Ignition updates to the Unofficial FAQ

It reminds me of my Mazda rotary with points. I could see the dwell begin to take up too much time as the RPM got higher.

Randolph wrote in news: snipped-for-privacy@junkmail.com:

The diode is INTERNAL to the transistor package. Probably on the same substrate as the xstr.

I found ICs that were specifically designed for ignition control and driving the Darlingtons,but none with the same pin count of the IC pictured,nor any similarity to its part number. I do not believe it's an optocoupler,but a full control IC.Probably with circuitry to square up(shape) the drive pulse,and provide enough drive current,and IIRC,the ICs monitored and regulated coil current.(that would enable faster switching)

Randolph wrote in news: snipped-for-privacy@junkmail.com:

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.

snipped-for-privacy@XReXXIgnit.usenet.us.com wrote in news:d7kice$pvn$ snipped-for-privacy@blue.rahul.net:

I had a '74 RX-4 Coupe!

You guys...I swear...

If the subject gets any more high-flown, it's gonna head for outer space.

This is excellent info. Now I've got to make another page: More detail for the Electronics Whiz.

"TeGGeR®" wrote in news:Xns9668EF412B3E6tegger@207.14.113.17:

There's two current paths;the B-E path and the C-E path(main path). Current flows the opposite direction of the emitter arrow,for both base and collector currents.

Jim Yanik wrote in news:Xns966963506AD29jyanikkuanet@129.250.170.86:

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 OTHER direction.

these two pages, is the current flow through the transistors correctly depicted? Nobody has answered that question yet.

"TeGGeR®" wrote in news:Xns96699C7F9EAA2tegger@207.14.113.17:

The electrons are what's doing the moving,and they flow from neg to pos.

Jim Yanik wrote in news:Xns9669BF9265BD5jyanikkuanet@129.250.170.83:

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 THREE paths.

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.

This graphic:

(all on one line; copy-and-paste as necessary) shows the signal path from base electrode to collector.

This one:

(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 information.

Ah, but wait. I just thought of something: alt.electronics. Back soon...

"TeGGeR®" wrote in news:Xns9669D79281ACtegger@207.14.113.17:

Just checked message counts. sci.electronics.misc sci.electronics.repair and alt.home.repair seem better choices, in case anyone wants to follow along...

I'm hoping to elicit responses from somebody like Sam Goldwasser.

There is a big part of the confusion - electrons flow from [-] to [+].

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.

Mike

Most materials have an electron flow, which goes from negative to positive. I've heard that some materials can have a proton flow. Both may exist in a vacuum.

Current flow arrows on diagrams go from positive to negative.

Bipolar transistors are current amplifiers. When a current flows through the base-emitter diode junction, a stronger current is allowed to flow from the collector to the emitter. The C-E junction is .2 to .4 volts when the B-E junction is saturated (~.65 V). The current gain for a power transistor is usually 10 to 100. Darlington pairs have that gain squared. Gains are not at all consistent so they're usually specified as a range.

MOSFETs are tiny voltage controlled amplifiers. Absolutely zero static current is required to turn them on or off; just the capacitance current. Because of their infinite current gain, millions may be paralleled on a single chip to satisfy any current load. Their voltage gain is very low - a typical gate threshold voltage is 4V and a typical gate saturation voltage is 10V. There's no voltage drop between the source and drain, only resistance. High voltage capability makes each MOSFET junction larger and dramatically increases resistance.

IGBTs are similar to bipolar transistors but with an insulated gate like a MOSFET. They have the high voltage capacity of bipolars but need no driving current like a MOSFET. They're very slow so they're usually limited to controlling industrial motors. (Honda hybrid cars use them for their motors.)

Kevin McMurtrie wrote in news: snipped-for-privacy@corp-radius.supernews.com:

So then my diagrams are correct. I assumed the base electrode to act as the switch, turning power on and off between the collector and the emitter.

Thanks.

don't get no proton flow unless you're into nuclear chemistry. in semiconductors, conduction is by way of negative electrons & positive "holes". you /can/ have [positive] ions move in the semiconductor lattice, but they are not a part of the primary conduction mechanism & result in mass transport & therefore degradation of the semiconductor - they are not a proton thing.

"TeGGeR®" wrote in news:Xns9669D79281ACtegger@207.14.113.17:

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.

Jim Yanik wrote in news:Xns966A6FCED7938jyanikkuanet@129.250.170.86:

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?

"TeGGeR®" wrote in news:Xns966AD347EB6D4tegger@207.14.113.17:

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[

The emitter is the neutral part of it, the part the collector gets switched to.

Maybe the easiest way to think of it is as a relay, where the emitter is one end of the winding and one of the contacts. The base is the other end of the winding and the collector is the other normally open contact. When current is run through the "winding" (from the base to the emitter) the collector closes the circuit to the emitter.

There are a few technical details like polarity (the collector and base both have to be positive with respect to the emitter) and the base resistance (so low the current has to be limited by external resistance), but the operation in an ignitor is just like a very fast relay. In other circuits it isn't used as a relay, and the collector current is varied more proportionately to the base current.

Mike

Jim Yanik wrote in news:Xns966AD818247EBjyanikkuanet@129.250.170.85:

More information here than I've gotten yet. Thanks.

"TeGGeR®" wrote in news:Xns966AE4D412B10tegger@207.14.113.17:

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.