I have a internally regulated vehicle alternator which uses a conventional battery light wired in-between the exciter terminal and positive.

Obviously when the alternator isn't turning the exciter terminal is ground so the bulb is on, when the alternator is turning and charging the terminal becomes live thus no voltage difference and the bulb goes out.

I want to get rid of the bulb and have everything on my engine controlled and monitored with my own MCU.

As the alternator needs to 'see' a current on the exciter terminal I have no option but to replace the bulb with a low value resistor. Nothing above 25Ω works so I'm using a 20Ω resistor. This works and allows the alternator to charge.

I now want to hook up a 3v3 MCU to monitor the state of this circuit and to replace the visual element of the bulb. But it doesn't seem to be as simple as it first seems.

This is a standard alternator circuit: enter image description here

This is the circuit I've tried: enter image description here

This allows the alternator to charge and outputs 2.8-3v to the MCU when the alternator is charging, but when is 0.7V when not charging. Obviously I can get the voltage a bit higher by using a 3.3v zener ( I only had a 3v to hand), but how can I get the 0.7 lower. Ive tried putting a pull down resistor in parallel with the zener creating a voltage divider but this affects the 3v when the alternator is charging.

I've also tried using a onto coupler but no matter what value of resistor I use I cannot get the opto to trigger.

It may seem that 0.7 is low enough to register low but I want to make sure that this circuit will work with all alternators. So I need the value to be as low as possible.

What is the best way to monitor the exciter terminal of a alternator with a MCU?

  • \$\begingroup\$ One easy way to do it would be to keep the bulb and use a hall effect current sensor like allegromicro.com/en/Products/Current-Sensor-ICs.aspx , which would not affect the alternator \$\endgroup\$ – BeB00 Sep 20 '18 at 17:29
  • \$\begingroup\$ I like that idea, but the bulb has to go, would the current sensor work in line with the D+ terminal and the load resistor? \$\endgroup\$ – B.Baker Sep 20 '18 at 17:38
  • \$\begingroup\$ Why does the bulb have to go? \$\endgroup\$ – BeB00 Sep 20 '18 at 17:39
  • \$\begingroup\$ For many reasons, the intention is to monitor and control various functions of mechanical engines, sometimes they are in luxury / expensive vehicles which have digital dashboards and having a lamp wired up would be ugly and considered a bodge. There are also many other reasons where wiring a bulb etc is not acceptable, there has to be a single wire directly between the D+ terminal and my device. \$\endgroup\$ – B.Baker Sep 20 '18 at 17:47
  • \$\begingroup\$ You are detecting alternator field current that is amplified in the alternator to create a charge current by detecting the error voltage from an internal reference. Do you want to measure EXcitation current or charge current? Either way, you need a 50mV to 75mV shunt at max current with an INA to amplify the voltage using a Kelvin-type shunt or even a length of flat wire. What do you expect to measure? Only the charge current can detect a bad 1 of 6 diodes? THe excitation current is not linear with charge current and varies with RPM as well as faulty diodes \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 20 '18 at 17:48

Alternator schematic

What happens at the indicator terminal is that there's a trio of diodes from the winding phases that provide the current for the field coil. When the alternator starts up, some current is needed through the field winding (on the rotor) for it to generate enough magnetic field to produce an output, so that's why you needed a low resistance to replace the bulb. Once the alternator is producing an output, the trio brings that terminal up to the same voltage as the battery, since the drops across the trio and main diodes are similar, so there's no potential across the bulb, and the trio is providing the field current. With the alternator not turning, the bulb (or bias resistor) current is still being dropped across the field winding, which is a few ohms so the voltage at the terminal is not zero. You could try putting a couple of diodes, (or a zener) in series with your 4k7 and another pulldown to ensure that the input is below the low state threshold on the MCU input.alternator charge signal

Other alternators have a regulator that switch the field internally based on the output of the windings under the residual magnetism of the rotor, these just have a transistor (usually a FET) that pull the indicator output down. These would function the same - they don't need the bulb or any other current source to aid starting up.

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The complexities inside the alternator do not come into play for your problem. This is a simple indicator. All you need is a replacement for the lamp (a resistor) to keep the alternator happy and the best way is to use an opto-isolator (which protects your MCU), what could be simpler.


simulate this circuit – Schematic created using CircuitLab

The logic level Optoisolator I suggested (TLP2361) will work for a 3.3V MCU and provides push-pull drive for the DIO pin and can be readily Gnd isolated too.
You could still use the indicator lamp if you so desired.

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  • \$\begingroup\$ This is what I originally thought, I did try it with a 4N32 but I could not get it to work. \$\endgroup\$ – B.Baker Sep 20 '18 at 19:00
  • \$\begingroup\$ @B.Baker Not all optos are the same. The 4N32 uses a bare-bones Darlington structure to receive from the LED emitter. The TLP2361 has conditioning and a proper output drive circuit to receive from its LED emitter. So a lot depends on the circuitry you applied to the 4N32's output side (which will not be anything like what Jack shows.) I can't say you did anything wrong or that the TLP2361 will perform better for you in this case. Just noting the difference to be dealt with. \$\endgroup\$ – jonk Sep 20 '18 at 19:10
  • \$\begingroup\$ @B.Baker I have a side-concern, though. The lamp provides a current-limiter function that helps protect the rotor coil regulator. For example, a direct application of the battery voltage without the lamp would likely wind up destroying it. Some care here is suggested, especially as it seems you are looking for a circuit that will apply reliably in a great many different scenarios; not all of them identical. The POTS phone system, for example, is actually any number of different switching systems with unique behaviors. Designing circuits to work with all of them was not an easy task, back then. \$\endgroup\$ – jonk Sep 20 '18 at 19:14
  • \$\begingroup\$ @jonk Not so, the circuit is obviously designed (by the automaker) so loss of an indicator lamp does NOT damage the alternator. Do design as you say would be asinine. \$\endgroup\$ – Jack Creasey Sep 20 '18 at 19:32
  • \$\begingroup\$ @jonk I suggested the TLP2361 for a reason, it is ideally suited for interface with a 3.3V MCU. \$\endgroup\$ – Jack Creasey Sep 20 '18 at 19:33

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