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Darlington pair

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Ha - thanks.
I can't just change the L1 - it's a Delphi Fuel injector's internal solenoid.
But this is of help. I already have the diode (but didn't show it) but I'm actually thinking of taking it out. The circuit will handle the 20 V spike. I will probaly put a cap in.b
 
If you look at your post 15, I was using a 36 v source. This gave ~20 v across the Load.
In post 20, we have a 20 V source but you seem to show 20 V across the load, despite the extra 100Ω to the base of MJ1.

I'm interested in the EMF across the solenoid because it affects timing (this isn't engine timing in the usual sense, of course, It's concerned with the amount of time fuel is delivered at its pressure of 6 tons per sq inch!!). It's timing allowing estimation of duration of Squirt!.

So, to repeat, with your 100Ω at MJ1, what would the L1 P.D. be?

Many thanks.
 
I've skimmed the entire thread, but maybe I missed it.
Do you have specs, or goals, that you are shooting for, like peak current, turn-off time, etc.?
Is the inductor actually 5mH and 4.5Ω?

EDIT: It appears to me that some (most? all?) fuel injectors are driven on the low side. Is yours special, or am I wrong?
 
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Hi Ron

driven on the low side

Not sure what "driven on the low side" is. But the injectors have a solenoid which lifts a shaft up, by acting on a magnetised flat top. So in X-section, this looks like a T. The stem has a neck. When the neck moves into line with a passage on either side, pressure is bled off, resulting in (explosive) movement of a valve to inject fuel (pressure is ~1000 atm).

My problem is (since real specs are secret) that I was opening the bleed ok, but not closing it fast enough. So too much fuel always went in. Two milliseconds is too slow. Cars work at fractions of a millisecond.

Part of the problem, I think the snubber (diode) "reverse shorting" the coil was slowing the field reversal. I calculate from L and R that this would take about 0.002 secs (2 ms) which is too long.

So I want to reduce the final voltage in the coil to about 12 V (just before switch off).
L is 2 mH, R is 4.5Ω. Line voltage is now 20 v.

So all I want just now, is "what is the voltage" (the output emf) at the emitter of MJ1 in the diagram, after including the 100Ω from BD711 emitter to MJ1's base; or what is the voltage across the coil (to ground).

:)

I can decide whether to remove the snubber or put in a cap later. I'm in agreement with those who think a Zener might blow.
 
A low side switch is between the load and ground. Your Darlington emitter follower is a high side switch, i.e., it is between the load and the positive supply voltage.
You told us the inductance and resistance. What is the target peak current?
There are a lot of misconceptions about snubbers. The Darlington will "snub" your flyback voltage at ≈-1.4V when the base goes to 0V, which is pretty darned close to what you get with a big Schottky diode (≈-0.5V). So, even without the diode, I think your turnoff will be slow.
If you use an avalanche-rated n-channel MOSFET, you can snub the flyback at 400V or higher. I have been playing with one on my workbench, and it really speeds up inductor discharge time (different application, similar inductance and resistance).
If you are forced to switch the high side (e.g., if the injector is screwed into the block or manifold), you can use a PMOS switch, but I don't know if avalanche-rated devices are available in p-channel, so your snub voltage would have to be lower, but it could be a lot better than what you will get with a Darlington. As a big plus, if you use a MOSFET, you won't have to deal with the high power dissipation, and attendant heat removal problems, that you get with a Darlington.
 
Mmm. Good stuff. I now understand the "low side" term.

We can't put it on the low side, the way we have the set-up.

I began this thread a year or more ago, because it seemed a MOSFET (beyond my meagre understanding) would be an improvement (thread: How can I use a CMOS circuit etc).

I'll think about what you are saying. I don't have the resources to develop an entirely new circuit, sad to say.


********************

What I haven't told you is this:

The circuit I drew is a current generator - OK? But in action, there are two ! Both are actuated at time zero, but the one we are considering ("Hold") acts longer. So one ("FIRE") gives a high current to open the Bleed system. You need a quick high magnetic field to lift the ferrite top across a small space. You only need a small current to hold it there. So you have a timer > current generator (Fire) giving a 25 V max pulse, falling at once to a lower voltage (Hold current). It is that voltage I am trying to determine and reckon it should be about 12 v. The current max in this case is 12/4.5 (V/Ω). Say 2.8 A.


Meanwhile I will think about your snubber comments (I am not an electronics engineer and you guys collectively have been great help. Attempts to find experts to help practically have been bad news)

PS It might seem odd that I keep referring to emf, not current, in dealing with a coil. But it's come down to emf both because R is known and constant, and because speed is involved.
 
Have you thought about uploading LTspice IV https://www.linear.com/designtools/software/#LTspice, it's a marvelous tool and can be very educational.
FETs are a very fun toy also and well worth learning the basics. The most important things IMHO are the turn on voltage and the gate capacitance, if a gate (capacitor) is charged and not discharged, the FET will stay on and it must be discharged via a resistor to the source (of the the FET)
I hope the attached drawing help.
What's the number of the injector? Is it for diesel or gas
 

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I value all this advice. Muchly!
However, I had hoped that the project would be one of building the drive for superchargers, fitting it all, machining the piston, combustion chamber, inlet and exhaust systems, understanding the ignition to improve power.
I am not an electronics engineer and tho' willing to build something if I know it works, I really don't want to try to master fully yet another area. (I have done a lot in my time, but there aren't enough hours in the day to take on learning what would evidentally be required). This sounds negative, but I really have been busy.

"Never give up"
 
@KINARFI

"What's the number of the injector?" It's for diesel

Delphi EJBR 02101Z

I have the manual but values of components aren't given.
 
We can help with that, if we know what you need.

Two current generators (actually voltage generators for my purpose, since the coil has a fixed resistance). iGen 1 is Fire; iGen 2 is Hold.

Both supplied from a 20 V, high capacity source.
Each is actuated by a low current, timed pulse, orginally from the flywheel.
Both timed pulses act at the same moment. But Fire acts briefly whereas Hold continues longer. Fire is about 0.5 ms but can be varied; Hold is about 1 ms but is more variable. (ms is 10^-3 sec).

The solenoid is 4.5 Ω, ~2 mH,.

I reckon we need ~5 A to Fire (=pull plate across space), ~2 A to Hold.

In the lower figure, the timer could be acting on the iGens attached below the coil.

What do you think? Do-able ?
:)

Malc
 

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I reckon we need ~5 A to Fire
Max current from a 20V source with a 4.5Ω coil = 4.4A :(.

Edit: If 20V is applied to that coil the current rises to only 3A in 0.5ms, 4A in 1ms.
 
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Two current generators (actually voltage generators for my purpose, since the coil has a fixed resistance). iGen 1 is Fire; iGen 2 is Hold.

Both supplied from a 20 V, high capacity source.
Each is actuated by a low current, timed pulse, orginally from the flywheel.
Both timed pulses act at the same moment. But Fire acts briefly whereas Hold continues longer. Fire is about 0.5 ms but can be varied; Hold is about 1 ms but is more variable. (ms is 10^-3 sec).

The solenoid is 4.5 Ω, ~2 mH,.

I reckon we need ~5 A to Fire (=pull plate across space), ~2 A to Hold.

In the lower figure, the timer could be acting on the iGens attached below the coil.

What do you think? Do-able ?
:)

Malc
In your attachment, you show a low-side switch. In post #26, you said the low side had to be grounded.
We can't put it on the low side, the way we have the set-up.
What's up with that?
 
1 ms. OK - it is very tricky. You're damned if you do, damned if you don't.

The point is, though, it actually doesn't matter if it's (relatively) slow to rise. Just bear with me here a moment: if the flywheel is turning at 30 r/s, in your 1 ms, it has turned 360/30 degrees, ie 12 degrees. So I just time Fire 12 degrees earlier than I want. But it is an interesting point because I might have been Firing later than I thought.

Anyway, how long it takes to build the current doesn't matter in one sense. My timing starts whenever it is open (=Fire). But I suspected the problem was slowness in closing - that's where you've been helping here.
 
If you want 5A in 0.5ms you will need a much higher voltage than 25V.
 
In your attachment, you show a low-side switch. In post #26, you said the low side had to be grounded.

What's up with that?

Mix up. If we started afresh, the low side could be switched. However, with my existing circuit, it could not be adapted - physically.
 
1 ms. OK - it is very tricky. You're damned if you do, damned if you don't.

The point is, though, it actually doesn't matter if it's (relatively) slow to rise. Just bear with me here a moment: if the flywheel is turning at 30 r/s, in your 1 ms, it has turned 360/30 degrees, ie 12 degrees. So I just time Fire 12 degrees earlier than I want. But it is an interesting point because I might have been Firing later than I thought.

Anyway, how long it takes to build the current doesn't matter in one sense. My timing starts whenever it is open (=Fire). But I suspected the problem was slowness in closing - that's where you've been helping here.
Just to keep things clear here-
30revs/sec*360°/rev=10,800°/sec. That's 10.8°/millisec. Not sure how you came up with 12°.
 
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