Datasheet

INA219

SBOS448F –AUGUST 2008– REVISED SEPTEMBER 2011

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The Current Register (04h) is then calculated by this result by the power LSB that is 20 times the

multiplying the shunt voltage contents by the 1 × 10

-3

current LSB, or 20 × 10

-3

, results in a power

Calibration Register and then dividing by 4096. For calculation of 5990 × 20mW/bit, which equals

this example, the shunt voltage of 2000 is multiplied 119.8W. This result matches what is expected for this

by the calibration register of 20480 and then divided register. A manual calculation for the power being

by 4096 to yield a Current Register of 2710h. delivered to the load would use 11.98V (12VCM –

20mV shunt drop) multiplied by the load current of

The Power Register (03h) is then be calculated by

10A to give a 119.8W result.

multiplying the Current Register of 10000 by the Bus

Voltage Register of 2995 and then dividing by 5000. Table 3 shows the steps for configuring, measuring,

For this example, the Power Register contents are and calculating the values for current and power for

1766h, or a decimal equivalent of 5990. Multiplying this device.

Table 3. Configure/Measure/Calculate Example

(1)

STEP # REGISTER NAME ADDRESS CONTENTS ADJ DEC LSB VALUE

Step 1 Configuration 00h 019Fh

Step 2 Shunt 01h 07D0h 2000 10µV 20mV

Step 3 Bus 02h 5D98h 0BB3 2995 4mV 11.98V

Step 4 Calibration 05h 5000h 20480

Step 5 Current 04h 2710h 10000 1mA 10.0A

Step 6 Power 03h 1766h 5990 20mW 119.8W

(1) Conditions: load = 10A, V

= 12V, R

SHUNT

= 2mΩ, V

SHUNT

FSR = 40mV, and V

BUS

= 16V.

Product Folder Link(s): INA219