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DRV8303 Measuring Current with Sense Amps

Hello,

Using the DRV8303, we are trying to measure the current across a low-side shunt at 1 mOhm.  The connection is the same as RS1 on page 21.  We have the REF pin at 3.3 Volts.  In all cases I am measuring current with the sense amp at 10 V/V Gain.  I have two questions related to measuring the current in this way.

1.  Equation (1) on page 12 does not include the offset voltage that you measure when you use the DC_CAL functionality.  If I set EN_GATE low, I find the voltage is 0.3 Volts (0.306 Volts as an example reading).  If I keep EN_GATE high, the voltage coming out of SO1 is close to 1.65 Volts (1.651209 Volts as an example reading).  Which value should be used, and how should it be used in the equation?

2.  Disregarding the offset, I use Equation (1) but the readings don't appear to be correct.  I have a fixed load of 14.52 Ohms across two phases, and I'm sending PWM through the phases at 50%.  The multimeter measures a voltage of 13.8 Volts, so I would expect to see a measurement that yields about 1 A with Equation (1).  The voltage coming out of the sense amp in this case is 1.632 Volts, so the current is calculated as 181 mA.  What am I doing wrong in this case?

  • Hi Jared,

    The amplifier is disabled when EN_GATE = LOW, so you can ignore that case.

    When EN_GATE = HIGH, the SO output should be VREF/2 (1.65 if VREF = 3.3V). You can use DC_CAL to short the amplifier outputs to determine how far the amplifier output is from the true VREF/2. This is the output offset and should be removed/add from your real measurements.

    Can you send just a fixed voltage and measure the current? Can you also get a scope capture of the input vs. output. PWMing with a resistor load will result in a square wave current. This is typically smoothed by the inductance of motor.

    Also, a 1mOhm shunt resistor is quite small to try and measure 1A, you signal will almost be in the noise floor.

    By my calculations,

    1.651209 - 1.632 = 0.019209 V

    Then divide by gain of 10 V/V to get the input differential voltage = 1.9209 mV

    Dividing this by 0.001 Ohm gives 1.9209 A