More fuel injector questions

[et...@whidbey.com]

I figure Clare will know the answer but maybe others will too. I got
my hands on an old Acura fuel injector last night. It is apparently
the high impedence type because it measures 14 ohms resistance and
draws only .7 amps @ 12 volts. This means I don't need a fancy chip,
just a 555 timer circuit. So I put one together using a relay (because
that's what I have on hand) instead of a mosfet to apply power to the
injector. I put a snubber diode across the injector just like I did
for the relay coil. All the injector circuits I have seen online use
either a pretty high voltage zener, 75 volts, or a resistor across the
injector. From what I have read the 75 volt zener seems to be the
most common and it lets the injector close slower to prevent damage to
the injector from slamming closed. I guess the circuits with a
resistor in parallel with the injector do the same thing. I don't have
any 75 volt zeners in my kit but do have lots of resistors in many
different values and watt ratings. Does using just a resistor make
sense? Maybe I misread the circuit.
Thanks,
Eric

[Tim Wescott]

The higher the voltage at the injector, the faster the magnetic field
will collapse and the faster it'll close.

75V means that they're using transistors rated for 100V or so, which
means spendy. It also means that the magnetic field is collapsing way
faster than it was generated. So they're at least tuning the turn-off
time.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

I'm looking for work -- see my website!

[et...@whidbey.com]

So can I just use a regular diode like I would with a relay but with a
resistor in series with the diode if I want to have the injector close
slower?
Eric

[Jim Wilkins]

<et...@whidbey.com> wrote in message
news:gqk28chag4lvqen85...@4ax.com...

A resistor in series with the snubber diode makes the injector coil's
magnetic field collapse -faster- by dissipating the stored energy that
drives that current. Whatever current was flowing through the coil
when you switched off the power continues to flow through the diode,
whose purpose is to prevent that current from generating a high
voltage surge, like an ignition coil or a water hammer in plumbing.
The resistor absorbs the energy from the current faster than the diode
drop alone. I've seen a relay stay closed for 15 seconds with only a
diode drop impeding the current.

-jsw

[cl...@snyder.on.ca]

A resistor won't do it - it will just make the injector draw more
power. The relay won't act fast enough. I'd use a diode backwards
across the coil - a 1n5404 or 5406 will be plenty. Might get away
with a 1n4004 (400 volt 1 amp)

[cl...@snyder.on.ca]

You could - but you really want it to close fast. A resistor could
limit the current so you can use a 1n4004 instead of a higher current
device.

[David R Brooks]

If you will be installing multiple injectors & drivers, you can also
use fast diodes returned to a common "supervoltage" rail, which has a
capacitor & Zener down to either ground, or the DC+ rail. The
supervoltage rail may float at +75V (say), being charged by the
"flyback" pulses from the injectors. This idea comes from an old Epson
application report, directed at dot-matrix printers (not so different,
if lower power.)
If you do this, you will quickly see just how much energy is wasted in
heating that Zener. There are ways to reclaim that energy, & return it
to the main DC rail, effectively using it to help fire the next injector
in sequence. (Sorry I can't discuss the details: they were the property
of my employer.)

[et...@whidbey.com]

Thanks Clare. I know that in an engine the injector needs to close
fast but in my application, applying tiny amounts of cutting lubricant
to a spinning tool, I can get by with a slower closing speed. What I
do want is for the injector to last as long as possible. So as long as
the injector isn't closing any faster than it would in a car I'm
happy. I wonder if the higher viscosity of the lube will be cushion
enough?
Eric

[Tim Wescott]

Well, I don't know how you usually connect a diode with a relay.

The _higher_ the voltage across the coil, the _faster_ the current will
decay.

[Tim Wescott]

A buck converter, from the 75V rail down to battery, with the feedback
coming from the supervoltage rail (because you're trying to regulate that
voltage and not the battery voltage).

Not completely easy-peasy, but it should be possible with a handfull of
off-the-shelf parts.

[Tim Wescott]

If you're talking about a circuit like this, then the HIGHER you make the
resistor value, the FASTER the current in the coil will decay, and the
FASTER the injector will close.

If you have a website or some other easy-to-access reference that says
differently, that was written by someone who actually knows their physics
(i.e., not just some columnist for a car magazine), then I'd like to see
it.


+12V o ----------------o-----.
| |
| |
.-. |
resistor | | |
| | |
'-' C| injector
| C| coil
| C|
| |
- |
diode ^ |
| |
| |
o-----'
|
|
|
||-+
||<-
o---------------||-+
|
|
===
GND
(created by AACircuit v1.28.7 beta 02/28/13 www.tech-chat.de)

[Jim Wilkins]

<et...@whidbey.com> wrote in message
news:nkg48cl72p1u5m5r1...@4ax.com...

I would obtain all four injectors from a junkyard engine and test the
worst looking for the minimum voltage and current to open it and the
maximum that burns it out, or at least see if it withstands 12V to 14V
continuously. Then I would measure the steady state flow rate versus
the fluid pressure, or height of the unpressurised reservoir if it's
enough.

Once you confirm that they are acceptable and know their limits you
can refine your design.

-jsw

[cl...@snyder.on.ca]

There is nothing that will be dammaged by slamming the injector
closed, and a slow close can allow the nozzle to drip rather than
sprat at the end of the injection - you want the lube on the work, not
running down the lube application device.

[cl...@snyder.on.ca]

Unless the reservuir is several hundred feet up, the pressure won't be
adequate. He will need a pump, and likely some sort of regulator. a
simple pop-off would be adequate and simple.

[Steve Walker]

On 1/20/2017 5:38 PM, cl...@snyder.on.ca wrote:
<SNIP>

>>
> Unless the reservuir is several hundred feet up, the pressure won't be
> adequate. He will need a pump, and likely some sort of regulator. a
> simple pop-off would be adequate and simple.
>

True. IIRC GM fuel injectors run at about 45 PSI. Any lower, and the
spray pattern will be poor. Another IIRC, disc type injectors have a
wider spray pattern. Using a pintle type will provide a narrower
pattern. If you can get injectors from a GM throttle body, they operate
on about 12 to 14 PSI.

[cl...@snyder.on.ca]

On Fri, 20 Jan 2017 19:40:16 -0500, Steve Walker <Im...@home.com>
wrote:

And flow a LOT more.