A new ignition circuit

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dsage

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Long leads cause all kinds of havoc with radiated interference being picked up by the wiring and sneak paths for high energy back to the battery. It usually results in the hall sensor blowing. Also, connect things exactly as the diagram shows (short and direct) There must be a direct lead from the engine block (spark plugs) back to the battery. Do not assume that a round-about ground connection will suffice. Also the hall sensor should have all three wires shown tightly twisted together and connected directly back to the circuit. Again - don't assume that because a wire eventually finds it's way to where it needs to go that that will suffice.
The high energy spark current / voltage will be looking for any path it can find back to the battery so it must be directed properly so it doesn't try to follow say your hall sensor wiring back to the battery.
I'm not sure what coil you plan to use. If it's a standard automotive coil of 12v then you'll need a ballast resistor of maybe 2 ohms and 25 watts. The current should probably be limited to 5 or six amps. There is no real need for a high power ignition system on a model. It's just a waste of power.
I (and many others) use four AA Nimh rechargeable batteries (approx 6v) in a holder and a Ford 12v COP coil (small). I don't use a ballast resistor because the batteries themselves limit the current. So this makes a nice compact arrangement.
BUT something must limit the current. It depends on what coil you're using how much it will try to draw.
Don't run the system on a power supply. Power supplies will not supply the pulse current reliably and it will likely give poor results.
 
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paulc

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Very good!
Is there a possible change to the PCB component list that would allow me run on 12 volts without the need for an external ballast resistor?
 

dsage

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No. The board is just a driver. It's up to you to limit the current. It was made that way to keep it small. The ballast resistor value can vary and in general are very large and can dissipate some heat. Also some model coils don't require ballast resistors. It was just easier to make it that way.
There are car coils that have the ballast resistor inside. You'd have to research a vehicle where one of those was used.
What is your reason for insisting on 12v?
The previously mentioned AA battery solution is a nice small arrangement and easy to hide.
 
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dsage

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I just checked the link you gave to my original saved spec sheet and it's the same.
I'm not sure what schematic you are using. I went back to post #144 to the schematic that John Gedde (who was collaborating on this design) and he used the correct symbol that looks pretty much like the spec sheet. Symbols vary a bit.
I know very early on I used the incorrect symbol for an IGBT because I couldn't find one in Eagle PCB. But the connections have always been correct.

Anyway - just pay attention to the connections and you'll be fine. If you're using the PCB layouts I posted a while back it should take care of itself.
 
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dsage

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I guess time moves on. Yes it appears the 2N4124 has gone obsolete. Very strange.
It's a pretty much bread and butter transistor.
The substitute should work.

BUT

That particular substitute has a difference in the pinout. The legs are situated looking at the flat face left to right:
CBE (collector base emitter)
The 2N4124 pinout is
EBC (emitter base collector)

You can use the substitute. You will just have to turn the transistor facing the other way leaving the center lead in the same place.

Double check my work. I'm pressed for time at the moment.

Download the data sheet for both transistors and you'll see the difference in orientation of the legs.
For some reason this issue crops up a lot with these transistor packages. But it's easily corrected as above.
Actually I'm surprised they recommended it as a direct replacement. Electrically it's suitable but physically if you inserted it in an existing circuit as a replacement for an original it would not work.

BTW. The link you supplied asks for cookies. I didn't allow it so I had to get the part number and go to Digikey to observe both transistors.
So again - check my work.
 
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dsage

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As long as it's OK spec wise I will sort the orientation using the datasheet. I just wanted to be sure it works with your circuit.

Many thanks.

Odd. I just checked Digikey again for the BC547 Farnell recommended. They list the BC547 as obsolete as well.
So use what you can get I guess. I'll have to find a replacement with a more stable supply.

Spec-wise the BC547 looks like it should be ok. Nothing special is required there. It's used as a switch. Either on or off.

I'll have to check my stock of parts. Maybe I have a BC547. If so I'll try it in an existing finished board for you.
I won't be able to do that for a bit.

But for the cost of them. I'd say go ahead and give the BC547 a try.
 
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dsage

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Update:

You're in luck !
I rummaged around through my never used parts and found a dozen or so BC547 transistors. Surprised me.
The BC547 works fine to replace the 2N4124.
BUT
You just have to reverse the collector emitter orientation as mentioned before.
 

cds4byu

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This link shows the 2N3904 as a replacement for the 2N4124. DigiKey has no 2N3904 in stock, but suggests a 2N4401BU as a replacement. They have 41,000 in stock at a price of $0.35 each.

Order page is here

Carl
 

Rocket Man

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WHERE is the circuit drawing? I buy 2N3904 in packs of 100 for $3 free postage.

100_3692.JPG
 

dsage

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This is post #236. (in the top right corner of this box).
Scroll up the page to the top and click on page 11. and down to #211.
Basic stuff navigating this forum.


(you can also scroll down to the bottom and select page 11 too).
 

dsage

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Quick glance. It looks good. The proof is in the pudding they say. Give it a go and see how you make out.
The only minor suggestion is to fatten up the traces from the battery to the coil and from the coil to the transistor. i.e the path the coil current will take.
If it's pcb material your using add a heavy layer of solder on the board or if it's wire use maybe #18 or so. You want minimal resistance in that path.
Also be sure to ground your engine block DIRECTLY back to the battery negative with a separate wire for spark current to return.
Also don't forget a ballast resistor in series with your coil if it requires one.
 
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Is there any way to modify this circuit to fire the plug when the circuit opens if using conventional points? I have an old model made by someone of some skill quite a while ago using contacts on the exhaust valve push rod that used a condenser and coil. Conventional ignition fires when the points open and the field in the coil collapses. If the engine stops with the contacts closed the coil will get very hot or burn up so I want to use this circuit as a kind of one shot. I really don't want to make any visible modifications to the engine. Ideas please. Bob
 

dsage

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If you build the driver according to the circuit diagram it DOES by design fire the plug when the points OPEN.
In fact the nice thing about this design is that it will not remain on and burn up your coil. If the engine stops with the points closed it will "time out" and stop driving the coil after a period of time.

To that point: If you are using the circuit on a full sized hit-miss engine running at typically very low rpms the time-out feature might become a problem especially on starting. The engine might kick back when you pull it over by hand. If so let me know. It's easily fixed by increasing the value of C2.
This has never been a problem with models.
 
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