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That's one way. Apply a known voltage to the injector and monitor the rate of rise of the current. Or use an inductance meter.is there a simple way to estimate the inductance of the solenoid? I suppose it would need a time base scope.
Is that 18V a brief pulse (1mS) or is it the steady-state voltage needed to fire it (i.e. with the current limited by the coil resistance)?Yet we needed 18 v minimum to get it to Fire. And about 15A in rehearsal without it in place.
Can't get there with those numbers. Current = voltage / resistance, so with 4 ohm the most you could hope for is 6 amps. The fact that it is 4 ohms would also suggest a lot of wire and pretty high inductance. Any part number on the injector?
Ah, so that doesn't get us much further forward. Can you zero your meter ohms range (there should be an adjuster somewhere)? Or connect the coil in series with a known resistor (say 100 Ohm, 2W) across a 12V battery and measure the voltage drop across the coil, then calculate its resistance from the measured volts.Of course, I don't know if 4 ohms is correct. If the Multimeter said 0 for zero, not 4 ohms, I'd be more certain. The meter says 4 and so does the coil - the coil might be close to zero ohm.
Hello:
May I ask: concerning the solenoid in a fuel injection unit, where the solenoid lifts a ferrite piston, allowing the ultra-high pressure fuel to inject, what happens when the current is cut?
In my circuit, about 16 amps at 20 v DC briefly (about 2 ms) activate the coil. The current comes from a Current Generator.
On switch-off, (I presume) there should be an induced spike pushing "current" from the "bottom" of the coil in the same direction to earth. We have a big diode permitting this current to return to the top of the coil.
I'm wondering how this might delay the valve closing. Given that the coil still goes to earth, maybe it
Yes, while current flows through the solenoid and is above the hold open threshold, the injector will remain open. The inductive energy is dissipated by the resistance of the coil and the voltage drop across the diode.