Design Steps

Determine the given output voltage range and supply conditions for the application circuit. For this example the following conditions are used:

• V_outMin = 5V
• V_outMax = 0V
• V_CC = 12V
• V_bias = 5V
• V_EE = 0V
• T_MIN = –40 °C
• T_MAX= 125°C

1. Using the TMP61 look up table in the Thermistor Design Tool (SBOC595), find the approximate R_{PTC_Min}, R_{PTC_Max}, and R_{PTC_Nom} for the given application supply conditions. R_{PTC_Min} and R_{PTC_Max} are selected near the local minimum or vertex of the PTC curve.
   1. R_{PTC_Min}(–40) = 6.537kΩ
   2. R_{PTC_Max}(125) = 17.852kΩ
   3. R_{PTC_Nom}(25) = 9.962kΩ
2. Per the Thermistor Design Tool (SBOC595), a resistor value of 10kΩ is selected for R1

   R1 = 10kΩ

3. A feedback resistor of 10kΩ will be selected for this example

   R4 = 10kΩ

4. V_NTC(T) will be treated as a linear function to solve for the output across a given output range. The V_out(T) equation can be treated as a line as V_NTC(T) is considered linear.
   Equation 1. $V_{out}\left(T\right)=V_{NTC}\left(T\right)\times \left(1+R4\times \left(\frac{R3+R5}{R3\times R5}\right)\right)-V_{bias}\times \left(\frac{R4}{R5}\right)$
   Equation 1. $Y\left(X\right) = X\times M +\hspace{0.17em}B$
5. The minimum and maximum V_NTC values must be calculated using the given conditions. The curve will be optimized using the PTC resistor end points.
   Equation 1. $V_{NTC\_Min}\left(-40\right)=V_{bias}\times \left(\frac{R2}{R_{PTC}\left(-40\right)+R2}\right) =5V\times \left(\frac{10k\Omega }{6.537k\Omega +10k\Omega }\right)$
   Equation 1. $V_{NTC\_Min}\left(-40\right)=3.02V$
   Equation 1. $V_{PTC\_Min}\left(125\right)=V_{bias}\times \left(\frac{R2}{R_{PTC}\left(125\right)+R2}\right) =5V\times \left(\frac{10k\Omega }{17.853k\Omega +10k\Omega }\right)$
   Equation 1. $V_{PTC\_Min}\left(125\right)=1.80V$
6. The ‘slope’ of the line, M, can be calculated using the desired output voltage range:
   1. V_outMin = 5V
   2. V_outMax = 0V
   Equation 1. $M = \left(\frac{V_{outMax}- V_{outMin}}{V_{NTC\_Max}- V_{NTC\_Min}}\right) = \left(\frac{0V- 5V}{1.80- 3.02V}\right) = 4.07$
7. The non-inverting gain term of the equation can be substituted as M to solve for R5. R5 will be solved using V_outMin and V_{NTC_Min}:
   Equation 1. $V_{outMin} = V_{NTC\_Min}\times \left(1+R4\times \left(\frac{R3+R5}{R3\times R5}\right)\right)-V_{bias}\times \left(\frac{R4}{R5}\right)$
   Equation 1. $V_{outMin} = V_{NTC\_Min}\times \left(M\right)-V_{bias}\times \left(\frac{R4}{R5}\right)$
   Equation 1. $5V = 3.02V\times \left(4.07\right)-5V\times \left(\frac{10k\Omega }{R5}\right)$
   Equation 1. $R5 = -\left(\frac{V_{bias}\times R4}{V_{outMin}- V_{NTC\_Min}\times M}\right) = -\left(\frac{5V\times 10k\Omega }{5V- 3.02V\times 4.07}\right) = 6.842k\Omega$
8. Solve for R3 using the slope of the linear approximation, M:
   Equation 1. $M =\left(1+R4\times \left(\frac{R3+R5}{R3\times R5}\right)\right)$
   Equation 1. $4.07 = \left(1+10k\Omega \times \left(\frac{R3+6.842k\Omega }{R3\times 6.842k\Omega }\right)\right)$
   Equation 1. $R3 = 6.217k\Omega$
9. Using the linear approximation the resistor network has been solved achieved the desired output voltage range.
   1. R1 = TMP61
   2. R2 = 10kΩ
   3. R3 = 6.217kΩ
   4. R4 = 10kΩ
   5. R5 = 6.842kΩ
10. The following table highlights the resistor values for different design cases: