Ignition coil charge time
Ed Lansinger
elansi01 at mpg.gmpt.gmeds.com
Fri Mar 24 21:01:03 GMT 1995
Peter Wales wrote:
[voltage seen at the collector of the transistor]
> |
>---| |
> | |
> | |
> |_____|-----|
>A B C D
>
>A is the before switch on time
>B is the driver transistor switched on but not drawing a lot of current so
>it is not saturated
>C is the coil saturated and the transistor drawing full current so the
>slight drop in voltage across the coil is from the Vce sat of the transistor
>D is the switch off of the transistor and the coil voltage decaying to start
>generating back EMF
>
>You should size your transistor to be able to drive the coil continuously at
>DC and have a Vce big enough to cover the back EMF. I use 450v 10amp devices
>and never have failures. The minimum pulse width for the full spark is B-C
>and remains a constant for that coil. C-D is overkill as the coil is
>magnetically saturated and no more energy will go into it.
Also, an apologetic ME stated that he keeps forgetting that "capacitors charge and coils
saturate" (sorry, my mail program just swallowed the message so I don't remember who).
Just to clarify (belabor?) a few points -
1. As Peter points out, leaving the transistor on for the period C-D is needless. This
is because the coil current has built to a maximum I=E/R, where E is the voltage drop
across the coil (Vbatt-Vce) and R is the resistance of the coil. Since I is at a
maximum, you can't store any more energy in the coil. However, this is not because the
coil is "magnetically saturated". More current may store more energy in the coil (i.e.
if you went to a 16V battery). Or it may not. In fact, it may have already reached the
point of maximum energy storage somewhere in A-B. The deciding factor is the
composition, construction, and size of the core of the coil. The magnetic material in
the coil core is what "saturates". Simply put, you cannot magnetize a piece of iron
past a certain strength (magnetic flux density). Another way of looking at it is that
the inductance L of an inductor with a ferro-magnetic core changes with the current
flowing through it and eventually maxes out at high enough currents.
2. Holding past C is unnecessary. But you might reasonably want to stop before C if you
have already built up enough energy in the coil to fire the spark reliably. This will
keep temperatures down and reduce draw on the electrical system. It may also allow you
to select a smaller and less expensive transistor.
3. With an ignition system that has a current-limiting feature, the point C on the
graph actually corresponds to desaturation of the transistor by the control circuit for
the purpose of reducing current flow. It's still pointless (and dramatically increases
heat buildup in the transistor) to hold any longer. In this case, though, the current
limit is not a function of E and R or any supposed "saturation" of the coil. I think
the picture will look different than Peter suggests for a system that has no current
limiting, as the transistor will go into saturation shortly after A and the voltage will
rise only slowly after that as current continues to build, not abrubptly as the diagram
suggests when current stops building (although the transition may still be discernable).
Saturation of a transistor occurs with sufficient base current, not when the collector
current stops rising. Maybe Peter's situation is different from what I have observed.
4. A nice way to measure things is to insert a .1 ohm (10W or better) resistor between
ground and the emitter of your drive transistor. You could also insert the resistor
between +12 and the hot terminal of the primary. Use a scope to watch the voltage drop
across the resistor. The voltage drop is proportional to the current flow. With a .1
ohm resistor, just multiply the voltage seen on the scope by 10 and you'll know exactly
how much current is flowing at any given time. You get nifty pictures like:
_____
/ !
/ !
/ !
/ !
/ !
/ !
_____/ !_____
^ ^ ^
coil peak spark fires
begins current
to
charge
If you measure the slope of the rising part you know how fast your coil is charging. If
you know that N amps of current is required for reliable ignition, you can then figure
out the correct charge time. This should be very close to the time predicted by the
equation provided by Matt Franklin and John Stein (given that you know the inductance of
the coil). Increasing the battery voltage will increase the charging rate and lower the
required charge time.
As Bill Lewis illustrated in his message, trial and error is frequently the best way of
figuring how much coil current (and therefore charge time) you need.
5. As Peter points out, you will want to choose a transistor with a sufficiently high
breakdown voltage (say, 450V). You will also want to protect your transistor from
higher breakdown voltages by adding a beefy Zener, Transil, or other
overvoltage-protection diode (say, 400V) between the collector and base so that if the
primary side voltage gets too high current conducts through the Zener to the base of the
transistor, turning it on and letting the coil energy dissipate with relative safety as
heat through the body of the transistor. (Also add a 47ohm resistor between the base
and emitter.) That's how you protect your transistors from running the coil with the
plug wires disconnected. Some ignition drive transistors have this protection built in.
If you've got enough heat sink area and you're within the Safe Operating Area of the
transistor you can run like this for as long as you like.
6. "Saturation" simply means that you've reached a limit of some sort despite the
continued application of a stimulus. It's not a term specific to inductors (or
electronics, for that matter).
-------------------------------------------------------
Ed Lansinger
General Motors Powertrain
Powertrain Control Center
Premium V Software & Calibration Group
Milford Proving Ground, Milford, MI
elansi01 at mpg.gmpt.gmeds.com 8-341-3049 (810) 684-3049
The opinions stated above are mine, not my employer's.
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