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What does the unit nVh stand for?

Kenneth Minicuci
Prodigy 20 points
Other Parts Discussed in Thread: BQ3055

Hello all, I am using the bq3055 Li-Ion battery pack manager for a school project and one of its features is continuously monitoring current so it can use the method of coulomb counting to monitor the capacity of the battery. As I was reading the datasheet for the bq3055 to try and figure out what exactly it meant by "continuously" monitoring the current I found on page 17 in the 'Charge and Discharge Counting' section it states "The bq3055 continuously integrates the signal over time, using an internal counter. The fundamental rate of the counter is 0.65 nVh."

As an EE student, I have never seen the unit nVh before. Can someone please explain to me what it means when it sates that the rate of the counter is 0.65 nVh?

Thank you very much in advance!

  • 0
    TI__Guru 73830 points

    Kenneth,

    nVh means nano-volt hours

    It refers to the resolution of the integrating counter in the coulomb counter interface.

    You can search the TI Website and find answers to many items such as this. I answered this one in an earlier post.

    Regards

    Tom

  • 0 in reply to ThomasCosby
    Prodigy 20 points

    Thank you Tom,

    I'm still a little unsure as to how it is used. This is my best guess:

    If my device constantly draws 2 Amps across a current shunt of 10 mOhms then the voltage across the shunt would be 20mV or 0.02 Volts. If the counter rate is 0.65 nVh then 0.02 Volts / (0.65x10^-9) Vh would give me about 30769230.77 with units of 1/hr. Multiplying this by 1 hr / 3600 sec will yield 8547 (1 / sec) or Hz.

    Does this mean that the bq3055 will measure that shunt voltage 8,547 times per second or am I taking that nVh number to mean something completely different. 

    My main goal was to figure out what the datasheet means by "continuously" monitoring the current for the coulomb counting method.

    Thanks for the help! And remember I'm just a student so take it easy on me if I'm completely wrong :)

    Regards,

    Kenneth