Yes, that is what I meant, and that is what the data sheet shows.
I did some more digging, and you are right, it is not an opto-coupler.
It is indeed a chip for controlling ignition coils. It is made by
Telefunken / TEMIC, and they had a family of 3: U2225B is used when the
input is from an inductive sensor, U2226B is intended for use with a
microcontroller / microprocessor and U2227B is used when the chip is
connected directly to an optical sensor. I have not had any luck finding
the data sheet for this part.
So then it wouldn't make much sense to try and show it.
Randolph, I'm having trouble understanding the current path through the
transistor. I found this page:
It helps me understand more, but I don't get which way the current goes
through the base electrode. I have a suspicion that my diagrams show the
current going the wrong way through the transistor.
I'm having trouble getting my mind around this.
I am aware that "flow" is _commonly_ considered to be from the positive to
negative terminals of the battery, but the electrons themselves go in the
On these two pages, is the current flow through the transistors correctly
depicted? Nobody has answered that question yet.
The electrons flow from POSITIVE TO NEGATIVE. The electrons go from where
they are (-) to where they're not: The "holes" (+).
It's the actual everyday signal that's commonly perceived to go from
negative to positive.
But we have THREE paths in a transistor ("transfer resistor"). For a non-
techie, this is non-intuitive. I do not get how TWO terminals can have
Please try to understand that I am not trying to be difficult, but that
this is not at all making sense to me.
I am hoping that someone, somewhere, will post with an explanation that
makes sense to my mind. In my professional life I have taught and trained
many, many individuals, and most have had certain things that just would
not "click" until the information was presented a certain way. I am seeking
that way, and I will persist until I find it. This is driving me crazy.
(all on one line; copy-and-paste as necessary)
shows the signal path from base electrode to collector.
(again, all on one line)
appears to show the path from emitter to collector.
I do not get this and I am trying madly to understand. Graham W would be
able to correct me in an instant. He has been the most persnicketly
critical observer and the most productive from my point of view. Graham,
where aaaaaare you?...
Graham was the ONLY one to suggest alterations to the Main Relay function
graphics. Graham was the ONLY one to inform me of certain HTML errors, the
correction of which make it easier for browsers to display the intended
Ah, but wait. I just thought of something: alt.electronics. Back soon...
Just checked message counts.
seem better choices, in case anyone wants to follow along...
I'm hoping to elicit responses from somebody like Sam Goldwasser.
The entire "hole" thing never helped me, either. I got a lot farther when I
started thinking of where the "positives" flowed, because both vacuum tubes
(which were still common when I was learning electronics) and NPN
transistors (which are the most common now but least common originally, both
for technical reasons) use negative ground. Trying to follow electron flow
distorts the idea of the ground, while thinking of "positives" flowing from
the power supply to ground worked great. (Also the "positives" flow in the
direction of the arrow on the emitter.)
For NPN transistors, here is the simple view. The emitter is grounded and
the collector has positive voltage applied to it. The transistor doesn't
conduct because the collector-base junction is reverse biased. Now positive
voltage is applied to the base. Below about 0.7 volts on the base nothing
much happens. As 0.7 volts is approached the base-emitter junction starts
drawing current, just like any other ordinary silicon diode. The
base-emitter current causes tens to hundreds of times that much current to
flow from the collector to emitter. As the base voltage rises to about 0.8
or 0.9 volts, the base-emitter current is so high that the collector current
can't go any higher - the voltage at the collector has dropped to only
0.1-0.2 volts, and the entire supply voltage (like the 12V battery) is
across whatever load is between the power supply and the collector. In the
ignition circuit, the collector has grounded the primary of the coil. This
condition is called "saturation" because increasing the base current doesn't
do anything to the collector any more.
It is important in switching circuits like the ignition to saturate the
transistor. If the collector voltage doesn't go very near ground, the
transistor has to dissipate the current times whatever voltage is left. If
the voltage is only twice the saturation voltage (say, 0.3 instead of 0.15)
the transistor has to dissipate twice the power.
think of a Y water pipe.One arm of the Y is smaller than the other.But the
total water flow thru the bottom of the Y divides and part passes thru the
left arm and part thru the right arm.You can control how much water passes
thru the right arm by adjusting the flow thru the left arm.(but the water
pipe does not have any current gain)
Just think of a vacuum tube;the cathode(negative terminal) is heated so it
will emit *electrons*,which are attracted to the positively charged anode
plate,thus;ELECTRON FLOW,from negative to positive.
So then my drawings are NOT correct. I need to show the emitter (closest to
the coil) "switched off", and not the collector (farthest from the coil).
Right? Or does it matter since the effect is the same?
For a NPN transistor,the collector should go to the coil,and the emitter to
ground. The other end of the coil goes to +12V.
The internal diode shunts the back EMF around the transistor to
ground,protecting the transistor.
I just looked at your schematic,and it appears correct.except that terminal
3 of the Igniter module does not go straight to the Darlington base,it goes
to the IC that controls the Darlington.You need a rectangle indicating the
control IC between the Pin 3 and the Darlington base.Pin 1(tach drive)
probably goes to the control IC,too,certainly not to ground,Pin 4.
(the emitter of the Darlington probably goes to the control IC,too,then
thru a small value resistor[<1 ohm] for current monitoring by the IC,then
Something additional I thought of after I sent the last post(sorry!);
The ECU does not ground the igniter module.It only sends a signal (to the
control IC inside the igniter)for the Darlington to ground the coil.If the
ECU were to be the ground for the igniter,that would mean that the entire
coil current(several amps) would have to travel through the long wire from
igniter to ECU,and the ECU itself would have to switch that high current to
ground,which is the purpose of the igniter.
----- Original Message -----
Sent: Saturday, June 04, 2005 8:37 AM
Subject: Re: Ignition updates to the Unofficial FAQ
That was how Jim got into the semantic trap to start with. The ECU supplies
a low current ground to pin 4, which grounds an input on the IC, but the
main ground - the one the coil current flows through - is the one shown in
the lower right corner of the ignitor; the metal body of the ignitor itself.
The current from pin 4 is undoubtedly in the range of 1 ma.
I've removed the innards of the igniter from the graphics and will add that
body round later on, changing the yellow and orange lines around to suit.
Since the workings are far more detailed than I had originally imagined, I
will leave the igniter as a "here be dragons" blank box until somebody else
decides to supply a correct schematic for me to copy.
Thanks very much to all.
I find it useful to think in terms of the equation
emitter current = base current + collector current
and consider that the E,B,&C are regions of the transistor with the base in
the middle. The B is much smaller than the E & C and is doped to the
opposite polarity. This creates two junctions that are essentially diodes
(hence the term BJT, bipolar junction transistor). When the BE junction is
forward biased (B more + than E for NPN, and the opposite for PNP
transistors), the carriers in the E region diffuse into the B region, but
only a small number can flow out of the B terminal relative to the number
than the larger E region can provide. The rest flow into the C region
(which is larger than the B region), assuming that the circuit is such that
there is someplace for them to go. Thus a relatively small base current can
control a much larger collector current.
As for the direction of current flow debate, it is largely context
dependent. Once you understand the application and that electrons move from
areas of excess (higher potential) to areas with relatively more holes
(lower potential), the semantic questions are of less consequence.
Now this is a very simplified description of how a transistor works, but I
think a correct one. Am I making sense, Jim. You can likely explain it all
much better than I if you are the Jim that I have seen post on s.c.r. over
the years. IIRC, your experience in electronics is substantial.
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