Design Goals

Design Description

Some applications require a circuit to measure small nominal currents while withstanding a high short-circuit current, such as a circuit breaker. This circuit design document describes an isolated current-sensing circuit that can accurately measure nominal load currents from ±10 mA to ±10 A, while withstanding a short-circuit current up to ±200 A. For the purposes of this circuit, assume the output is used with a 3.3-V single-ended ADC, such as one that can be found integrated into an MSP430. The isolation between the line current being measured and the ADC is achieved using an isolated amplifier (AMC1302). With a 1-mΩ shunt resistor, the expected minimum nominal current produces a ±10-µV signal, a signal which is too small to resolve accurately near a zero voltage input due to the delta-sigma modulator dead zone. To remedy this, the circuit uses a 2-channel operational amplifier (TLV9002) to both amplify the signal by a gain of 5 V/V and set the common-mode voltage to 1 V; this not only brings the minimum nominal current out of the dead zone, but also brings the maximum nominal current up to match the full-scale linear input range of the isolated amplifier. The full-scale linear input range of the isolated amplifier is ±50 mV, with a differential output swing of ±2.05 V, on an output common-mode voltage of 1.44 V and a fixed internal gain of 41 V/V. On the output side of the isolated amplifier, a second 2-channel operational amplifier (TLV9002) is used, where: the first channel is used to set the single-ended common-mode voltage to 1.65 V and the second channel transforms the differential output signal from the isolated amplifier into a single-ended that can be used with a 3.3-V single-ended ADC.

Design Steps

1. Determine the appropriate shunt resistor value based on the maximum nominal current.
   $R_{SHUNT} = \frac{V_{inMax}}{I_{inMax}} = \frac{50mV}{10A} = 5 m\Omega$
2. Since this shunt resistor must be able to withstand a 200-A short-circuit current, further reduce the shunt resistor resistance by a factor of 5 which is compensated for in step 6. Determine the shunt resistor power dissipation during maximum nominal current operation.
   ${Power R}_{SHUNT} = {I_{inMax}}^2 \times R_{SHUNT} = 100 A^2 \times 1 m\Omega = 0.1 W$

   Determine the shunt resistor power dissipation during minimum nominal current operation.

   ${Power R}_{SHUNT} = {I_{inMin}}^2 \times R_{SHUNT} = 0.1 {mA}^2 \times 1 m\mathrm{\Omega } = 0.1 \mu W$
3. Determine the shunt resistor power dissipation during short circuit. Be sure to verify that the selected short-term overload specification (typically 5 × nominal) is capable of withstanding the power dissipated by the short circuit.
   ${Power R}_{SHUNT} = {I_{inShort}}^2 \times R_{SHUNT} = \mathrm{40,000} A^2 \times 1 m\mathrm{\Omega } = 40 W$

   Select a shunt resistor with power dissipation reduced by a factor of 5. So, if the short-term overload requirement is 40 W, shunt P_dissipation = 8 W. For more details, see the Design considerations for isolated current sensing analog design journal.

4. Channel 1 of the TLV9002_IN is used to set the 1-V common-mode voltage of the single-ended output of channel 2 of the TLV9002_IN. The 1-V output of channel 1 is also sent to the positive input of the AMC1302. With a 5-V supply, a simple resistor divider can be used to divide 5 V down to 1 V. Using 4 kΩ for R₁, R₂ can be calculated using the following equation.
   $R_2 = \frac{V_{CM} \times R_1}{V_{DD} - V_{CM}} = \frac{1.00 V \times 4000 \mathrm{\Omega }}{5.00 V - 1.00 V} = 1000 \mathrm{\Omega }$
5. Channel 2 of the TLV9002_IN is used to amplify the voltage from the shunt resistor such that the full-scale input voltage range of the AMC1302 is utilized when measuring the maximum nominal current range. With a shunt resistor of 1 mΩ and a maximum nominal current of ±10 A, the output voltage from the shunt resistor is ±10 mV. Since the maximum input voltage of the AMC1302 is ±50 mV, the output of the shunt resistor must be amplified by 5 V/V. While keeping R3|R4 at 1 kΩ, the resistor value of R5|R6 can be found with the following equation.
   $Gain \left(\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$V$}\right.\right)=\frac{R_{\mathrm{5,6}}}{R_{\mathrm{3,4}}} ;{ R}_{\mathrm{5,6}}=Gain \left(\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$V$}\right.\right) \times R_{\mathrm{3,4}}=5 \raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$V$}\right. \times 1 k\mathrm{\Omega }=5 k\mathrm{\Omega }$
6. Verify the absolute maximum voltage limits of the input of the AMC1302 are not violated during a short-circuit based on the chosen shunt resistor. A 200-A short-circuit current results in a 1-V differential voltage being applied to the AMC1302. Since the input common mode is set to 1 V, a maximum of 2 V is applied to the negative input of the AMC1302 with respect to GND1.
   $V_{inAMC} = 200 A \times 0.001 \mathrm{\Omega }\mathrm{ }\times \mathrm{ }5\mathrm{ }\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$V$}\right. = 1 V$

   The absolute maximum input voltage for the AMC1302 is 500 mV greater than the high-side supply voltage (as stated in the AMC1302 Precision, ±50-mV Input, Reinforced Isolated Amplifier data sheet). With a 5-V high-side supply voltage, the absolute maximum input voltage rating is not violated.

7. Channel 1 of the TLV9002_OUT is used to set the 1.65-V common-mode voltage of the single-ended output of channel 2 of TLV9002_OUT. With a 3.3-V supply, a simple resistor divider can be used to divide 3.3 V down to 1.65 V. Using 1 kΩ for R₇, R₈ can be calculated using the following equation.
   $R_8 = \frac{V_{CM} \times R_7}{V_{DD} - V_{CM}} = \frac{1.65 V \times 1000 \mathrm{\Omega }}{3.3 V - 1.65 V} = 1000 \mathrm{\Omega }$
8. While the TLV9002 is a rail-to-rail operational amplifier, the output of a TLV9002 can only swing a maximum of 55 mV from the supply rails. Due to this, the single-ended output of TLV9002_OUT can swing from 55 mV to 3.245 V (3.19 V_pk-pk).
9. The V_OUTP and V_OUTN outputs of the AMC1302 are 2.05 V_pk-pk, 180 degrees out of phase, and have a common-mode voltage of 1.44 V. Therefore, the differential output is ±2.05 V or 4.1 V_pk-pk.

   To stay within the output limitations of TLV9002_OUT, the output of the AMC1302 needs to be attenuated by a factor of 3.2 / 4.1. When R₉ = R₁₀ and R₁₁ = R₁₂, the following transfer function for the differential to single-ended stage can be used to calculate R₁₁ and R₁₂.

   ${\mathrm{V}}_{\mathrm{O}\mathrm{U}\mathrm{T}}=\mathrm{ }\left({\mathrm{V}}_{OUTP}-{\mathrm{V}}_{OUTN}\right)\times \left(\frac{{\mathrm{R}}_{\mathrm{11,12}}}{{\mathrm{R}}_{\mathrm{9,10}}}\right)+{\mathrm{V}}_{\mathrm{C}\mathrm{M}}\mathrm{ }$
10. Using the previously-calculated output swing of the TLV9002_OUT and setting R₉ and R₁₀ to be 10 kΩ, R₁₁ and R₁₂ can be calculated to be 7.8 kΩ using the following equation.
    $3.2=\mathrm{ }\left(2.465\mathrm{V}-415\mathrm{m}\mathrm{V}\right)\times \left(\frac{{\mathrm{R}}_{\mathrm{11,12}}}{10k\mathrm{\Omega }}\right)+1.65$

    Using standard 0.1% resistor values, a 7.8 kΩ resistor can be used. This provides the maximum output swing within the limitations of the TLV9002.

11. Capacitors C₁ and C₂ are placed in parallel to resistors R₁₁ and R₁₂ to limit high-frequency signals. When R₁₁ = R₁₂ and C₁ = C₂ , the cutoff frequency can be calculated using the following equation.
    ${\mathrm{f}}_{\mathrm{c}}=\mathrm{ }\frac{1}{2\times \mathrm{\pi }\times {\mathrm{R}}_{\mathrm{11,12}}\times {\mathrm{C}}_{\mathrm{1,2}}}$

    When the C₁ = C₂ = 1 nF and R₁₁ = R₁₂ = 7800 Ω, the cutoff frequency can be calculated to be 20.414 kHz.

    ${\mathrm{f}}_{\mathrm{c}}=\mathrm{ }\frac{1}{2\times \mathrm{\pi }\times 7800\mathrm{\Omega }\times 1\mathrm{n}\mathrm{F}}=20.414\mathrm{k}\mathrm{H}\mathrm{z}$

Design Simulations

DC Simulation Results

Figure 24 shows the simulated DC characteristics of the voltage across the shunt, the differential input/output of the AMC1302, and the single-ended output of the TLV9002 amplifier from –10 A to 10 A.

Simulation Results

Short-Circuit Event Simulation shows a simulation of the circuit during a short-circuit event by demonstrating how the inputs and outputs react at ±200 A. The red and blue lines going through the graphs mark the points where the output of the AMC1302 begins clipping. From that point on, the purpose of the circuit is to continue operating after the short-circuit event. In the section, the values of the gain and shunt-resistance on the high-side of the AMC1302 were chosen to avoid damage during this event. The following simulation validates those choices: the maximum input voltage entering the AMC1302 at the short-circuit event is ±1 V, which is lower than the absolute maximum ratings of the part. Therefore, the simulation confirms that the circuit continues operation after the short-circuit event passes.

Short-Circuit Event Simulation

Sine-Wave Simulation Results

Sine-Wave Simulation shows the output of the shunt, the differential input and output of the AMC1302, and the single-ended output of the TLV9002 in response to a sine wave with an amplitude from –10 A to 10 A. The differential output of the AMC1302 is ±2.05 V_pk-pk as expected, and the single-ended output is 3.19 V_pk-pk and swings from 55 mV to 3.245 V.

Sine-Wave Simulation

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