Posted by Tim Stevens on 09/08/2018 19:42:37:
Silly-old-duffer, John Haine
I'm not sure I understand your point(s) – I am looking at a device which moves wires through a (permanent, unchanging) magnetic field, or more likely, moves the field around the wires. Nothing is done with the output except supplying a (high resistance) voltmeter. So, doesn't Faraday's Law apply in any event, without creating pulses and then turning them back into a proportional output – the voltage from the alternator relating directly to the rate of cutting of lines of force? No generation of pulses, no average, no integration, and the only circuit is a few rectifiers and perhaps (but I doubt it) a smoothing capacitor.
And for the avoidance of doubt, yes, I know there will be a small non-linearity because of the diodes, but I'm sure we can take account of that in designing the scale for the voltmeter.
So, what am I missing, please?
Tim
Hi Tim,
We're discussing a simpler way of generating the rpm voltage that doesn't need the car to drive an alternator. The advantage is elimination of the need for a £ brushless motor and belt drive. Nothing wrong with your idea apart from there may be a cheaper easier alternative
At it's simplest, when a stream of DC pulses are used to charge a capacitor through a resistor (Vin) , the DC voltage across the capacitor at Vout is proportional to the pulse rate.
So if a sensor is arranged to emit a pulse on each rotation of your cars flywheel, then the pulse rate will be low when the engine is idling and the capacitor will only partially charge causing the voltmeter to read low. When the engine is accelerated the pulse rate increases, causing the capacitor to collect more charge. Then the voltmeter reads a higher voltage.
As the voltage appearing across the capacitor is proportional to engine rpm, the voltmeter can be calibrated directly in rpm.
John's pointed out it's not quite that simple because the input pulses all need to be the same width. That's achieved with some simple electronics. At least I think it is! I'm hoping to find a tested circuit rather than design one from scratch.
Dave