ina219 datasheet

R
2mW
SHUNT
10A
Load
+12V
VCM
GND
V
I
V
IN+
V
IN-
Power Register
I C-/
SMBUS-
Compatible
Interface
2
Current Register
Voltage Register
SDA
SCL
A0
A1
0.1µF10µF
+3.3V to +5V
V (Supply Voltage)
S
´
INA219
SBOS448G –AUGUST 2008–REVISED DECEMBER 2015 www.ti.com
Typical Application (continued)
For this example, the minimum-current LSB is calculated to be 457.78 µA/bit, assuming a maximum expected
current of 15 A using Equation 2. This value is rounded up to 1 mA/bit and is chosen for the current LSB. Setting
the current LSB to this value allows for sufficient precision while serving to simplify the math as well. Using
Equation 1 results in a calibration value of 20480 (5000h). This value is then programmed into the Calibration
register.
Figure 29. Example Circuit Configuration
The bus voltage is internally measured at the IN– pin to calculate the voltage level delivered to the load. The Bus
Voltage register bits are not right-aligned; therefore, they must be shifted right by three bits. Multiply the shifted
contents by the 4-mV LSB to compute the bus voltage measured by the device in volts. The shifted value of the
Bus Voltage register contents is equal to BB3h, the decimal equivalent of 2995. This value of 2995 is multiplied
by the 4-mV LSB, and results in a value of 11.98 V. As shown, the voltage at the IN– pin is 11.98 V. For a 40-
mV, full-scale range, this small difference is not a significant deviation from the 12-V common-mode voltage.
However, at larger full-scale ranges, this deviation can be much larger.
The Current register content is internally calculated using Equation 4, and the result of 10000 (2710h) is
automatically loaded into the register. Current in amperes is equal to 1 mA/bit times 10000, and results in a 10-A
load current.
The Power register content is internally calculated using Equation 5 and the result of 5990 (1766h) is
automatically loaded into the register. Multiplying this result by the Power register LSB 20 × 10
–3
(20 times 1 ×
10
–3
current LSB using Equation 3), results in a power calculation of 5990 × 20 mW/bit, and equals 119.8 W.
This result matches what is expected for this register. A calculation for the power delivered to the load uses
11.98 V (12 VCM – 20-mV shunt drop) multiplied by the load current of 10 A to give a 119.8-W result.
9.2.2.1 Register Results for the Example Circuit
Table 8 shows the register readings for the Calibration example.
Table 8. Register Results
(1)
REGISTER NAME ADDRESS CONTENTS ADJ DEC LSB VALUE
Configuration 00h 019Fh
Shunt 01h 07D0h 2000 10 µV 20 mV
Bus 02h 5D98h 0BB3 2995 4 mV 11.98 V
Calibration 05h 5000h 20480
Current 04h 2710h 10000 1 mA 10.0 A
Power 03h 1766h 5990 20 mW 119.8 W
(1) Conditions: load = 10 A, V
CM
= 12 V, R
SHUNT
= 2 mΩ, V
SHUNT
FSR = 40 mV, and V
BUS
= V
IN-
, BRNG = 0 (VBUS range = 16 V).
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