Design Goals

Design Description

This thermal switch solution will signal low (to a GPIO pin) when a certain temperature is exceeded thus alerting when conditions are no longer optimal or device-safe. This circuit incorporates an NTC thermistor with a comparator configured in a non-inverting fashion.

Design Notes

1. The resistance of an NTC thermistor drops as temperature increases.
2. The TLV7041 has an open drain output, so a pull-up resistor is required.
3. Configurations where the thermistor is placed near the high side of the divider can be done; however, the comparator will have to be used in an inverting fashion to still have the output switch low.
4. Best practice involves placing a positive feedback resistor to add external hysteresis, for simplicity, it is not done in this example.

Design Steps

1. Select an NTC thermistor, preferably one with a high nominal resistance, R₀, (resistance value when ambient temperature, T_A, is 25°C) since the TLV7041 has a very low input bias current. This will help lower power consumption, thus reducing the likelihood of reading a slightly higher temperature due to thermal dissipation in the thermistor. The thermistor chosen has its R₀ and its material constant, β, listed below.
   ${\mathrm{R}}_0=100\mathrm{k\Omega }$
   $\mathrm{\beta }=3977\mathrm{K}$
2. Select R₁. For high temperature switching points, R₁ should be 10 times smaller than the nominal resistance of the thermistor. This causes a larger voltage difference per temperature change around the temperature switching point, which helps guarantee the output will switch at the desired temperature value.
   ${\mathrm{R}}_1=\frac{{\mathrm{R}}_0}{10}$
   ${\mathrm{R}}_1=\frac{100\mathrm{k\Omega }}{10}=10\mathrm{k\Omega } \text{(Standard Value)}$
3. Select R₂. Again, this can be a high resistance value.
   ${\mathrm{R}}_2=1\mathrm{M\Omega } \text{(Standard Value)}$
4. Solve for the resistance of the thermistor, R_thermistor, at the desired temperature switching point. Using the β formula is an effective approximation for thermistor resistance across the temperature range of -20 °C to 120 °C. Alternatively, the Steinhart-Hart equation can be used, but several device-specific constants must be provided by the thermistor vendor. Note that temperature values are in Kelvin. Here T₀ = 25 °C = 298.15K.
   ${\mathrm{R}}_{\mathrm{thermistor}}\left({\mathrm{T}}_{\mathrm{sp}}\right)={\mathrm{R}}_0\times {\mathrm{e}}^{\mathrm{\beta }\times (\frac{1}{{\mathrm{T}}_{\mathrm{sp}}}-\frac{1}{{\mathrm{T}}_0})}$
   ${\mathrm{R}}_{\mathrm{thermistor}}(100°\mathrm{C})=100\mathrm{k\Omega }\times {\mathrm{e}}^{3977\mathrm{K}\times (\frac{1}{373.15\mathrm{K}}-\frac{1}{298.15\mathrm{K}})}$
   ${\mathrm{R}}_{\mathrm{thermistor}}(100°\mathrm{C})=6.85 \mathrm{k\Omega }$
5. Solve for V_thermistor at T_sp.
   ${\mathrm{V}}_{\mathrm{thermistor}}\left({\mathrm{T}}_{\mathrm{sp}}\right)={\mathrm{V}}_{\mathrm{cc}}\times \frac{{\mathrm{R}}_{\mathrm{thermistor}}\left({\mathrm{T}}_{\mathrm{sp}}\right)}{{\mathrm{R}}_1+{\mathrm{R}}_{\mathrm{thermistor}}\left({\mathrm{T}}_{\mathrm{sp}}\right)}$
   ${\mathrm{V}}_{\mathrm{thermistor}}(100°\mathrm{C})=5\mathrm{V}\times \frac{6.85\mathrm{k\Omega }}{10\mathrm{k\Omega }+6.85\mathrm{k\Omega }}=2.03\mathrm{V}$
6. Solve for R₃ with the threshold voltage, V_TH, equal to V_thermistor. This ensures that V_thermistor will always be larger than V_TH until the temperature switching point is exceeded.
   ${\mathrm{R}}_3=\frac{{\mathrm{R}}_2\times {\mathrm{V}}_{\mathrm{TH}}}{{\mathrm{V}}_{\mathrm{cc}}-{\mathrm{V}}_{\mathrm{TH}}}$
   ${\mathrm{R}}_3=\frac{1\mathrm{M\Omega }\times 2.03\mathrm{V}}{5\mathrm{V}-2.03\mathrm{V}}=685\mathrm{k\Omega }$
   ${\mathrm{R}}_3=680\mathrm{k\Omega } \text{(Standard Value)}$
7. Select an appropriate pull up resistor, R₄. Here, V_pu = 3.3V (digital high for a microcontroller).
   ${\mathrm{R}}_4=51\mathrm{k\Omega } \text{(Standard Value)}$

Design Simulations

DC Temperature Simulation Results

Design References

Texas Instruments, SLVMCS1 simulation, circuit file

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