roarin_mouse

How fuel injectors work
Fuel injectors are electrically controlled valves comprised basically of the valve, a spring to hold it closed and an electro-magnet to pull it open when the computer wants fuel to flow. Electromagnets are also called “solenoids”. When voltage is applied to a solenoid, current begins to increase and as the current increases, a proportional magnetic field is produced. If the voltage remains connected, the current will continue to increase until I = V/R (Ohm’s Law) and after that time, the current remains constant.

High impedance or “saturated” injectors have enough resistance to limit the maximum current to less than one Ampere. The maximum energy that can result is (I^2L)/2 or about 4mJ for an injector with 8 mH of inductance and 14 Ohms of resistance. The resistance, serving only to limit the current, consumes power in the form of heat (P = I^2R). That is, they burn about 14 Watts per injector whenever they are active.

It takes a lot less energy to hold an injector open than it does to initially open it. Low-impedance injectors take advantage of this. They have about 2 Ohms of resistance and 5 mH of inductance. Ideally, 0 Ohms would be better but this is unrealizable. Because of the low resistance, it is the driver’s job to control the current. This is done by reducing the applied voltage as soon as 4A is reached. Beyond that point, only 1 Amp is required to hold the injector open. Now with 4A initially flowing through the injector, we have 40mJ of energy to pull the injector open. After that, at 1A, we need only 2.5mJ to hold it open. The power, in the form of heat is 32W “peak” initially but decreases to only 2 Watts during the “hold” phase. With a good driver, low impedance fuel injectors are not only more responsive (10 times the energy) but use much less power to operate.

roarin_mouse

The time required for current to reach a given level can be expressed with the following expression:
t = (-L/R) (ln(1-IR/V))

If we plug in 0.005 (5mH) for L, 2 for R, 4 for I, and 14.5 for V, we have:
t = (-0.005/2) (ln(1-(8/14.5)) = 0.00206 or 2.06ms

Psycho_91Camaro

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Doctor J

The description and math offered seem reasonable at first blush. The time calc may be a little misleading though.

Saturated injector experimental data was presented here some time back:
TurboBuicks.com - Another MYTH debunked! (fuel injector DC)

Pay particular attention to the wave forms shown in the reference papers on p. 5 and the graphs on p. 7.


The issue with saturated injectors is that they display two flow regions; partly-open and fully-open. Though
not explicitly shown, I'd expect peak-and-hold injectors to exhibit the same behavior on a shorter time scale.


Re the reference thread:
As shown in the charts, the power needed to pull the pintle off its seat doesn't require the coil winding to
be fully saturated - looking at the graph on p. 7 the injector begins to open @ about 1 ms. It then provides
modulated flow until ~1.5 ms, when it is fully open (bench test).

The level of coil energy(time) needed to pull the pintle off its seat seems to me to be related to rail pressure
which resists the opening, and Vbatt which provides the energy to open. Both vary by application, hence
actual fuel flow and injector opening will not be easily derived from equations only - at least at small pw's.


For peak-and-hold injectors, the faster opening time is mostly governed by the lower inductance coils
used in their construction - the rate of current to the coil presumably follows a curve something like
(Vbatt / L )
up to saturation at the current limit. This is illustrated in the Motorola chart here (refers to ignition coils):
http://www.corvetteforum.net/c4/doct...ilcurrentc.jpg


Thus a low-inductance (P & H) coil can 'charge' faster, owing to a higher current rate, with stored energy
still expressed as [ 1/2 L * I *I ]. Since the coil has been partially de-energized before the end of the cycle,
I'd expect the P & H injector could close faster than a saturated injector too.

FWIW

roarin_mouse

Actually, the timing is correct. This incorporates only the electrical aspects of the injectors. Obviously there is a transitional period (mechanically) when the injector is in between states (opening, closing). I also left out the current induced by the motion of the pintle. The model I used is a series RL network (for the injector) consisting of 2 Ohms and 5mH. I hope that this helps clarify what I was saying.