Electronics > Beginners

Will this work (optocoupler over shunt)?

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Mark Hennessy:
This can be done with a few turns of wire around a reed switch. I like this idea because it minimises losses, is nice and simple, and reed switches are cheap.

About 25 years ago I found a module made by Audi that monitors brake lights using this technique - I probably still have it somewhere. The only added complexity was a couple of transistors to latch the faulty state on IIRC.

Depending on your application, you might need a transistor to invert the logic, but if you don't need that, you might not need any other parts. Here's a reed switch that is good for 500mA, so should drive the relay coil (or possibly even the load you want to switch with the relay?) directly:

https://uk.farnell.com/hamlin/mdsr-4-22-38/switch-reed-spst-no-0-5a-200v/dp/2103636

spec:
While it is obviously good idea to reduce the senor voltage drop, the OP seemed quite happy with the voltage drop of 2V in his original circuit. Don't forget that the battery voltage varies from car to car and according to battery state of charge and temperature. There is another point too, if the bulb is incandescent the filament resistance has a strongly positive temperature coefficient and the absolute  cold resistance of the filament will have a wide spread from one bulb to another and one manufaturer to another. So it is very important to put all these theories in perspective, especially as the human perception of light is logarithmic. 

The -2.2mV deg C temp coefficient of a Si forward biased junction is not significant in this application where the difference being measured is 0A and 2A. The effect of this coefficient is often overstated: it only amounts to a change of voltage of -0.37% Deg C

The only thing the OP asked is will his circuit work, and afraid to say that the answer is no. But, if a low forward drop were required, it would be a simple design task to achieve that. One cheap opamp or comparator would do the job for next to nothing.

The current activated reed relay approach is clever, but it does introduce an electro-mechanical part, and a hall effect sensor would be another approach as already mentioned. The infrared detector is the Rolls Royce approach, because, rather than by proxy, it measures the actual parameter of interest, the light output.

ArthurDent:
Here's a circuit that has been around for some time. I haven't tried it but it should work. The advantage of current sensing in the supply going to the bulb is that it can be done remotely whereas a light sensing circuit needs the sensor next to the bulb which may not be possible as in auto taillights.

If you can put a photo diode or photo resistor near the bulb the circuit becomes very simple.

http://www.discovercircuits.com/DJ-Circuits/bad-bulb.htm

spec:

--- Quote from: ArthurDent on December 05, 2018, 04:50:46 pm --- The advantage of current sensing in the supply going to the bulb is that it can be done remotely whereas a light sensing circuit needs the sensor next to the bulb which may not be possible as in auto taillights.
--- End quote ---
Good point.

spec:
UPDATE #1  2018_12_06 (decoupling and filtering added)

Attached below is the schematic for a bulb current monitor with a voltage overhead of 50mV. It uses a colossally expensive precision dual comparator costing 33 pence. ::)

The 25mR resistor could just be a four terminal resistor made from a piece of wire.

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