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ic 4013 & timer

malc9141

Nov 19, 2009
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What you have said is what I thought, but now it's completely clear. Thank you.


I'm not sure what configuration you have
I'm going to leave the solenoid configuration for the moment and ask about plan B (or was it A) .

Plan B was to to feed 36 v in, perhaps turning down current. I'm sure we have XS current (because I know of an injector test rig which operates at about 5 amp but doesn't need to open and close as fast as we need). We can run at 15 amp.

We argued about this, current and voltage, and you have introduced an additional idea. To explain our argument, because the valve needs to open too fast to let the bleed drop the pressure, I said we needed high voltage to build a quick current (The Man estimated the inductance is only about 2.5 uH . So he couldn't believe we needed more voltage. But at 12.5 v it rarely worked at all while at 18 v it was reliable in test mode).

But now (since your remarks) we are wondering if it's opening OK but not shutting fast enough. Possibly big-step-forward.

Now will this help? Plan B: apply 36 v to the current generator. Put a 10 amp fuse in to the lead to the solenoid. Protect the timer circuit by taking the 36 v rail, and between that rail and the 7812, putting a 2.5 resistor (0.5W) and then the Zener diode (10 v 1.5 W) and then the 7812 with its two caps. Then the ic 4013.

Question: is that OK?

Now, the current in the solenoid will collapse faster. But I'd better wind back the applied current, I suspect. So, second Question: is that right?

This has avoided using the kick-back idea to generate voltage which might lead me to do something beyond my understanding. The thing at present works remarkably but not perfectly so it's best to alter things the least.

Malc





I suspect
 

Resqueline

Jul 31, 2009
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With 2.5uH and 10V applied the current will rise with 4A/us, with 35V applied it'll rise with 14A/us. With the standard protection diode scheme it'll take almost as long to ramp down the current.
But also remember that with microsecond pulses the driver transistor driver will have to be well designed.

The regulator plan B is correct. I see you choose 2.5 ohms which is a lot lower than my choice but probably still ok. I'd be more comfortable with 5 Ohms though.
The fuse can be a lot smaller because the pulses are narrow & far between. I believe 5 or even 3A should hold up.

Second question: With the higher voltage the current will rise faster but it won't collapse faster unless you employ the other protection diode scheme.
Since you have a constant-current driver you need not adjust the current much higher than what just makes the solenoid open. The high voltage will still be applied initially, until the current reaches the set point whereupon the voltage decreases. Any higher current will not make it open faster (in theory) but will lead to higher current consumption, more heating of the solenoid, and quite possibly a longer turn-off time.
 
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malc9141

Nov 19, 2009
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This is all extremely helpful. Far beyond what I had expected. I think the current decline idea might be crucial.
BTW, I think I did get a 10 ohm R - not 2.5 as I said.

You say it won't collapse faster unless you employ the other protection diode scheme.

this being ---

To get a high dc voltage you just put this diode from the solenoid drive line and to a 40V capacitor instead. Then you need a 36V zener in parallell with the capacitor to limit the voltage. The solenoid supply is taken from the capacitor instead of the B+. Put a diode from B+ and to the capacitor to get the circuit going in the first place.
The advantage with this configuration is that the solenoid turns off faster than with a diode directly across it.


Mmm. Can't see this but do accept it. I suppose the cap/zener must blunt the flow of current to the Top (+) of the solenoid. But won't this cause a nasty negative surge in the circuit that provides the current - as if the diode wasn't there???

I really am interested in your idea of collapsing the current quickly.


Maybe I haven't gained anything with the extra input voltage to the current generator? Incidentally, I'm assuming it is well designed. The Man always thought carefully and there was always a clear conversation.

Malc




To get a high dc voltage you just put this diode from the solenoid drive line and to a 40V capacitor instead. Then you need a 36V zener in parallell with the capacitor to limit the voltage. The solenoid supply is taken from the capacitor instead of the B+. Put a diode from B+ and to the capacitor to get the circuit going in the first place.
The advantage with this configuration is that the solenoid turns off faster than with a diode directly across it.
 

malc9141

Nov 19, 2009
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Still trying to understand but missing something.

1) At present, with our layout, a pulse of B+ goes into solenoid at "Top" and travels through coil to earth. At Switch Off, a +ve EMF develops at "Bottom" of solenoid. We shunt that back to via a diode to cancel the instant -ve EMF above the solenoid.

2) If I put a 40 v cap (let's give it +/- terminals) the + faces the current generator.
What happens with the Fire pulse? A current goes via the zener (in // with cap) to the Top of the solenoid as usual. Does some of that neutralise the -ve charge on the downside of the cap??? But how can we allow a +ve pulse into the -ve side of the cap. Isn't that forbidden?
Anyway, then, when the Fire pulse stops, a +ve EMF develops at the bottom of the solenoid and flows (via the original diode) to the top of the solenoid (which is now -ve). Does this neutralise the instant -ve charge from the Top EMF developing in the cap-ve side??

3) I can't see why this allows the solenoid to discharge fully any faster - ie get rid of the bottom +ve switch-off EMF - than if the switch-off + ve EMF goes straight to earth, or is part diverted to the Top.

Sorry to be banging on; you've been very helpful and it's you and me and my expert engineer. But at this end, we have no electronics background. I'm using ancient school physics!

BTW, I bought and read a marvellous book, Great Physicists by W H Cropper ISBN
978-0-19-973516-7 which tests the imagination. If you're in a Library, it's worth a look!

Malc
 

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Ok, so you have a top-side driver. That makes for a negative spike at turnoff. My description above was for a low-side driver.
But if the driver transistor is a PNP (or P-channel) and so has collector towards the solenoid you can still make a high-voltage "bleed/supply" (a negative rail) utilizing the Bosch idea.
The driver transistor will be exposed to a higher voltage yes, and the sum of the power rails must not exeed the voltage rating of the transistor.

If the driver is an NPN (or N-channel) then it has emitter towards the solenoid and then you can't do it any other way than what you have now.
But reducing the driver current will also reduce the turnoff time.

The thing with coils is that at time zero at turn-on the current starts at zero and increases linearly. It may eventually level out due to its resistance. At turn-off the coil has accumulated a charge, consisting of the energy product of current times time (the area under the curve). When turned off it wants to dispense with this energy, and the higher the (output) voltage the shorter is the time needed to account for this energy "packet" [Joules].
And also the higher you have set the current the longer it will take to dissipate this energy.

Imagine Einstein, inventing the concept of lasers in 1917, and only in 1960 we we able to actually make one.. That's creative thinking!
 
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malc9141

Nov 19, 2009
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Whew.. maybe getting there.
The drive transistor is a ZTX653 which (looking at the Man's diagram) seems to be an NPN as it has an arrowhead joining the line to the solenoid.

So I'll try just upping the voltage and seeing how little current I can get away with (I know i & v are in no way interchangeable - not saying that - the current 'lifts the valve').

Einstein. No doubt about it, but there were others, too. I thought Lasers came out of wartime radar theory. I don't worry that Einstein has light as particles (with mass) yet travelling at c, seeming to contradict his own rule. Of course, his leap there was that he simply didn't mind whether they were particles or waves.
A thought (as someone at one time doing physical chemistry), constants are often dimensionless (satisfying) but sometimes dimension-ful. In physical chem, you have constants with dimensions which are conditional, eg mass ^-1.
Now wouldn't it be interesting if Einstein's constant c (metres/sec) was conditional? Only true in our present universe. Might have different values at different temp and pressure! Of course, I have no doubt this idea is a bit daft.

Anyway, my job is to make this b* engine perform! Thanks again.

Malc
 

malc9141

Nov 19, 2009
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No, sorry, the transistor is MJ1503 but it's still NPN if the arrowhead pointing out to the output means that.
 

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Jul 31, 2009
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Ok, it seems you can't shorten turnoff time other than using as low drive current as possible then.
Good luck experimenting!
 
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