I understood the average difference between the input and output voltages is called common voltage .But still i have a doubt when it comes to current sense amplifier. i a m confused like is it the voltage drop across the shunt resistor .which we will feed as input to the current sense amplifier.
i am using 36V 2500 mAh battery pack its maximum voltage is 42 v when the battery is fully charged and its minimum 30V when the battery is fully discharged.
Iam using a 6.6 mohm shunt resistor it will give maximum voltage drop of 100 mv for max 15 Amps of current . is this voltage drop is called as common mode voltage.
The current sense amplifier iam plannin to use is AD8207 whose input supply is 3.3v . so when i search in google about choosing the current sense amplifier i saw that when we use High side bidirectional current sensing amplifier the common mode voltage should be equal to the bus voltage Why. and the AD8207 current sense amplifier whose common mode voltage is equal to my nominal battery to load bus voltage that is 36v but it is not equal to my maximum bus voltage 42 v . so can i use AD8205 as current sensing amplifier in my circuit.
What is bus voltage is it the nominal voltage of Battery pack given to the load or maximum voltage of battery pack given to the load.
So according to my understanding the common mode voltage is the voltage drop between the shunt resistor and it is extremely low compared to the bus voltage and why we need a current sense amplifier whose common mode voltage is equal to the bus voltage .
My battery to Load specification:
Battery pack nominal voltage : 36v
maximum discharge voltage : 42v
minimum discharge voltage : 30v
Nominal current : 2.5 A
maximum current that can be used in the bus (between battery to load): 15 A
shunt resistor value based on 100 mv dropout voltage : 6.6 mohm
What is exactly a common mode voltage? and based on my bus voltage and my common mode voltage.Can i use AD8207 in my application.
based on above battery to load specification can i use AD807 current sensing amplifier with 3.3 v of input voltage and 20 V/V of gain.
Please find the attached datasheet of AD8207 :enter link description here
Best Answer
In your situation it is helped if we put numbers on your diagram: -
The common mode voltage when 10 amps are flowing through the 6.6 milli ohm resistor is: -
$$\dfrac{36 + 35.934}{2} = 35.967 \text{ volts}$$
No, it's the average voltage of both inputs - it's got nothing to do with "average difference" - that term makes no sense.
No, the specified common-mode input voltage should be greater than the maximum possible bus voltage (and preferably, by some margin).
No, it's the voltage that is common to both inputs with respect to ground or 0 volts.