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Current Shunt Resistor

Calculate shunt resistor voltage drop, amplifier output, power dissipation, and ADC resolution for current sensing.

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Formula

V_sh = I × R_sh, P = I² × R_sh

R_sh— Shunt resistance (Ω)
I— Measured current (A)

How It Works

A current shunt is a low-resistance precision resistor placed in series with a current path to produce a measurable voltage drop proportional to current: V_sh = I × R_sh. This small voltage (typically 1–100 mV at full scale) is then amplified by a differential amplifier or dedicated current-sense amplifier (CSA) IC for ADC sampling. The key trade-offs are power dissipation (P = I² × R_sh, which must not overheat the shunt or affect the circuit efficiency), amplifier gain (higher gain allows smaller R_sh but increases noise), and ADC resolution. For a 12-bit ADC with full-scale V_ref, the current resolution is ΔI = (V_ref/4096)/(G × R_sh). Common shunt values are 1–100 mΩ for high-current applications (>1 A) and 1–10 Ω for low-current (<100 mA) measurement. Dedicated ICs (INA219, INA240, MAX9934) integrate the amplifier and often include a sigma-delta ADC for direct digital current readout over I²C.

Worked Example

Problem
Measure 0–20 A using a 5 mΩ shunt with an INA240 (gain = 20 V/V) and 3.3 V, 12-bit ADC. Find shunt voltage, amplifier output at full scale, and resolution.
Solution
1. Shunt resistance: R_sh = 5 mΩ = 0.005 Ω 2. Full-scale shunt voltage: V_sh = 20 A × 0.005 Ω = 100 mV 3. Amplified output: V_amp = 100 mV × 20 = 2.0 V (within 3.3 V ADC range ✓) 4. Power dissipation at 20 A: P = 20² × 0.005 = 2 W — use a 3 W rated shunt 5. ADC resolution: ΔI = (3.3/4096)/(20 × 0.005) = 0.806 mV/0.1 V/A = 8.06 mA/LSB Result: The 5 mΩ shunt gives 100 mV full-scale, 2.0 V amplified output, and 8 mA per ADC step.

Practical Tips

  • ✓Use a dedicated current-sense amplifier IC rather than a discrete instrumentation amplifier — INA240, INA219, and MAX9934 are designed for bidirectional current sensing with integrated EMI filters.
  • ✓For battery management, use low-side sensing (shunt between load ground and system ground) to avoid common-mode voltage issues when the supply rail varies.
  • ✓Add a small RC filter at the amplifier input (e.g., 10 Ω + 100 nF differential) to suppress high-frequency switching noise from PWM motor drivers.

Common Mistakes

  • ✗Placing the shunt on the high side and using a single-supply ground-referenced amplifier — high-side sensing requires a rail-to-rail or high-voltage difference amplifier; ground-referenced amplifiers only work for low-side shunts.
  • ✗Ignoring Kelvin connections — ordinary PCB traces to the shunt add series resistance that appears as measurement error; use a 4-terminal (Kelvin) shunt resistor and route voltage-sense traces from the shunt terminals directly.
  • ✗Undersizing shunt wattage — at high currents the I² term dominates; a 10 mΩ shunt at 10 A dissipates 1 W and will drift significantly if only a 0.1 W resistor is used.

Frequently Asked Questions

Low-side sensing places the shunt between the load and ground — simple single-supply amplifiers work, but ground disturbances affect the measurement and the load ground floats. High-side sensing places the shunt between the supply and the load — provides a true common-ground measurement and detects leakage currents, but requires a high-common-mode-voltage amplifier (e.g., INA240 rated to 80 V).
Choose R_sh to produce 50–100 mV full-scale voltage (small enough to minimise power loss, large enough for good SNR). Then select the amplifier gain so the amplified voltage fills the ADC input range. Verify power dissipation P = I²R_sh is within the shunt's power rating with a 2× safety margin.
Only if it has a low temperature coefficient (TCR < 50 ppm/°C) and is rated for the required power. Standard 1% metal-film resistors have TCR ~100 ppm/°C; precision shunt resistors (e.g., Vishay WSL or BVS series) have TCR < 50 ppm/°C and four-terminal Kelvin connections for accurate sensing.

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