I'm trying to build as simple a circuit as possible that will pull a logic line low every time an automotive ignition coil 'fires'. My first inclination is to simply attach the 'negative' terminal of the coil to the base of of an NPN BJT through a current-limiting resistor, with the emitter tied directly to ground. So there would be 5V on the collector and roughly 300V on the base resistor when being triggered. Somehow, though, I feel like I'm missing something fundamental that won't let this work. Perhaps it's just me being spooked by the 'high' voltage. I suppose I could use a voltage divider at the base, but I still feel like I'm missing something. Looking for confirmation regarding the proposed functionality, or of my idiocy. Thanks.

Edit: I don't intend to pass any current through the transistor, besides what little is required by the device itself. The idea is that the transistor will go into saturation when the ground is removed from the coil (when the counter-EMF brings the coil's 'negative' terminal up to around 300 volts). Here's a (rather amateurish) schematic of what I'm thinking:

enter image description here

Switch 'S1' in that schematic is just a placeholder for the IGBT/Points triggering the ignition coil. Sorry again for the confusion, and thank you, everyone, for all the replies!

  • \$\begingroup\$ Please add a circuit how things are wired in your proposal, inductor, breaker and battery voltage. \$\endgroup\$
    – jippie
    May 22, 2013 at 6:08
  • \$\begingroup\$ Your description sounds like you are trying to pass the primary coil current through the base-emitter junction via a resistor that would severely limit the primary current and would not work. Please post a circuit diagram. \$\endgroup\$ May 22, 2013 at 6:28
  • \$\begingroup\$ A BJT is in essence a current driven device, hence base voltage is, for most purposes, irrelevant, as long as the actual voltage expressed on the base-emitter junction is within device ratings. However, without a schematic, determining whether the approach is feasible or practical, is very difficult. Please upload a schematic on any public image hosting site, and provide the link as an edit to the question: Someone with edit permissions will incorporate the diagram into your question. \$\endgroup\$ May 22, 2013 at 7:22
  • \$\begingroup\$ I didn't see the schematic when I wrote my response. What you have depicted looks like it ought to work, but consider that when S1 opens, whatever the primary current happens to be, that will all be driven into the base of the transistor. The transistor probably won't need more than 1 or 2 mA to saturate, and for any decent output from the coil, the primary current is likely up in the hundreds of mA, maybe even an amp or more. This will be hard on the transistor. Suggest adding a resistor from base to ground.. \$\endgroup\$
    – JustJeff
    May 24, 2013 at 10:12
  • \$\begingroup\$ The value depends on the primary coil current at S1 opening and the transistor, but I'd guess it would be 100 ohms or less. Assuming this is a one-off experiment, you could start low and work upward. \$\endgroup\$
    – JustJeff
    May 24, 2013 at 10:15

5 Answers 5


enter image description hereRepicating your circuit in LT spice]![Base current and base voltage waveforms1In all the cases you cannot say that the base voltage is irrelevant for the transistor's operation or as you said transistor;s saturation operation.

If I consider your case the the transistor is mainly dependent on the base voltage and also if you have not taken care of the transistor for its ability to sustain the huge base current it might even blow off.

Now consider when the switch is suddenly opened the voltage induced across the inductor is huge and it is seen by the 10k resistor. This gives rise to a huge base current which might even blow the transitor. So you need to check the maximum Vbe which can be given to the base of the transistor before switching on the circuit.

Solution: If you can place a zener of some voltage lesser than the supply voltage then I think your circuit will work.

One input: Placing inductor at the base of the transistor is always not a good thing to do. Better try to place it in the collector with a free wheeling diode across it.

If you think your transistor can sustain a huge base voltage of 100s of volts and base current of 100s of mA then you can go with the circuit you have else No.

  • \$\begingroup\$ Thanks for the reply. It's amazing how clear you feel a question is, until you see the answers you get. Not only did I neglect to mention that the coil I'm connecting to never sees above 300 volts, but I didn't even really ask the right question. It should have been "is base voltage irrelevant when base current is limited within acceptable levels' or something of the sort. At any rate, I decided to modify my circuit with a voltage divider on the base (while keeping the 10K current limiting resistor). PS - Now I feel like an idiot for not modeling it in LTspice myself... \$\endgroup\$ Jun 1, 2013 at 4:57
  • \$\begingroup\$ No problem chris! Its good that you have put that this question here. I even did not put the answer directly. I just gave a thought over the inductor and could think of the above mentioned situation. No harm in posting a question however simple it is, folks will get to know few things out of it. \$\endgroup\$
    – IamDp
    Jun 1, 2013 at 15:59

If I read that right, your proposed circuit will insert a resistor and a B-E junction between the coil and ground. That won't work very well.

Ignoring the transistor for a moment, think about what that resistor is going to do to the current in the ignition coil. The current will be greatly reduced, and you'll be lucky if the engine fires at all. (Assuming it's part of an engine).

But there's hope, if you just want a logic line that says 'hey this thing is firing'. The basic principle in engine ignition systems is to apply the battery voltage to the ignition coil most of the time, and then interrupt that current at the moment the cylinder is meant to fire. The sudden interruption of the current is what induces the high voltage on the output side.

Now it could matter whether the system driving the coil is an electronic ignition or an old time "point" system. I can't speak to the complexities of the electronic systems as I'm just not familiar with them, but I'd speculate that they simply use some kind of FET or SCR to do what the mechanical contacts do in the old point system, which is just to interrupt that battery input.

So with the point system (and possibly with electronic ignition) a couple of things happen when the current going into the coil primary is interrupted. One is of course the high voltage spike on the secondary. But also the input to the primary, which sits at battery voltage most of the time, will suddenly swing to a large negative voltage. This happens because the primary inductance wants to prevent changes in the current, so the current kind of just keeps going in for a brief time, which makes that input look like a large negative voltage. Not as large as the secondary voltage, but undoubtedly much higher than the battery voltage. That is what you can monitor, without interfering with the coil's performance.

Apologies for not having a schematic tool ready. Here's an outline of the idea: Attach the NPN emitter to ground. The collector will be your logic output, and will go to battery+ through R3 and ground through R4. Finally, the base will connect to battery+ through R1 and to the ignition coil input via R2.

The idea is to have the NPN on saturated while the coil is 'resting', and briefly shut off while the coil is being fired. R1 biases the NPN into saturation. R2 allows a whiff of that primary negative kick into the base to cut the NPN off. R3 and R4 form a divider that brings e.g. 12V down to 5V when the NPN is off.

If the battery is 12V and the NPN is something like an old 2n2222, you'd choose R3 to limit collector current to a safe value, probably around 2k2. Pick R4 at 1k5 and you'll get around 4.9V when the transistor is off, and of course around 0.2 when it is on. Now R1 needs to be small enough to saturate the transistor from the battery voltage, so 5k6 there would put about 2mA into the base, which should do it.

R2 is the trickiest part. Depending on the coil that negative spike in the primary could be nearly anything. You want things such that R2 steals just the current R1 supplies, in order to cut off the transistor. Make R2 too large, it won't shut the transistor off. Make it too small, and it will pull current the Wrong Way through the B-E junction and your transistor will pop. Probably advisable to stick a diode across the B-E junction, cathode to B, anode to E; this will 'clamp the base voltage to ground' and keep the transistor safe. (but don't use some old power rectifier, you need something with fast response time) Probably the way to determine R2 is experimentally; start large, like 1M, and work down until you get reliable operation.

  • \$\begingroup\$ should emphasize, those negative-going pulses you get from the primary are going to be short, probably microseconds or less. A diode built for 50 or 60Hz power rectification wouldn't suffice. \$\endgroup\$
    – JustJeff
    May 23, 2013 at 10:23
  • \$\begingroup\$ of course, that was all written with me thinking of the switch being on the high side of the primary. \$\endgroup\$
    – JustJeff
    May 25, 2013 at 14:06

How about something like this:


simulate this circuit – Schematic created using CircuitLab

When the SW1 is closed, no current flows across R1 because the spark plug gap is an open circuit. When SW1 opens, L1 can only dissipate its energy across the spark gap, and to that end it develops some large voltage in the tens of thousands of ohms. The voltage rises until a current is able to flow across the gap. The current develops a voltage across the current sensing resistor R1. This voltage powers the circuit formed by the large resistor R2 and optocoupler, lighting the internal LED which activates the phototransistor.

The R1 and R2 values mean nothing; I have no idea about choosing R1 and R2 so that the optocoupler activates over the possible range of currents that flow to the spark plug! R1 limits the voltage in the optocoupler input circuit by virtue of its small size (V = IR). R2 has to be some sensible value with respect to the voltage developed by R1 to limit the current.

The optocoupler solves the problem of coupling unknown voltage levels between the circuits. They do not even have to share a common ground. The R1/R2 circuits just lights an LED, and the other circuit responds to light.

  • \$\begingroup\$ Thanks for the response! Unfortunately, I don't think this approach is terribly viable. The schematic I provided shows only a single inductor (the primary side of the ignition coil), but it's the secondary of the transformer where the spark plug is connected (and trying to put a 100Mohm resistor in the path of 30KV just sounds like asking for trouble). \$\endgroup\$ Jun 1, 2013 at 5:04
  • \$\begingroup\$ @Chris It's mΩ (milli-) not MΩ! It's meant to be a tiny current- sensing resistor. \$\endgroup\$
    – Kaz
    Jun 1, 2013 at 5:08
  • \$\begingroup\$ Oh geez, sorry! Not used to seeing milliohms I guess. Sort of like seeing MFD on an old capacitor and trying to get your brain to see "MicroFarad". It's still problematic, trying to insert something between the secondary terminal and the spark plug (even cracks in a plug wire's insulation will lead to corona and other undesirable events). Definitely an interesting approach though, I may test it just to see. Thanks! \$\endgroup\$ Jun 1, 2013 at 16:25

to protect the B-E of transistor from excessive negative voltage, add a 1N4007(breakdown voltage=1KV) diode across B-E(anode-E,cathode-B).


No, that ought to work. You should check your resistor wattage and your transistor current, etc, but otherwise that will work.

  • \$\begingroup\$ If you need to check the resistor wattage when it directly feeds the base of a transistor, what sort of current do you think will be passing into the base of the transistor and which component is more likely to "die" first? Given that you haven't offered a value for this resistor, what current do you think the base might need to turn on the transistor? \$\endgroup\$
    – Andy aka
    May 22, 2013 at 7:12
  • \$\begingroup\$ He wants to pull down a 10k resistor connected to 5V, so I'd say about 25 mA. \$\endgroup\$ May 22, 2013 at 22:42
  • \$\begingroup\$ What the hell are you smoking? 10K into 5V is 500 uA, and that's collector-emitter current. If your transistor has decent hFE, your base current would be ~1/50 - 1/100 of that, or about 25-50 microamps. \$\endgroup\$ May 23, 2013 at 4:19
  • \$\begingroup\$ Sorry, meant uA, not mA. \$\endgroup\$ May 23, 2013 at 22:56

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