NPN BJT in saturation, base voltage irrelevant?

Question

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:

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!

Answer 1

1In 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.

Answer 2

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.

Answer 3

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.

Answer 4

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).

Answer 5

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