I am software guy and this is my first time here. So pardon me for my lack of understanding.

I have a device with analog input, that can read analog input in the form of DC Voltage within the range of 9V-DC to 24V-DC. The source of analog output emits the analog output in the form of DC voltage in the range of 40V-DC to 100V-DC.

How do I receive the analog value in my device?

I find on the internet that there is a DC-DC converter but those are usually fixed voltage input and output. And as you can see, my system requires variable range input and output.

For ex:

  • 100V converts to 24V
  • 60V converts to 16.5V
  • 40V converts to 9V

After I read the value successfully, at the firmware level, I can display the value of 24V into 100V. It's just that I cannot read the 100V value directly because the device has a higher limit of being able to read up to 24V only.

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    \$\begingroup\$ The values you mentioned are not linear. If you want to have a linear output, don't use a DC/DC converter, just look for a "voltage divider". \$\endgroup\$ – Huisman Apr 6 '19 at 12:00
  • 2
    \$\begingroup\$ Do also make sure not to exceed the voltage ratings of the used resistors. You might want to use 3 resistors instead of 2. \$\endgroup\$ – Huisman Apr 6 '19 at 12:13
  • \$\begingroup\$ @Huisman Why not put that in the answer section? \$\endgroup\$ – Unknown123 Apr 6 '19 at 19:08

I would divide by 5 with a voltage divider and make the final correction in firmware. Is it really this non-linear? If yes, use Excel curve fit, assuming that you have floating point available in your firmwareVoltage Divider.

Resistors, 1/2W, 200V. In stock at Digikey. SFR16S0004022FR500 SFR16S0001002FR500

This assumes that your input impedance is high. If it is low and predictable, you can adjust the voltage divider appropriately. If it is not predictable, I would use a high-voltage op-amp.

The power dissipation in the 40k will be 0.16W. You dont want to go anywhere near the rated power of a resistor, 2X margin or more is preferred. My father taught me 40+ years ago, the rated power is just before it will catch on fire (exaggerating somewhat).

  • 1
    \$\begingroup\$ This approach couples the tolerance of the divider network to the integral non linearity of the sensor, so an in situ full scale and 0 scale calibration may be necessary to correct for divider error. They will also change with temperature so low ppm resistors should be chosen. Note that absent FPU, fixed point is usually adequate for sensor calibration curves \$\endgroup\$ – crasic Apr 6 '19 at 22:37

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