Can someone fix this code? Controlling voltage of a Digital Potentiometer(AD8400) using LM317 voltage regulator through Arduino Uno (such that the current remains constant)

I'm working on a project where my setup should receive a steady Current(I). I'm using AD8400 Digital Potentiometer and LM317 voltage regulator connected to an Arduino Uno. This circuit is connected to my setup which is connected to a multimeter. The multimeter displays the voltage(V) and the Current(I) values, where the Current(I) fluctuates because of the built up resistance, so I was trying to write a code to control the voltage of the digital potentiometer(AD8400) such that the Current reading on the multimeter remains constant (say 4mA).

Here is a link to the datasheet for the device I am trying to use: https://www.analog.com/media/en/technical-documentation/data-sheets/AD8400_8402_8403.pdf

Here is a circuit diagram describing the circuit connections of LM317 voltage regulator:

In the circuit diagram the Potentiometer is represented by 'R'.

I wrote this code to control the voltage but there is some issue in it. And I couldn't figure out how to change the voltage such that the Current(I) remains constant.

clear all;
a=arduino('com3','Uno','Libraries','SPI');

'arduino' requires MATLAB Support Package for Arduino Hardware.

d_pot = device(a, 'SPIChipSelectPin', 'D10','bitrate',1);
d=a,d_pot;
value=200;reverse=0;
for i=1:10
     if value>=200
        reverse=1;
     end
    if value<=200
         reverse=0;
    end
    if reverse==0
        value=value+1;
    else
        value=value-1;
    end
    writeRead(d,50,value);
    pause(0.1)
end

Any input would be really appreciated. Thank you!

Answer 1

[Walter Roberson]

     if value>=200
        reverse=1;
     end
    if value<=200
         reverse=0;
    end

If value is exactly 200, then you set reverse to 1 and then you set it back to 0.

[Vinayak]

That is already done, there is an issue in the code which I am not able to resolve & i need an extra part for this code to vary the voltage such that my Current(I) remains almost constant, which i couldn't figure out.

[Walter Roberson]

https://www.mathworks.com/help/supportpkg/arduinoio/ref/writeread.html

writeRead() for an arduino SPI device normally expects two parameters, one being the device and the other being a command; if the third parameter is present it would have to be a precision such as 'uint8'.

As shown in https://www.mathworks.com/help/supportpkg/arduinoio/ug/communicate-with-an-spi-eeprom-device.html if there are additional values to be sent then you would make them part of the second parameter, such as

writeCmd = 0b00000010;
address = [0 0];
dataToWrite = [writeCmd,address double('Hello')];
writeRead(eeprom, dataToWrite)

If the command is 50, then that implies that you should

writeRead(d, [50 value])

[Walter Roberson]

In order to maintain a constant current, then each time you change the voltage, you would also have to change the digitial pot resistance. Increase the voltage, decrease the resistence, so that their product stays constant (though it might perhaps have to be more complicated than that due to constant internal resistences.)

[Vinayak]

Thanks for your advice. What if I keep the voltage constant & vary only resistance such that the Current remains almost constant? That sounds less complicated, doesn't it? Can you give me a logic for that?

[Walter Roberson]

The AD8400 appears to also be SPI, so programming it would appear to be much the same.

The manual at https://www.analog.com/media/en/technical-documentation/data-sheets/ad8400_8402_8403.pdf implies you might need to send a 10 bit data value -- though if you are using the single-output version then it looks to me as if the two high bits would always be 00

The device has a built-in resistance that is added to D/256 * resistance_span -- so if you wanted to control the resistance completely linearly you would need to subtract the constant resistance from the target resistance, and divide by the resistance span and multiply by 256 to get the value to send.

[Vinayak]

Thank you, your solution was helpful.