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Transistor assisted ignition circuit

qwerty009

Dec 8, 2013
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Hi, i'm working with one old 2 cylinder 4 stroke stationary engine - as a winter hobby.
Originally it has magneto ignition, but since it's quite hard to start i thought i'd try battery +ignitions coil with transistor assisted firing.

I purchased velleman kit K2543 and altho it works it has lot of shortcomings.
allumage.png

Main shortcoming being that it dissipates LOT of heat from resistors - so i'm looking to improve it so i can put in into closed plastic box.

Since i'm no pro when it comes to electronics i have some questions how to do that...

1) I tried to substitute TIP162 with IGBT but for some strange reason IGBT didn't switch as it should have. When points were closed and Base of 2n2219 is effectively grounded it should not conduct and therefore GATE of IGBT should have received full 12V and be turned ON (saturated). But what happened is that there was some ~7V on it gate and it wasn't fully on and it dissipated lot of heat. What might have caused this?
With TIP162 it works fine (tip 162 doesn't overheat). If i can use IGBT i can substitude 150ohm resistors with single 4,7k resistor or similar.

2) Why are there 2 zener diodes in series? Why not single Zener?

I also found one more circuit that has one interesting thing that i don't understand.
ignition.gif

D1 and D3. If they are connected like they are and there will be voltage spike let's say 200V and zener will conduct then voltage spike will got to Gate of transistor - won't it destroy it?
Other thing that i don't understand is that if there are inductive spikes then how won't they reach first transistor?
 
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Harald Kapp

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1) A positive voltage at the base of the TIP162 opens the transistor and Vbe is approx. 0.6V. The transistor conducts and Vce is ~0.2V /give or take a few mV). Therefore Vce <Vbe and the zener diodes do not conduct.
With an IGBT, you need Vbe >> 0.6V to make the transistor conducting. In that case, however, the zener diodes, seeing Vbe >>Vce are conducting and current will flow to the output. The exact voltage seen at the output depends on the load and the characteristics of the IGBT. This also answers your question 3).
2) two zeners inseries have a pass voltage of ~1.2V for a positive voltage on the base of the TIP162. Since Vbe is only ~0.6V, the diodes will not conduct. cf. answer 3).
3) If there's an overvoltage on the output, D3 will conduct, making T2 conductive, too, thus reducing the overvoltage. Without D1 the positive gate voltae would pass through then conductive D3 to the output. cf. answer 1).
 

qwerty009

Dec 8, 2013
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1) If i understand correctly (regarding upper circuit with TIP162). In order to make it work with IGBT i must reverse first zener diode so their anodes would be together?
Doing so protection would still work and diodes wouldn't conduct when gate voltage is applied? (Would this configuration also work with tip162? (npn darlington)


2) There are 2 diodes after 2n2219. One that goes to Base of tip 162 is there to block over-voltage spike to 2n2219 if zener conducts? What does other one do that is connected to ground?

3) If i would take 2 18V zeners, connect their anodes together and put them between Base of tip162 and Gnd would that circuit still work and be effective in protecting transistor? Or what would be zeners optimal voltage? I found one ingnition igbt datsheet and there are zener clampls between GE & GC http://www.irf.com/product-info/datasheets/data/irgb14c40lpbf.pdf altho their values are not specified (i understand one of those clamps is 370-430V or both of those have same value)?
 
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duke37

Jan 9, 2011
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I think the reason for two zeners in series is to get enough voltage.
The TIP162 has a maximum collector voltage of 400V. two 150V in series will limit the collector voltage to 300V. Two 150V zeners may be easier to find than one 300V zener.
By connecting the zeners back to the base of the TIP162, the zener ciurrent limits the turn off of the transistor so limiting the collector voltage.
If the zeners were connected to ground, they may have to dissipate a large power. The energy has to go somewhere, if it does not go into the spark, it is better to go into the transistor which will be on a heat sink.
 

Harald Kapp

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1) No. Reversing one zener diode will reduce the threshold voltage to 150V. You'd have to add a diode as in the second circuit shown by you.

2) These diodes protect the TIP from negatibe voltages at the base. Otherwise there is a risk that a negative base voltage may turn on the transistor due to reverse breakdown.

3) 18V is way too much. The B-E breakdown voltage is typically somewhere between 6-10V (see datasheet). Using standard diodes is more robust than a zener diode replacement.
 

qwerty009

Dec 8, 2013
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Tnx guys things have become lot clearer :)

In circuit 2
ignition.gif

I understand that here D2 protects against negative voltage spike and there is no need for diodes between B-E (if i would substitute this to circuit nr 1 and remove those 2 diodes)?

Oh, and theres capacitor in circuit 1. What does it do exactly? Does it protect transistor once it closes and voltage on the coil spikes? If transistor can take it is it better to have no capacitor (in terms of spark strength)?
 
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