Design Goals

Design Description

Comparators are used to differentiate between two different signal levels. With noise, signal variation, or slow-moving signals, undesirable transitions at the output can be observed with a constant threshold. Setting upper and lower hysteresis thresholds eliminates these undesirable output transitions. This circuit example will focus on the steps required to design the positive feedback resistor network necessary to obtain the desired hysteresis for an inverting comparator application.

Design Notes

1. The accuracy of the hysteresis threshold voltages are related to the tolerance of the resistors used in the circuit, the selected comparator’s input offset voltage specification, and any internal hysteresis of the device.
2. The TLV7041 has an open-drain output stage, so a pull-up resistor is needed.

Design Steps

1. Select the lower biasing resistor, R₂. This resistor can be modified for any design. In this case, it is assumed that power conservation is necessary, therefore, R₂ is selected to be large.
   ${\mathrm{R}}_2\mathrm{ }=\mathrm{ }500\mathrm{k}\Omega$
2. Select the switching thresholds for when the comparator will transition from high to low (V_L) and low to high (V_H). V_L is the necessary input voltage for the comparator output to transition low and V_H is the required input voltage for the comparator to output high.
   
   ${\text{ V}}_{\mathrm{L}}{\text{=2.2V and V}}_{\mathrm{H}}\text{= 2.5V}$
3. Analyze the circuit when the input voltage is V_H. At this point, V_o=3 V=V_PU and the transition to a logic low is initiated in the comparator output. Using Kirchhoff's Current Law, solve for an equation for R₁.
   
   $\frac{{\mathrm{V}}_{\mathrm{PU}}-{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_3+{\mathrm{R}}_4}+\frac{{\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_1}=\frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_2}\Rightarrow {\mathrm{R}}_1=\frac{{\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{H}}}{{\displaystyle \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_2}-\frac{{\mathrm{V}}_{\mathrm{PU}}-{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_3+{\mathrm{R}}_4}}}$
4. Analyze the circuit when the input voltage is V_L. At this point, V_o=0 V and the transition to a logic high is initiated in the comparator output. Using Kirchhoff's Current Law, solve for an equation for R₁.
   
   $\frac{{\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{R}}_1}=\frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{R}}_2}+\frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{R}}_4}\Rightarrow {\mathrm{R}}_1=\frac{{\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{L}}\times \left({\displaystyle \frac{{\mathrm{R}}_2+{\mathrm{R}}_4}{{\mathrm{R}}_2{\mathrm{R}}_4}}\right)}$
5. After defining some constants, set the two equations for R₁ equal to obtain a quadratic equation for R₄.
   1. $\mathbf{Constants}\mathbf{:}$
      $\mathrm{A} = \frac{{\mathrm{V}}_{\mathrm{REF}}}{{\mathrm{V}}_{\mathrm{L}}}-1$
      $\mathrm{B} = {\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{H}}$
      $\mathrm{C} = \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_2}$
      $\mathrm{D} = {\mathrm{V}}_{\mathrm{PU}}-{\mathrm{V}}_{\mathrm{H}}$
      $\mathbf{Simplified}\mathbf{ }\mathbf{Quadratic}\mathbf{ }\mathbf{for}\mathbf{ }{\mathbf{R}}_{\mathbf{4}}\mathbf{:}$
      $\left( \frac{\mathrm{B}}{\mathrm{A}}-\mathrm{C}\times {\mathrm{R}}_2\right) \times {{\mathrm{R}}_4}^2+\left[ \frac{\mathrm{B}}{\mathrm{A}}\times \left( {\mathrm{R}}_2+{\mathrm{R}}_3\right) -\mathrm{C}\times {\mathrm{R}}_2\times {\mathrm{R}}_3+\mathrm{D}\times {\mathrm{R}}_2\right] \times {\mathrm{R}}_4 +\left( \frac{\mathrm{B}}{\mathrm{A}}\times {\mathrm{R}}_2\times {\mathrm{R}}_3\right) =0$
   2. If the output stage is push-pull, then make the following modifications to the above equations:
      ${\mathrm{R}}_3 = 0$
      ${\mathrm{V}}_{\mathrm{PU}} = {\mathrm{V}}_{\mathrm{CC}}$
      $\mathrm{D} = {\mathrm{V}}_{\mathrm{CC}}-{\mathrm{V}}_{\mathrm{H}}$
6. Solve the quadratic equation for R₄ and pick the most logical result.
   ${\mathrm{R}}_4 = 808.88\mathrm{k\Omega } \cong 809\mathrm{k\Omega }$
7. Calculate R₁ by substituting the value for the A constant into the equation for R₁ found in step 4.
   ${\mathrm{R}}_1 =\frac{{\mathrm{V}}_{\mathrm{REF}}-{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{L}}\times \left({\displaystyle \frac{{\mathrm{R}}_2+{\mathrm{R}}_4}{{\mathrm{R}}_2{\mathrm{R}}_4}}\right)}= \left(\frac{{\mathrm{V}}_{\mathrm{REF}}}{{\mathrm{V}}_{\mathrm{L}}}-1\right)\times \left(\frac{{\mathrm{R}}_2\times {\mathrm{R}}_4}{{\mathrm{R}}_2+{\mathrm{R}}_4}\right)=\mathrm{A}\times \left(\frac{{\mathrm{R}}_2\times {\mathrm{R}}_4}{{\mathrm{R}}_2+{\mathrm{R}}_4}\right)$
   ${\mathrm{R}}_1 = 112.36\mathrm{k\Omega } \cong 112\mathrm{k\Omega }$

DC Transfer Simulation Results

Transient Simulation Results

Design References

See Analog Engineer's Circuit Cookbooks for TI's comprehensive circuit library.

See Comparator with Hysteresis Reference Design TIPD144.

See Circuit SPICE Simulation File SLVMCQ0, Inverting Comparator with Hysteresis Circuit Reference Design.

For more information on many comparator topics including hysteresis, propagation delay and input common mode range please see TI Precision Labs – Op amps.

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