• State
  • Locked Locked
  • Replies 1 reply
  • Subscribers 63 subscribers
  • Views 169 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

bq24650 lead-acid battery

Gamaliel De La Cruz
Prodigy 10 points
Other Parts Discussed in Thread: BQ24650

hi, i want to design a circuit with bq24650 and a lead-acid battery, but i saw the design example in this application note http://www.ti.com/lit/pdf/SLVA437, i checked this, and i not understand this.
if the result of RT is 887 k in the desig you put 947k and if R23 the result is 100k you have 10k and you put the 25k in this resistor?

  • 0
    TI__Guru**** 167196 points

    The 947k ohm computed value was lowered to the closest 1% standard value resistor. Otherwise I would have had to use multiple resistors in series to get the exact value.  Figure 3 shows the result of using the lowered value.

    Resistor R23 must be large enough to protect the FET but not too large that it lowers the FET's gate voltage below at least 2 V above its 2.5-V maximum threshold voltage.  The equation computes the minimum value of 10kohm.  I did not compute the upper end limit for R23 but it could be anywhere in between.