In my task needed measure the load current of not more than 2A, the range of 200 mA to 1.5 A. Load power supply voltage: 12 volts.
Voltage drop across the shunt with a maximum load current of 2A: Vsense
, where RS
- resistance of the shunt, ILOAD
- the load current. Vsense
= 0.1 * 2 = 0.2V.
Heat loss in the shunt with a load of 2A amount: PD = I2
R = 22
* 0.1 = 0.4 Watts.
IC output current: Iout(max)
, for ZXCT1010 value Gt
is 0.01. Therefore Iout(max)
= 0.01 * 0.2 = 0.002 A.
Calculate the resistance at maximum load current of 2A. RG = Vout
= 5 V / 0.002 A = 2500 kOhm. The closest value of the resistor: 2.4K. For given value the output voltage of the IC is: Vout
= 2400 * 0.002 = 4.8 V.
If the resistance RG
= 2.4K and the minimum current of load will be 200 mA, the voltage from the IC is: Vout
) = 2400 * (0.01 * 0.1 *0.2) = 0.48 V. Thus with a load current of 200 mA, the voltage supplied to the ADC is 0.48V, when the load current of 2 A, respectively, 4.8V.
To measure the voltage used voltage divider. To get the output voltage of 4V with 12V input, we use the calculator voltage divider
, divider values to be 1K and 500 Ohms.