Ed and Joe,
The diode in the EC2 allows the current
from the collapse of the injector magnetic field to flow to the positive supply
rail (~14V); it doesn’t oppose this.
A resistor allowing this current to flow would also result in a close
delay since the current flowing is what maintains the magnetic field during
this delay. What is needed is
a way to decrease the rate of voltage rise just after the EC2 pulse ends so
arcing in the A/B selection relay is suppressed. After the relay contacts open enough
that an arc is no longer possible (which shouldn’t take long) an open
circuit condition now would allow the injector to close quickly. The arcing may or may not be a problem
any given time the relay is opened since the timing of the end of the EC2 pulse
and the opening of the relay are independent and arbitrary. One possible solution is a RC snubber rather than the resistor that Joe proposed. I think Tracy is working on checking this out. I have installed this in my plane and it
works. Tracy, however can do a much more thorough job of evaluating this and be
sure the change is reliable. Let’s
give him a chance to do this before we do something we wish we hadn’t.
Steve
-----Original Message-----
From: Rotary motors in aircraft
[mailto:flyrotary@lancaironline.net] On
Behalf Of Ed Anderson
Sent: Sunday, March 11, 2007 9:16 AM
To: Rotary motors in aircraft
Subject: [FlyRotary] A solution?
was : The truth??? / Injector flow rate mystery solved
Sounds like a
reasonable approach to me Joe. A pull-down resistor
would be relatively easy for me to install - I have the
resistor pack required for the peak-hold type injectors. So I could
easily place four additional resistors in that box.
If I understand you (please correct
me if I don't), the pull down resistor should go between the injector and the
EC2 sinking terminals. That way the current induced when the intended
pulse terminates and the magnetic field collapses will have a path to ground
rather than being opposed by the diode in the Ec2.
The value of said resistor could be
around 100 ohms. Since the induced voltage could reach from 50 - 100+
volts an 100 ohm resistor could flow from
0.5 - 1 Amp (for a very short
duration). As far as affecting the 12 Volt signal it would only draw
12/100 = 0.12 amp or 120 milliamps. That would be pulled through the
injectors at all times. The injector resistance is probably (peak
and hold case) around 3 ohms. So the injector would draw 12/3 = 4 amps
(DC case - its undoubtedly less due to the A/C impedance of the coil). It
make take some experimenting - but 100 ohm looks like a good place to start.
The wattage should probably be
around 5 - 10 watts just to be on the safe side.
So certainly looks like a suggestion
that would work, Joe.