According Ohm's law, obviosly. 0.1 Ohm resistor under 1A current will give 0.1V voltage drop. Multiplying it by operation amplifier with fixed ratio 10:1 will give 1V to ADC. If adc accuracy is 10 bits, code value will be 2^10*1V/5V code.
In order to minimize power rating in the shunt and voltage drop, thus 75mV shunts are commonly used at rated current. 100mV is also common.
Common mode noise by careful design of traces should be reduced with shield tracks so that differential gain does not amplify common mode noise.
CMRR always drops with rising frequency so beware of datasheet specs and add ferrite CM choke if SMPS noise is high.
On the high side Differential Op Amps must have Rail to rail input common mode range.
On the low side, ground reference makes that easy as more Op Amps have CM range to ground or Vee.
My power supply voltage is 0 to 12 Volt and the maximum 3 A current .
if i use 0.1 ohm resistor the maximum voltage drop is 0.1*3= .0.3 volt.
How to scale it without using a op amp ?
0.3/1023 =.293810^-3 millivolt .
can i use as the following example like ?
maximum value 1024* 2.93 =300*10 = 3000 Milliampere . ( just for an example )
Can i make a 0.1 ohm resistor ?
How does calculate the wattage of the shunt resistor in my application ?
10 bit ADC with 5V power supply will give you 5mV accuracy (much less in real because of noise)
0.1 Ohm resistor under the 3A current will drop 0.3V. So, basicaly, you can measure it with ADC. Actual accuracy will be 5mV/0.1 = 50mA, but in reality about 200mA.
Next thing. Under maximum current 3A and voltage drop 0.3V shunt resistor will have to dessipate 0.9W of heat. Are you ready for that? )))
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By the way, you can use ADC with 1.024V reference voltage. Thus, your accuracy will be 1mV and current can be theoreticaly measured +/- 10-20mA if layout will be good enought.