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I am trying to program a digital resistor and set it's wiper resistance. The general circuit looks a little like this where "pot" represents the Maxim DS1801 digital resistor and DUE is, of-course the 3.3v Arduino DUE.

POT       DUE
CLk ----> SCK
D    ----> MOSI
RST ----> CS (ChipSelect) (PIN10)

Now so far, I have sucessfully been able to interface with the DS1801, but I am having issues determining how to properly set the wiper resistance values. The DS1801 datasheet says the it accepts a 16bit input and from application example online it looks like the values range from 0 to 64. Here is my code:

// include the SPI library:
#include <SPI.h>


// set pin 10 as the slave select for the digital pot:
const int slaveSelectPin = 10;
int mute = 64;
int loud = 63;
int soft = 0;

void setup() {
  // set the slaveSelectPin as an output:
  pinMode (slaveSelectPin, OUTPUT);
  // initialize SPI:
  SPI.setBitOrder(LSBFIRST);
  SPI.setDataMode(SPI_MODE0);
  SPI.setClockDivider(slaveSelectPin,SPI_CLOCK_DIV8);
  SPI.begin();
  delay(10);
   digitalPotWrite(mute, mute);
}

void loop() {

  // Raise the volume from off to loud
for(int i = 0; i <= 63; i++){
  digitalPotWrite(i, i);
  delay(5);
    }
  }

void digitalPotWrite(int left, int right) {
  // take the SS pin low to select the chip:
  digitalWrite(slaveSelectPin, HIGH);
  //  send in the address and value via SPI:
  SPI.transfer(left);
  SPI.transfer(right);
  // take the SS pin high to de-select the chip:
  digitalWrite(slaveSelectPin, LOW);
}

I THINK my issue lies with setting the clock rate for the DS1801. The data sheet say the chip runs at 10MHz which is why I wrote

SPI.setClockDivider(slaveSelectPin,SPI_CLOCK_DIV8);

to divide the 84MHz default speed down to 10MHz.

Anyone have suggestions on why my code is not working??? At the moment, I have LED's connected to chip with series resistors to limit current below 2mA. The resistors just blink randomly with the code posted above.

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  • 10MHz is too high to have flapping around in the breeze. Show your circuit setup. Apr 22, 2015 at 22:56
  • I'm not exactly sure what you mean by that? lol ... Okay, I'll go snap a picture of the circuit and update my post. Why would 10MHz be too high if that's what it says in the datasheet? I don't doubt you, I'm just wondering thats all! :) Apr 23, 2015 at 3:21
  • Poor wiring can turn a 10MHz signal into a dog's breakfast, so extra care must be taken when routing such. Apr 23, 2015 at 3:33
  • Alright, I threw up some pictures, hopefully you can make that out.... essentially, blue is the bit line (D - MOSI), white is the 10MHz clock, and green is Chip Select (or RESET as the datasheet calls it) Apr 23, 2015 at 3:45

2 Answers 2

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There is no way in hell you're going to get a 10MHz signal down that line. Shorten it to 3 inches total maximum and reduce the clock rate to 2MHz or lower. It is possible to get a 10MHz signal through, but you'd need to switch to twisted pair wiring and proper termination to even begin to have a chance to get it to work.

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Well, it turns out that apparently my issue was not having the AGND pin hooked up. In addition to that, I was trying to switch the wiper resistance value too quickly. By putting at least a 10ms delay between calls, it works much better.

Also, it does seem to work even at 10MHz just fine...

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