Ignition coil driver circuit

With this circuit I am trying to make it generic so that I can plug any NPN power transistor and any PNP transistor into it and still have it work.

I have soldered it up as far the BC327. The collector of this transistor is soldred to a small plug which I can divert to an LED so that I can see the oscillation with a 1000u timing capacitor in place.

The 555 circuit oscillates just fine but I am loosing the oscillation when I pass it through the BC327 and then through the LED.

Does any one know of a worked example of the calculations required for a transistor arrangement the same or similar to this?

I am not getting my head around this. For example is te 150R resistor a base resistor for the Darlington or is it a collector resistor for the BC327?

If you calculate the base resistor for the BC327 to give you a CE current of 600mA, which should be enough base current to saturate a 2SC5446, then what effect does the 150R (or some other value) have on the base current for the 2SC5446?

**broken link removed**

Q1 & Q2 form a Sziklai pair, a cousin of the Darlington pair. It has the effect of inverting polarity of the 555 output.

Gain is only slightly less than a Darlington.

**broken link removed**

The 150 ohm resistor limits current through both transistors, particularly Q2's bias terminal.

If you install a different NPN darlington, make sure it is rated for its bias to carry whatever amount of mA is going through.

BradtheRad said:

Q1 & Q2 form a Sziklai pair, a cousin of the Darlington pair. It has the effect of inverting polarity of the 555 output.

Click to expand...

Thankyou very much Brad. This is the missing link that has been preventing me from finding the right transistor examples to follow. Up to now I have been reading about other amplifier type examples whose calculations are oibviously not applicable to this sort of arrangement.

I have also done a bit more reading on PNPs and, as I currently understand it, I have to find a resistance value for the BC327 base that blocks current flow when the 555 pin 3 is high.
I calculated a value of 330R via the standard transistor formulas with the purpose of getting 600mA from its collector that will then drive the larger 2SC5446. But since I now know that the two transistors form a Sziklai pair this assumption is wrong.

As far as I can deduce, A BC327 base resistance value of 330R is allowing current to sink into the 555 regardless of whether pin 3 is high or low. I am getting about 9.5V from 555 pin 3 and its supply rail is about 11V so it would seem it is entirely possible for current to sink to the 555 even when its output pin is high.

The value probably has to be 2200R as in the jaycar jacobs ladder kit but possibly a bit more or less depending on the specs of the 555 I use. It will be a matter of trial and error. Am I on the right track here Brad?

Or is it possible to calculate the resistance value, at least roughly? If so how? What if I was to replace BC327 with one of the other PNP BCs?

BradtheRad said:

Gain is only slightly less than a Darlington.

**broken link removed**

The 150 ohm resistor limits current through both transistors, particularly Q2's bias terminal.

If you install a different NPN darlington, make sure it is rated for its bias to carry whatever amount of mA is going through.

Click to expand...

The 2SC5446, from a tv circuit board, has a max CE current rating of 20A or so and of the order of 5A for its BE current rating. So I think it will probably take what ever I throw at it from the BC327. Perhaps I don't need an resistor at all between the two transistors.

Need to read up on Sziklai pairs first however.

Your discussion is missing the point, I fear. The problem of standard 555 (not CMOS version) is that the output voltage is not guaranteed to switch the PNP transistor off without an additional load resistor. Typically, the transistor switches off, but very slowly.

You can change the value of 2.2k resistor over a wide range without solving this problem anyhow. The solution is to place an additional resistor, preferable connected as a voltage divider, between BC327 base and emitter. The value is rather uncritical, any value between 1k and 10k will fit in combination with the 2.2k series resistor.

It should be noted additionally, that the darlington transistor has internal base-emitter resistors that are not shown in the simplified schematic symbol. Without this resistors, switch-off would be inacceptable slow and create huge transistor losses. It may be nevertheless suggested in some cases to add an external base-emitter resistor to speed up switching or "drain" the leakage current of the driver circuit.

The transistors in this circuit form a quasi complementary Sziklai pair.

An ordinary Sziklai pair looks like this: **broken link removed**

Correct?

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FvM said:

Your discussion is missing the point, I fear. The problem of standard 555 (not CMOS version) is that the output voltage is not guaranteed to switch the PNP transistor off without an additional load resistor. Typically, the transistor switches off, but very slowly.

Click to expand...

Can you elaborate on this a little FvM? I thought that it was the current that was of primary concern in switching a transistor on and off.

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I have the BC327 output hooked up to my diagnostic LED and it is oscillating OK with an inverted signal. So 10% duty cycle is being converted to 90% duty cycle.

But if I can improve the performance of the circuit with this additional resistor then I will.

Perhaps making the BC327 switch off faster will more clearly delineate the off state of the LED from its on state. At present the LED turns off for a just discernible period with a 1000u timing capacitor.

So what exactly are we trying to acheive by converting the BC327 base resistor to a voltage divider arrangement with the emitter?

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So I tried a 10k resistor from the BC327 emitter to the base and there might be a slight improvement in the turn off of my diagnostic LED. It is a bit hard to tell from an LED though.

boylesg said:

As far as I can deduce, A BC327 base resistance value of 330R is allowing current to sink into the 555 regardless of whether pin 3 is high or low. I am getting about 9.5V from 555 pin 3 and its supply rail is about 11V so it would seem it is entirely possible for current to sink to the 555 even when its output pin is high.

Click to expand...

If the PNP is supplied by 11V, and its base sees 9.5V from the 555 pin 3, then the PNP will not shut off.

The 555 pin 3 must provide supply V (or within .5V of it), so that it can shut off the PNP transistor.

FvM also mentioned this.

I guess we just naturally expect the output to reach supply V. It only makes sense for it to do so. However the 555 IC has been manufactured in several versions over the decades. There could be versions that are designed differently.

The value probably has to be 2200R as in the jaycar jacobs ladder kit but possibly a bit more or less depending on the specs of the 555 I use. It will be a matter of trial and error. Am I on the right track here Brad?

Click to expand...

General advice is to start off by trying a low current setting, safe for components. If it doesn't do the job then increase current gradually while staying within limits. We don't always know the limits until we exceed them, however. Yes, it's trial and error like you said.

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boylesg said:

The transistors in this circuit form a quasi complementary Sziklai pair.

An ordinary Sziklai pair looks like this: **broken link removed**

Correct?

Click to expand...

There are two configurations. The above diagram is an NPN driving a PNP.

Your schematic in post 1 has a PNP driving an NPN (or rather an NPN darlington).

Add the two darlington arrangements and that makes a bunch of possibilities.

We choose one or the other depending on what is the polarity of the incoming signal, and what output polarity we need, etc.

The maximum collector current for BC327 is -800mA.

The maximum base current for 2SC5446 is 9A.

So I would not need any resistor at all between this and the BC327. However I am clearly not going to be able to saturate the 2SC5446 with a BC327 from the electrical characteristics in the datasheet for 2SC5446 which specify the the base current required for saturation is around 3.5A.

Compared to a saturation current of 0.6A for the original MUH10012 darlington which is well within the limits of the BC327.

So it looks as though I will have to select another of my salvaged tv power transistors that has a lower base saturation current.

2SC4458 looks better suited.

Possibly 2SD2499, 2SD2498 or BU2508F, although these ones have base saturation currents of about 0.8A which is at the limit of the BC327.

Even if I replace the BC327 with a tranny with a higher collector current rating I will start to push the current limits of my voltage regulator. Unless I connect the emitter of my BC327 replacement to the unregulated 12/24V rail rather than the regulated 12V rail.

How am I doing FvM? Am I talking rubbish here or am I on the right track?

I didn't notice your intention to use 2SC5446 instead of a darlington because I primarly looked at the schematic.

My comments have been referring to the original BC327 + darlington configuration. BC327 is however too weak to drive a TV horizontal transistor wit respective low current gain.

Regarding series resistors, besides other aspects, they have to keep save operation conditions for the involved transistors. E.g. the Q1 collector resistor must not dimensioned below 12 or 15 ohm with BC327 not to exceed 800 mA. I also agree to your consideration that higher drive currents for the output transistor are unsuitable due to the high supply current.

The discussed compound or "Skizlai" pair solution isn't applicable for the present circuit because it won't allow the output voltage to swing above 12.

FvM said:

The discussed compound or "Skizlai" pair solution isn't applicable for the present circuit because it won't allow the output voltage to swing above 12.

Click to expand...

Well I have at least got the BC327 oscillating correctly which was not the case on the original printed circuit board - must have stuffed it up with a solder blob some where.

Next step is to get the MU10012 and another HR output transistor with similar specs oscillating with an automotive globe attached across it in place of the ignition coil.

The ciruit design was in a retail kit from Jaycar, so I would hope that the Skizlai pair in it is appropriate and works.




Using these calculations: http://www.kpsec.freeuk.com/trancirc.htm, and working backwards from the MUH10012 saturation base current of about 600mA, I can't for the life of me understand how they get a BC327 base resistance of 2.2k. I came up with about 160R.

For the BC327.....

Ib = Ic / hFE min
Ib = 0.6 / 40 = 0.015A

Rb = Vcc * hFE / 5 * Ic
= 12 * 40 / 5 * 0.6
= 480 / 3
= 160R
What am I doing wrong here?

I also noticed that I will probably be able to drive some other higher rated HR output transistors with a saturation base current of 0.8A with a BD140 rather than a BC327.

The 555 draws relatively little current and if the BD140 draws only 0.8A, then the total is likely to still be within the 1A limit for the 12V regulator supplying them.

But unless I get the above calulcation right and come up with the same value (2.2k) for the BC327, then I have little hope of calculating the correct value for a BD140.

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How do you calculate the resistance value between a Sziklai pair?

The original circuit isn't calculated for a base current of 600mA, both resistors have to be changed in this case.

boylesg said:

For the BC327.....

Ib = Ic / hFE min
Ib = 0.6 / 40 = 0.015A

Click to expand...

Although your math might be correct, real gain figures are known to wander far from the spec tables. Your real transistor could have much higher gain than 40 (the minimum spec).

This suggests there may not be any one proper figure to use in such calculations.

BradtheRad said:

Although your math might be correct, real gain figures are known to wander far from the spec tables. Your real transistor could have much higher gain than 40 (the minimum spec).

This suggests there may not be any one proper figure to use in such calculations.

Click to expand...

I think I can see your point here. I ended up finding the datasheet for the ST version of the darlington (the one I actually have) and the specs are some what different to those in the datasheet I originally had:

IC = 8 A IB = 100 mA
IC = 10 A IB = 250 mA
IC = 12 A IB = 300 mA

In the other datasheet IB was around 600mA so clearly the this device was not the same as the physical device I have. I could have sworn the name of the device was the same as what is on my real device......I can't check now because I have overwritten my original datasheet.

I think I might also be getting confused between hfe values and hfe groups.....the datasheets are quite confusing on this and there is not that much consistency.

Clearly this is why I am failing to come up with the same resistance value that they did - I am glad I have sorted that out at least.

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Would some one mind explaining the following to me. What is the hFE min of both transistors? What do the sets if figures under the headings min, typ and max represent? hFE values or clasiifcation values?


BD140:
Min Typ Max
hFE* DC Current Gain IC = -5 mA VCE = -2 V 25
IC = -150 mA VCE = -2 V 40 250
IC = -0.5 A VCE = -2 V 25


hFE hFE Groups IC = -150 mA VCE = -2 V
for BD136/BD140 group-10 63 160
for BD136/BD140 group-16 100 250



BC327:

hFE1 DC Current Gain VCE= -1V, IC= -100mA 100 630
hFE2 VCE= -1V, IC= -300mA 40


hFE Classification
Classification 16 25 40
hFE1 100 ~ 250 160 ~ 400 250 ~ 630
hFE2 60- 100- 170-