I am creating a circuit which will be used in an Engine Management system.

Its function is to: a) either drive a 12v ignition coil directly (using a normal IGBT) b) or if required, offer instead a 5v TTL 'trigger' output

I have sketched out a diagram.

I believe 'switch' section A can be accomplished by simple logic gates to choose where to direct the MCUs trigger signal - either it goes to driver type B or driver type C.

B - I plan to use a simple AND gate to give the 50mA 5V output when in "TTL" mode

C, is just a simple IGBT (there is a logic driver built in to it etc, that part is fine)

Juntion D is the issue (as only 1 pin can be used for the ignition outputs): The user may either be connecting to this to a simple 12V ignition coil (which will be using the IGBT part to operate, switching it ground etc


An external igniter that requires just the 5v trigger to be there

So the output system (especially the part in (B) need to not be damaged by the connection of a 12v ignition coil, and so forth.

The user cannot open the ECU to for example change physical jumpers.

Looking forward to your ideas

enter image description here


This gives you a TTL-compatible output signal that can also be pulled up to 12 V or much higher by an external load.


simulate this circuit – Schematic created using CircuitLab

You can then use the non-inverted drive signal in TTL mode but the inverted drive signal in IGBT mode. This can be done by modifying the signal at source or by XORing your relay control line with the drive signal.

The inverter needs to be able to drive your 50 mA at 5 V. From your question, you've found suitable logic gates for this anyway.

D1 protects the inverter from voltages above its supply, such as the coil 12 V if you drive that. It should drop around 0.7 V when the output is driving high. R1 limits the inverter output current during switching, when the IGBT may also be conducting. It should drop 0.5 V at 50 mA.

R2 ensures that the tiny leakage current through D1, when reverse-biased with 12 V, is leaked down to GND. This stops a potentially damaging voltage being put on the inverter output. The inverter will probably have an internal clamping diode between its output and its positive supply rail but I don't know which logic family you're using, hence R2. The absolute worst-case reverse-biased current for a 1N4007 is 50 uA, so the 10 K would only produce a 0.5 V drop if it was the only sink for that current, which it isn't.

The D1 and R1 voltage drops are approx. 0.7 V and 0. 5V when driving high at 50 mA. Then there's the drop in the logic gate output at 50 mA to consider. This should still give you a good TTL logic high (>2.4 V) out of the circuit.

If you decide that you want a stronger drive to a higher TTL high voltage than this circuit will give, you could use the below circuit instead.


simulate this circuit

Here, the high-drive is provided by a PNP running from a 6 V rail. The higher rail compensates for the 0.7 V and 0.3 V drops in D1 and Q2 respectively. So you get a strong 5 V output at 50 mA. If your load is fine with a TTL high of 4 V at 50 mA, you can use a 5 V rail, or select between 5 V and 6 V or whatever.

R3 and R4 are a potential divider, set to deliver Q2 a Vbe of 0.7 V to turn it on when the logic buffer output is 4 V or less. This means that Q2 is only on when Q1 is off. R3 and R4 draw a base current of around 600 uA, allowing for at least 60 mA output even with a low Q2 hFE of 100.

Incidentally, I did look at using a comparator for Q2 but the fewer freely-available, fast parts about with a >50 mA output were a few pounds. The buffer, PNP and two resistors plus sourcing and assembly costs will be much less. You'd previously that cost is very important here.

  • \$\begingroup\$ Yes,m some ignitors require 25mA plus to trigger (they have darlington arrays inside them) so to yeild this you're lookig at 100-200ohms - thats a lot of heat being wasted internally. Good idea though if we could get away with 5mA! :) \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 16:25
  • \$\begingroup\$ Well, its 250mW at 50mA \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 16:28
  • \$\begingroup\$ Sure :-) So the TTL output's probably not sinking that much, then...any idea what the TTL sink current max. is? \$\endgroup\$
    – TonyM
    Jan 25 '18 at 16:41
  • \$\begingroup\$ That looks like an elegant solution - D1 would cause a voltage drop though - coukd you recommend a logiv family that would trigger at 3v3 (I assumecmos levels) and survive the flyback of the coil? \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 17:07
  • \$\begingroup\$ Tony could you explain this for me please: R2 ensures that the tiny leakage current through D1, when reverse-biased with 12 V, is leaked down to GND. \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 17:11

Juntion D is the issue (as only 1 pin can be used for the ignition outputs)

Why don't you put a relay on the output like this: -

enter image description here

So, both IGBT and TTL device receive an input but the relay on the output decides which one is selected.

Your original proposal is going to blow the lid off the TTL driver unless you have a very cunning plan that avoids this.

  • \$\begingroup\$ A relay is a no go for a system like this unfortunately :( \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 16:14
  • \$\begingroup\$ You ought to explain why. \$\endgroup\$
    – Andy aka
    Jan 25 '18 at 16:26
  • \$\begingroup\$ becuase having mechanical systems inside an ECU is generally not good - relays wear etc - an ECU is solid state (in all cases I have ever come accross, ever) \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 16:28
  • \$\begingroup\$ @MattyT2017 have you ever heard of solid state relays? \$\endgroup\$
    – Andy aka
    Jan 25 '18 at 17:14
  • 1
    \$\begingroup\$ Yes, very expensive :) \$\endgroup\$
    – MattyT2017
    Jan 25 '18 at 17:19

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