After reading several papers regarding the kelvin connection and the difference between 4 wires and 2 wires connection. I end-up by choosing to use a 2 pads current sense resistor for my PCB. The problem I am encountering right now is that the datasheet recommends a shunt resistor of 0.01 ohm, which gives a voltage drop of 5 mV (with 500 mA and 4.2 V).
However, the ADC operates with a range of -125 mV to 125 mV. That's why I opted for a 0.1 ohm resistor, the voltage drop will be 50 mV, which makes the ADC more sensitive to voltage variations. Do you suggest me to go for a 0.1 or 0.01Ohm resistor ?
What current do you want to measure?
1A, 10A, 100A?
If up to 1A, then 0.1 ohm should be fine.
Above 1A you need to take the wattage of the resistor into account (to prevent overheating).
Sounds like an inconveniently small signal to work with. I'd use a larger shunt resistor so you have a bigger signal, assuming that this is feasible in your application.
What kind of circuit would this shunt resistor be part of? Can you upload a schematic?
Probably, but it depends on the context. Please see above.
Thanks so much.
Refer to page 44 of the datasheet you linked to, which includes the following on Rsense:
8.1.2.1.4 STEP 4: Determine the Sense Resistor Value.
To ensure accurate current measurement, the input voltage generated across the current sense resistor should not exceed +/–125 mV. For applications with a very high dynamic range, it is allowable to extend this range to absolute maximum of +/–300 mV for overload conditions where a protector device will be taking independent protective action. In such an overloaded state, current reporting and gauging accuracy will not function correctly.
The value of the current sense resistor should be entered into both CC Gain and CC Delta parameters in the Data Flash Calibration section of the Evaluation Software.
I.e. you can use a larger Rsense if you want, to increase accuracy of the measurement. If you allow the voltage to exceed the 125mV range, accuracy is reduced, but this is not necessarily a problem as long as any overcurrent condition is adequately handled.
There's no pressing need to adhere religiously to the 0.1R value that's shown in the typical application diagrams in the datasheet. You're free to choose within the range suggested. You determine this on the basis of the anticipated maximum (peak) current flow into and out of the battery pack.
Thank you for pointing this out. Actually, when reading the datasheet of a random battery such as this one : Battery datasheet. I read that the allowed battery charging voltage is 4.2V +/-0.05v. On the other hand, the output of my charger is 4.2V and 0.5A. So, if I use a resistor of 0.1 Ohm, it will generate a voltage drop of 0.05V and thus, I will just be on the charging limit of the battery! Don't you think that using a small resistor with an opamp would be a better idea ? it's just a suggestion!
With an empty battery, the charge voltage will need to be significantly lower in order to limit the charge current to a sensible value. Only when the battery is nearly full, the voltage can coast to a stable value of around 4.2V and the current will trail off towards (ultimately) near zero. The voltage drop due to the sense circuit is across a fixed resistor, so it's current-dependent. It will be near zero as the charge current approaches near zero.
The fuel gauge chip you're using has its own amplifier on board. The datasheet does not suggest to apply external amplification and I would most definitely not go there for reasons of complexity, proneness to engineering errors and associated failures and mostly for the total lack of necessity to go there.
The recommended datasheet is 0.01Ohm which generates a 5mv of voltage drop for a 14bits ADC that ranges from -125mV to 125mV. Do you think that 5mV is enough to sense the small current variations?
Depends on your resolution requirements.
As said before, you can adjust the sense resistor if you want to increase current readout resolution - at the cost of a small voltage drop. I don't think there will be a problem with charging the cells if you limit the Rsense voltage drop to let's say 100mV or so.
So, If I use like a 0.1 Ohm resistor, with 4.2V and 0.5A, I will get a 50mv of voltage drop, right ? Then to calculate the resistor power rating needed, it will be 0.25*0.1=0.025W, do you suggest me to go for a resistor with 250mW or 500mW of power rating? the already found packages for 100 mW are 0603 which are very small to handle the kelvin connection ! like 250mW won't be an overkill for my circuit ? or neither 500mw?
No, a higher rating is not a problem. In fact, a somewhat higher rating is nice because it'll improve heat dissipation in the resistor, so it doesn't warm up, which in turn prevents it from shifting in value. 250mW or 500mW would either be fine.
Even for the discharging process? because I am using the same shunt resistor for calculating the discharging current. The discharging is made via three resistor (only one at time), 47 Ohms,100 Ohms and 2K Ohms. I don't know if my calculations are correct but here's how I proceed :
P=R^I*I
I=sqrt(1w/47 Ohm)=0.14A
I=sqrt(1w/100 Ohm)=0.1A
I=sqrt(1w/2k Ohm)=0.02A
Does this mean that the maximum current that will flow from the battery in fast discharging is in the first case with 0.14A without warming the resistors ? it won't be bigger than that ?
That doesn't make sense. A 1W resistor will not magically dissipate 1W always regardless of what happens around it. Ohm's law still applies!
So you've got 3 discharge resistors of 47R, 100R and 2k2. If you connect them in parallel (all 4 MOSFETs conducting), this makes for a virtual single resistor of approx. 31.5R With a fully charged battery with a voltage of 4.2V this gives a discharge current of 133mA.
133mA is smaller than the 500mA charge current you based your shunt resistor calculation on. So you should be safe even if you don't do any further math. If you want to be sure, then do the math anyway: P = I^2 * R = 0.133^2 * 0.1 = ca. 1.75mW Yep, you're still safe! But you already knew that, because you were OK with 500mA and 133mA is less than that, and dissipated power reduces at a power of two.
