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Electronics (volts / resistance) question


Chris Wilson

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I want to make a couple of 9 position rotary switches for boost control and fuel or ignition trims. My ecu has a 5 volt output and an analogue input. Motec give values for each position of their own (expensive but very nice) switch in a datasheet. I *THINK* there's enough info to work out what resistor they have used on each pin, given position 0 is zero volts and position 9 is 5 volts. It doesn't have to be exact, but no doubt they've used common values. I can create a custom table if it's not totally exact. Is there enough info here to say what ohm resistors they have used? I assume half watt ones will be more than sufficient? Thanks.

 

http://www.gatesgarth.com/switch.pdf if it isn't attached below properly.

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The value of the resistors are not as important as you might think - you're making a string of resistors between 5V and ground which will give you evenly spaced voltage values in between, regardless of the resistance.

 

With the values in the pdf, it looks as if the lowest position will be 0.5V and the highest is 4.5V (the switch voltage output will be inbetween the "jumps" in the table).

 

You can create an equivalent switch by getting 10 of the same value resistors (I'd start with something in the region of 220 Ohms each) and connecting the 1st position pin to the 2nd position pin with a resistor, again with the 2nd to the 3rd, 3rd to the 4th etc. After this, connect the 1st postition pin to ground with another resistor and the 9th position pin to 5V with the last resistor. Finally, connect the wiper pin to your ECU input.

 

With this setup, you will have 0.5V across each resistor. With 220 Ohm resistors, the power dissipated in each resistor will be ((0.5V * 0.5V) / 220 Ohms = 0.001W. Half-Watt resistors would be overkill. 1/8th Watt resistors would be sufficient.

 

Let me know if I'm not clear; I'm happy to draw a picture if it helps.

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The resistor values don't matter insofar as making the voltage divider work, but it might be worth checking the impedance of the input to the ECU which might be in the tech spec somewhere. The maximum resistance between ground and the ECU input (in your case 1k980 Ohms with the switch in position 9) needs to be a lot smaller than the input impedance.

 

For example, neglecting the input impedance, at position 9 your output voltage would be exactly 4.5V. With a input impedance of 1kOhms this would drop to 3.76V. If the imput impedance was 10kOhms it would be 4.41V

 

Theres a handy calculator here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

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