Design Goals

Design Description

This bidirectional current sensing solution uses a current-sense amplifier and a high speed dual comparator with a rail-to-rail input common mode range to create over-current (OC) alert signals at the comparator outputs (OUTA and OUTB) if the input current (IG1) rises above 100A or falls below -35A. In this implementation, both over-current alert signals are active high, so when the 100A or -35A thresholds are crossed, the comparator outputs goes high. External hysteresis is implemented on both comparators so that the comparator outputs return to logic low states when the current reduces by 10% (90A and -31A). While the circuit below has shunt resistor R8 connected to ground, the same circuit is applicable for high side current sensing up to the common mode voltage range of the INA.

Design Notes

1. Select a comparator with rail-to-rail input common mode range.
2. Select a current sense amplifier with low offset voltage and a common mode input range that matches the requirements of the system.

Design Steps

1. To determine the comparator threshold voltages, first calculate the INA240A1 output voltages that correspond to the desired current thresholds. The calculations depend on the gain of the INA240 (20, 50, 100, 200 for A1, A2, A3, A4, respectively), the input current (IG1) and sense resistor (R8), and the reference voltage when the input current is 0 (VREF). Per section 8.3.2 in the INA240 data sheet, R8 is a function of the differential input voltage and the maximum input current to the INA240. Given that the input current in this system swings above 100A, by keeping R8 small, the power dissipation across R8 is lessened.
   $\mathrm{INA\_OUT}=\mathrm{VREF}+\mathrm{G}\times \left(\mathrm{INP}-\mathrm{INN}\right)$
   $\mathrm{INP}-\mathrm{INN}=\mathrm{IG}1\times \mathrm{R}8$
   $\mathrm{VREF}=\frac{\left(\mathrm{V}+\right)-0}{2}=\frac{3.3\mathrm{V}}{2}=1.65\mathrm{V}$

   
   Using these equations and the desired current thresholds, the following table is generated:

   DESCRIPTION		IG1	INA-OUT
   VH, CHB	Overcurrent threshold in forward direction	100A	1.65V + 20 × (100A × 0.33mΩ) = 2.31V
   VL, CHB	Recovery threshold in forward direction	90A	1.65V + 20 × (90A × 0.33mΩ) = 2.244V
   VH, CHA	Overcurrent threshold in reverse direction	-35A	1.65V + 20 × (-35A × 0.33mΩ) = 1.419V
   VL, CHA	Recovery threshold in reverse direction	-31.5A	1.65V + 20 × (-31.5A × 0.33mΩ) = 1.4421V

   First, focus on the top comparator (channel A), which is in an inverting comparator configuration. This comparator swings to a logic high when the current in the reverse direction exceeds -35A, and returns to a logic low when the current in the reverse direction recovers to -31.5A. These current levels correspond to voltage levels of 1.419V and 1.4421V, respectively.

2. Assume a value for R2 (the bottom resistor in the resistor divider). In this circuit, 10kΩ is chosen.
3. Derive two equations for R1 in terms of V+, V_L, V_H, R₂, R₃ by analyzing the circuit when INNA = V­_L and when INNA = V_H:
   ${\mathrm{R}}_1=\left(\frac{{\mathrm{V}}_{+}}{{\mathrm{V}}_{\mathrm{L}}}-1\right)\left(\frac{{\mathrm{R}}_2{\mathrm{R}}_3}{{\mathrm{R}}_2+{\mathrm{R}}_3}\right)$
   ${\mathrm{R}}_1=\frac{{\mathrm{V}}_{+}-{\mathrm{V}}_{\mathrm{H}}}{{\displaystyle \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_2}}-{\displaystyle \frac{{\mathrm{V}}_{+}-{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{R}}_3}}}$
4. Set these two equations equal to each other and then solve for R₃.
   $\left(\frac{{\mathrm{V}}_{+}-{\mathrm{V}}_{\mathrm{H}}}{{\displaystyle \frac{{\mathrm{V}}_{+}}{{\mathrm{V}}_{\mathrm{L}}}}}-{\mathrm{V}}_{\mathrm{H}}\right){{\mathrm{R}}_3}^2+\left(\frac{{\mathrm{V}}_{+}-{\mathrm{V}}_{\mathrm{H}}}{{\displaystyle \frac{{\mathrm{V}}_{+}}{{\mathrm{V}}_{\mathrm{L}}}}}+{\mathrm{V}}_{+}-{\mathrm{V}}_{\mathrm{H}}\right){\mathrm{R}}_2{\mathrm{R}}_3=0$
   $\left(\frac{3.3-1.4421}{{\displaystyle \frac{3.3}{1.419}}}-1.4421\right){{\mathrm{R}}_3}^2+\left(\frac{3.3-1.4421}{{\displaystyle \frac{3.3}{1.419}}}+3.3-1.4421\right)\left(10\mathrm{k}\right){\mathrm{R}}_3=0$
   ${\mathrm{R}}_3=0, {\mathrm{R}}_3=804.29\mathrm{k\Omega }$

   The standard 1% resistor value closest to this is 806kΩ.

5. Solve for R₁ using any of the two equations derived in 3:
   ${\mathrm{R}}_1=\left(\frac{{\mathrm{V}}_{+}}{{\mathrm{V}}_{\mathrm{L}}}-1\right)\left(\frac{{\mathrm{R}}_2{\mathrm{R}}_3}{{\mathrm{R}}_2+{\mathrm{R}}_3}\right)$
   ${\mathrm{R}}_1=\left(\frac{3.3}{1.419}-1\right)\left(\frac{\left(10 \mathrm{k\Omega }\right)\left(806 \mathrm{k\Omega }\right)}{10 \mathrm{k\Omega }+806 \mathrm{k\Omega }}\right)$
   ${\mathrm{R}}_1=13.093\mathrm{k\Omega }$

   The standard 1% resistor value closest to this is 13kΩ.

   The next step is to focus on the bottom comparator (channel B), which is in a non-inverting configuration. This comparator swings to a logic high when the current in the forward direction exceeds 100A, and returns to a logic low when the current in the forward direction recovers to 90A. These current levels correspond to voltage levels of 2.31V and 2.244V, respectively.

   

   High-side current sensing with comparator circuit derives two equations for V_TH (the voltage on the non-inverting pin) when the comparator output is in a logic low state and a high-impedance state (SBOA306 uses an open-drain comparator). These equations are then set equal to each other creating a quadratic equation to solve for R6. Since TLV3202 is a push-pull device, the output goes to a logic high state instead of a high-impedance state. Thus, the pull-up resistor value is 0 and V_PU is V₊

6. Rewrite the quadratic equation to match this circuit:
   
   $0={\mathrm{V}}_{+}\times {{\mathrm{R}}_6}^2+\left({\mathrm{V}}_{+}\times {\mathrm{R}}_7+{\mathrm{V}}_{\mathrm{L}}\times \left({\mathrm{R}}_7\right)-{\mathrm{V}}_{\mathrm{H}}\times {\mathrm{R}}_7\right)\times {\mathrm{R}}_6+\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{H}}\right)\times \left({{\mathrm{R}}_7}^2\right)$
   $0=3.3\times {{\mathrm{R}}_6}^2+\left(3.3\times {\mathrm{R}}_7+2.244\times \left({\mathrm{R}}_7\right)-2.31\times {\mathrm{R}}_7\right)\times {\mathrm{R}}_6+\left(2.244-2.31\right)\times \left({{\mathrm{R}}_7}^2\right)$
7. Choose a value for R₇. This resistor dictates the load current of the comparator, and therefore is large. For this circuit, R₇ is assumed to be 200kΩ.
   $0=3.3\times {{\mathrm{R}}_6}^2+\left(3.3\times 200\mathrm{k}+2.244\times \left(200\mathrm{k}\right)-2.31\times 200\mathrm{k}\right)\times {\mathrm{R}}_6+\left(2.244-2.31\right)\times {\left(200\mathrm{k}\right)}^2$
   ${\mathrm{R}}_6=4.47\mathrm{k\Omega }$

   The standard 1% resistor value closest to this is 4.42kΩ.

8. Calculate V_TH using R₆.
   ${\mathrm{V}}_{\mathrm{TH}}={\mathrm{V}}_{\mathrm{H}}\times \left(\frac{{\mathrm{R}}_7}{{\mathrm{R}}_6+{\mathrm{R}}_7}\right)=2.31\times \frac{200\mathrm{k}}{4.42\mathrm{k}+200\mathrm{k}}=2.26\mathrm{V}$
9. Choose a value for R₅. In this case, R₅ is chosen to be 10kΩ.
   ${\mathrm{V}}_{\mathrm{TH}}={\mathrm{V}}_{\mathrm{H}}\times \left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}\right)=9.802\mathrm{V}$
10. Solve for R₄.
    ${\mathrm{R}}_4=\frac{{\mathrm{R}}_5\times ({\mathrm{V}}_{\mathrm{s}}-{\mathrm{V}}_{\mathrm{TH}})}{{\mathrm{V}}_{\mathrm{TH}}}=\frac{10\mathrm{k}\times (3.3-2.6)}{2.26}= 4.602 \mathrm{k\Omega }$

    The standard 1% resistor value closest to this is 4.64kΩ.

Design Simulations

Transient Simulation Results

The below simulation results use a -70A to 130A, 100Hz sine wave for IG1.

Channel A

Channel B

Design References

Texas Instruments, SBOMB05 SPICE file, circuit software

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