K-type thermocouple table

The problem I see is that a K-type thermocouple has a negative range up to -6mV.

If you omit the negative range, and use only the positive range , then it is 0.000V to 33..275mV

0.033V/4095=8.0586e-6 V (8.05uV)

So each count would be **8.05uV** IF you COMPRESS THE FULL SCALE TO 0.033V/5.00V = 1/151th of 5V

This means you would need a **divide by 151 voltage divider** on the DAC output:

**divide by 151 voltage divider**

**Vo = Vin* R2/(R1+R2)**

Let Vo = 0.033mV = **0.000033V = 0.000033**

Let Vin = 5.00V

Let R2 = 1000 ohms

**then**,

0.000033V **= 5.00* 1000/(R1+1000)**

0.000033V **5.00 1000 **5.00 * 1000**

**------------** **= ------** ***** ** ------------ = ------------

** 1 1 ****(R1+1000) ****(R1+1000)**

0.000033V **5.00 * 1000**

**----------- **=** ** ------------

** 1 ****(R1+1000)**

**Cross Multiply**

0.000033__*(R1+1000) = 5.00*1000__

0.000033__*R1 + 1000***0.000033 __ = 1000*5.00**

0.000033__*R1 + 0.033 = 5000__

0.000033__*R1 = 5000-__**0.033**

**4,999.967**

**R1** **= ------------**

0.000033

**R1 = ****151,514,151 ohms**

**Proof:**

**Vo = Vin* R2/(R1+R2)**

Let Vo = 0.033mV = **0.000033V = 0.000033**

Let Vin = 5.00V

Let R1 = **151514151**

Let R2 = 1000 ohms

**Vo = 5.00* 1000/(****151514151****+1000)**

** = 5000/(**151,515,151)**

** Vo = 0.000033 V (0.033mV) (maximum K-type thermocouple voltage)

If I read the table correctly, 0.000 degrees = 0.397mV (0.397mV = 0.000397

**0.000397**/8.0586e-6 V = **49** DAC counts

So, given a voltage divider | where R1 = 151514151 ohms , 1 % , R2 = 1000 ohms , 1%,

and a DAC output of 49 counts (5V/4095= 0.001221/per count)

0.001221 V * 49=**0.059829** V

0.059829/151 = 3.9621893926529688119092092602026e-4 V = 0.0003962189 V (~**0.000397V** = **0.397mV**)