could try this. M1 has a brief period of high dissipation, but I think it will be so short that it will not do any damage. The positive feedback through M2 causes it to turn off rapidly once the voltage across the load drops by a couple of volts. If you think your solenoid will drop out prematurely under that condition, then it won't work.

One advantage of this is that, once the circuit has timed out and the switch is still off, it draws zero current. The 555 will always draw a little current.

 

Well the output of that certainly looks good Ron, it stays near enough 12v and drops off hard, which would suit my solonoid output.

Q - what are the timing min/max on this? Also just checking that when the trigger is high there is no output - output only when trigger cut? That's what I need, no output till the +12v trigger is cut.

Drawing current really isn't an issue, it's only to be used when the engine is on (so really more like 14.4v) and it has a 90amp alternator so I don't think any circuit I make electronically will worry it....

So Ron, if it was you, would you head down the 555 driving a realy route or this route? I think there as many componants in this as in the 555 circuit.

Thanks again,
Greg.

You could try this. M1 has a brief period of high dissipation, but I think it will be so short that it will not do any damage. The positive feedback through M2 causes it to turn off rapidly once the voltage across the load drops by a couple of volts. If you think your solenoid will drop out prematurely under that condition, then it won't work.

One advantage of this is that, once the circuit has timed out and the switch is still off, it draws zero current. The 555 will always draw a little current.

 

As drawn, the timing range is nominally 6 secs to 63 secs. It will be dependent on the individual MOSFETs you choose, and the tolerance of the timing components, but it should cover the range you want. If not, you can always make C1 bigger and R6 smaller to change the range.
Before I answer your 555 question, let me ask one. Are you powering the entire circuit with switched 12V? If the circuit is powered when the engine is not running, the 555 will always draw current. Also, your 555 circuit needs a few more parts.

The circuit only comes on when the switch opens. See the attached waveforms.

 

Other than a cap across the supply what parts do the 555 circuit need, I thought I had it finished?

The circuit is only used when the ignition is on/engine running. It's therefore powered by ignition switched +12v. The trigger is also drawn from a car 12v supply via a switch to another circuit, which when I turn that circuit off I need this circuit to provide it's output.

I hope that answers the power question - any kind of power drain really isn't an issue.

Looking forward to your comments, I'm itching to actually make something on the breadboard I now have in anticipation!

Greg.

As drawn, the timing range is nominally 6 secs to 63 secs. It will be dependent on the individual MOSFETs you choose, and the tolerance of the timing components, but it should cover the range you want. If not, you can always make C1 bigger and R6 smaller to change the range.
Before I answer your 555 question, let me ask one. Are you powering the entire circuit with switched 12V? If the circuit is powered when the engine is not running, the 555 will always draw current. Also, your 555 circuit needs a few more parts.

The circuit only comes on when the switch opens. See the attached waveforms.

 

Other than a cap across the supply what parts do the 555 circuit need, I thought I had it finished?

The circuit is only used when the ignition is on/engine running. It's therefore powered by ignition switched +12v. The trigger is also drawn from a car 12v supply via a switch to another circuit, which when I turn that circuit off I need this circuit to provide it's output.

I hope that answers the power question - any kind of power drain really isn't an issue.

Looking forward to your comments, I'm itching to actually make something on the breadboard I now have in anticipation!

Greg.

Your 555 circuit will trigger on contact bounce when the switch closes (not good), and it puts a spike on pin 3 that goes 12V above Vcc. The spec limit is 0.3V above Vcc.
My circuit will also trigger (briefly, only during the few milliseconds of bounce) on contact bounce. I'll show you how I think both can be fixed.

 

The OP is looking to maintain the relay pulled in for roughly 60 seconds after the supply is turned off. I don't think the 555 will maintain that level of current through a relay for anywhere near that long after its supply is cut. Or, maybe I misunderstand what he means by powering it through the ignition switch. John

 

The OP is looking to maintain the relay pulled in for roughly 60 seconds after the supply is turned off. I don't think the 555 will maintain that level of current through a relay for anywhere near that long after its supply is cut. Or, maybe I misunderstand what he means by powering it through the ignition switch. John

The circuit is powered by 12V which is only present when the ignition switch is on. The switch that triggers the circuit is separate from the ignition switch. He only needs the circuit to work when the ignition switch is on.

 

Sorry. With clear eyes and clear head, I re-read the comment immediately above and see what he means. Too many hours yesterday on tax issues. John