The method suggested in the article is to use external resistor to convert current to voltage. Then amplify the voltage using a programmable gain apmlifier and feeding it to an ADC.
Another way of doing it is to use TIA (Trans-Impedance Amplifier ) which will convert current to voltage and can also introduce programmable gain. In this case, the use of external resistor is avoided but the accuracy of the output result is dependent on the accuracy of the TIA gain. This technique can be used when the number of current sources to be measured is limited to 4, because a max of 4 TIAs can be used in a design.
thank you for your replay.
Do you know in which range with TIA (Trans-Impedance Amplifier ) method is possible to measure the current?
The output Vout of TIA is dependent upon the feedback resistor used in its implementation. The relation is as shown in the equation below:
Vout = Vref - Iin × R
where Vref is the reference input to the TIA, Iin is the current input and R is the feedback resistor in TIA.
The feedback resistor can be configured to any of the following values:
20K, 30K, 40K, 80K, 120K, 250K, 500K and 1000K ohms.
If you consider that the supply as 5V, the maximum current can be measured if R is set to minimum, that is, 20K. Hence, the maximum value of Iin will be 5V / 20K = 0.25mA
If you want to measure higher currents, then you can use external feedback resistor and internal Opamps (4 are available). This is one of the option available.
To understand the implementation of the TIA component and to know the APIs to be used, you can refer to the component datasheet either from Creator or from the following link http://www.cypress.com/?rID=48921