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BQ35100: BQ35100: Current Sense Resistor Selection formula doubt

bitou
Intellectual 390 points
Part Number: BQ35100

Hi,

I think the formula provided in BQ35100 Lithium Primary Battery Fuel Gauge and End-Of-Service Monitor (SLUSCM6E–JUNE 2016–REVISED APRIL 2019), the formula provided to calculate RSENSE is:

RSENSE (mOhms) = VSRMax/PeakLoadCurrent(mA)

I find this weird, the right formula should be U(V)=R(Ohms)*I(A)

So how is it possible to get R(10^-3) = U(10^-3) / I(10^-3) ==> R(10^-3) = U/I which is for me impossible

Considering that it is not mentioned to use mV in the formula, I also tried with R(10^-3) = U / I(10^-3) ==> U = R(10^-3) * I(10^-3)  which is for me impossible as well

So, on our board we have can measure the current through Rsense with a peak of current ~24mA. If we want 24mA to be at full range (125mV)

I caclulate resistor value as R = 0.125/0.024 = 5.2 Ohms and not mOhms


Am I right?

If I am right, in the datasheet, you suggest for 100mOhms for 125mV so Peak Load Current is 0.125/0.1 = 1.25A ?

From these two posts

says 125mV/100mA = 1.25 ohm

e2e.ti.com/.../3228273

And

says 100mV/200mA = 500mOhm for full range

e2e.ti.com/.../3349869

OR

For our 24mA, could we still use 0.1Ohm? So it mean that at our peak current, we are at ~629/32768 for the ADC of BQ35100 (15bits). Is it enough?

What do you suggest?



Thanks for the clarification

  • 0
    TI__Genius 10370 points

    bitou,

    If your max current is only 24mA you will need to increase the senseR size. 15 bits is not enough to accurately get a good measurement. I would not aim to use the full range of the ADC, but more like 70% of the full range. This will allow for some variance. 

    This is an incredibly low value and if you plan to use EOS mode i doubt you will be able to generate enough IR voltage loss on the battery (unless it is very small) to make this work.

    Thanks,

    Eric Vos

  • 0 in reply to Vos
    Intellectual 390 points

    Thanks for your answer

    Our problem is that we can get a current of around 250mA BUT for a short time and we can use BQ35100 because it is too short (less than 32mS). If we want to make it last longer, it would consume too much energy for our battery so we can't use it

    The other option, as you said, is too increase senseR. We have two intensity options, use 24mA or ~40mA (two values we can easily control) but, using 0.5R, we would use only 10-16% of the range and you said that it was not enough.

    To get 70% of the range (87.5mV), using 24mA, we would need a senseR of 87.5/24 = 3.6R. It is possible BUT, when the board use 250mA, the drop on senseR is too big and make the board VCC down quickly.

    To find a compromise, what is the minimum range you would recommend we should use?

    Thanks,

  • 0 in reply to bitou
    TI__Genius 10370 points

    bitou,

    How often does your board wake up (250mA spike)? The longer spikes are only needed when the gauge is gauging the battery. when talking about the smaller ranges 10-15% is possible it would work. My recommendation is to have this be as large as possible. 

    I would recommend you look at extending the 250mA spikes slightly longer when you chose to gauge an active pulse. This will give you the all around the best result. 

    1) Smaller senseR = smaller IR drop

    2) Larger current = more ADC bits

    3) Larger current = Larger battery voltage drop

    All thing that would be better for the gauge. 

    A third alternative would be to put a "learning resistor" to ground that you can use to gauge the battery. This is arguably the best option becuase the current will be very stable. 

    Thanks,
    Eric Vos

  • 0 in reply to Vos
    Intellectual 390 points

    Thanks for your answer

    > A third alternative would be to put a "learning resistor" to ground that you can use to gauge the battery. This is arguably the best option becuase the current will be very stable.

    What do you call a "learning resistor" to ground? Could you detail please? Thanks

    Our resistor is located between the ground of our board and the ground of the battery

  • 0 in reply to bitou
    TI__Mastermind 24870 points

    Hello Bitou,

    Learning resistor is the load you need to put on the battery to force a known current when you are gauging.