SBOA505 December   2020 OPA2991-Q1 , TLV197-Q1 , TLV2197-Q1 , TLV4197-Q1 , TMP61 , TMP61-Q1 , TMP63 , TMP63-Q1 , TMP64-Q1

 

  1.   Design Goals
  2.   Design Description
  3.   Design Notes
  4.   Design Information
  5.   Design Steps
  6.   Design Simulations
  7.   DC Simulation Results
  8.   Sensor Circuit Accuracy Using a Linear Approximation
  9.   Improving Accuracy of the Sensor Circuit
  10.   Design References
  11.   Design Featured Op Amp
  12.   Design Alternate Op Amp

Design Steps

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

  • VoutMin = 5V
  • VoutMax = 0V
  • VCC = 12V
  • Vbias = 5V
  • VEE = 0V
  • TMIN = –40 °C
  • TMAX= 125°C
  1. Using the TMP61 look up table in the Thermistor Design Tool (SBOC595), find the approximate RPTC_Min, RPTC_Max, and RPTC_Nom for the given application supply conditions. RPTC_Min and RPTC_Max are selected near the local minimum or vertex of the PTC curve.
    1. RPTC_Min(–40) = 6.537kΩ
    2. RPTC_Max(125) = 17.852kΩ
    3. RPTC_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. VNTC(T) will be treated as a linear function to solve for the output across a given output range. The Vout(T) equation can be treated as a line as VNTC(T) is considered linear.
    Equation 1. VoutT=VNTCT×1+R4×R3+R5R3×R5-Vbias×R4R5
    Equation 1. YX = X×M +B
  5. The minimum and maximum VNTC values must be calculated using the given conditions. The curve will be optimized using the PTC resistor end points.
    Equation 1. VNTC_Min-40=Vbias×R2RPTC-40+R2 =5V×10kΩ6.537kΩ+10kΩ
    Equation 1. VNTC_Min-40=3.02V
    Equation 1. VPTC_Min125=Vbias×R2RPTC125+R2 =5V×10kΩ17.853kΩ+10kΩ
    Equation 1. VPTC_Min125=1.80V
  6. The ‘slope’ of the line, M, can be calculated using the desired output voltage range:
    1. VoutMin = 5V
    2. VoutMax = 0V
    Equation 1. M = VoutMax- VoutMinVNTC_Max- VNTC_Min = 0V- 5V1.80- 3.02V = 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 VoutMin and VNTC_Min:
    Equation 1. VoutMin = VNTC_Min×1+R4×R3+R5R3×R5-Vbias×R4R5
    Equation 1. VoutMin = VNTC_Min×M-Vbias×R4R5
    Equation 1. 5V = 3.02V×4.07-5V×10kΩR5
    Equation 1. R5 = -Vbias×R4VoutMin- VNTC_Min×M = -5V×10kΩ5V- 3.02V×4.07 = 6.842kΩ
  8. Solve for R3 using the slope of the linear approximation, M:
    Equation 1. M =1+R4×R3+R5R3×R5
    Equation 1. 4.07 = 1+10kΩ×R3+6.842kΩR3×6.842kΩ 
    Equation 1. R3 = 6.217kΩ
  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:

Temperature

Output VoltageSupply
TMinTMaxVoutMinVoutMaxVbiasVccVee

–40°C

125°C

3V

0.33V

3.3V

3.3V

0V

Resistor

Values:

R1 = TMP61

R2 = 10kΩ

R3 = 8.1kΩ

R4 = 10kΩ

R5 = 9.1kΩ

–40°C125°C

1.7V

0.1V

1.8V

1.8V

0V

Resistor

Values:

R1 = TMP61R2 = 10kΩR3 = 5.5kΩR4 = 10kΩ

R5 = 6.6kΩ