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Author Topic: Faster pulse = higher resistance slower pulse = lower resistance  (Read 494 times)
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Hi
I know this should be simple but I cant for the life of me work it out.
Trying to work out the code for this.
Imagine a wheel having a reed switch on it. Every revolution it would would give a pulse as the reed switch it activated.
Eg. Its doing 1 revolution ever 1 second. This would make make my digital POT to = 100k
2 revolution would = 200k 3 = 300k and so on.
when the wheel dropped back to 2 revolution it would = 200k and when dropped to 1 revolution it would = 100k..

Any ideas?

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Divide and conquer.  There are two tasks here, detecting the rotation rate and programming the digital pot - tackle each separately.

For the first task you'll need to wire up the reed switch with a pull-up resistor (perhaps built-in pull up), and implement pulse debouncing and rate-counting - for such low pulse rates timing the pulses to measure period is perhaps the best method.

For the digital pot you'll have to provide the interface it needs - any particular digital pot in mind?  Do you need a pot BTW - ie what is it controlling?
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Hi Mark.
Thanks for replying.
I have a couple IC's on order the first POT is - IC,potentiometer,digital,100kohm,SC70-5,ISL90460TIE527Z
http://australia.rs-online.com/web/p/products/518-431/

Basically the resister will swipe high and low with just pulsing a couple different pins.
This is how I was going to control the resistance by the speed of the pulse. (revolution of the wheel)
I'm trying to program a arduino to do this.
Before I create the circuit for the IC, im trying to get it to work first with virtual breadboard.
I can get the values to climb but not decrease. I don't know programming enough.


I want to do something similar to this
     
-skip to the end
But by controlling the resistance by the speed of the wheel.


My other IC Pot I might use is Digital Potentiometer 32POS 10KOhmdigital




cheers
« Last Edit: August 20, 2012, 03:17:35 pm by yipyap777 » Logged

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