• State
  • Locked Locked
  • Replies 3 replies
  • Subscribers 64 subscribers
  • Views 1209 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

Methods of circuit failure in BQ24450 circuit

liam liam
Prodigy 20 points
Other Parts Discussed in Thread: BQ24450

Hi there

I am using the BQ 24450 to charge a 6V lead acid battery from either a 10.8V OC, 5W solar panel or a 15V power supply. The battery will be connected 24 / 7. My calculations for the circuit at as follows.

Vfloat 6.731466 Volts
Vboost 7.079974 Volts
Vthreshold 5.143104 Volts

When connected to the power supply, the I limit jumper will take the current limit 2.2R path (low charge current). When connected from the solar panel, it will take the 0.33R current limit path.

The three MOSFETs are all P type FDD4243 from fairchild, with plenty of heatsinking for each. http://www.fairchildsemi.com/ds/FD/FDD4243.pdf

My question: what possible failure modes can you think of in the circuit below? (even if unlikely). Your input would be greatly appreciated.....

Thanks

  • 0
    TI__Mastermind 31120 points

    If the design is done properly (components are rated correctly and good theral management), then I do not think there will be a problem.

    1) I would check the temperature of the components in the worst case operating conditions. 

    2) Sometimes circuits have trouble operating well at their boundires so I would see how the unit works with the input voltage down around the battery voltage.

  • 0 in reply to Charles Mauney
    Prodigy 20 points

    Thank you for your reply.

    the MOSFET's are rated to 150 degrees celsius.

    Absolute worst case scenarios:

    From a power supply:

    - Input voltage: 15V, it's a regulated switch mode +/- 1% on the spec but lets say 5% so 15.75V
    - Current limit position will be using the 2.2R  current sense resistor. say it's 5% out, so 2.178R giving approx 115mA charge rate.
    - Battery is at 5.14V, the threshold.
    - Input diode (not shown on schematic) drops the current minimum of 0.3 under load

    ((15.75 - 0.3  - 5.14) = 10.31V drop worst case

    0.115 * 10.31V = 1.18565 watts of heat to dissipate - a relatively small amount.

    From a 10.8V solar panel 5 - 100000 W:

    - Input voltage: 10.8V Open circuit (lets use this figure for worst case)
    - Current limit will be using the 0.33R current sense resistor. say it's 5% out, so 0.3267R giving approx 0.765228 A charge rate ( you wouldn't get this from a 5W panel though)
    - Battery is at 5.14V, the threshold
    - Input diode (not shown on schematic) drops the current minimum of 0.3 under load

    10.8V - 0.3 - 5.14 = 5.36V drop worst case

    0.765228 * 5.36 = 4.101622 W

    I don't think these numbers are that high given that the three mosfets are thermally well connected... I will calculate the temperature rise given the surface area shortly.

    In practice, with no airflow and 10.8V coming in I can't get it hotter than 70 degrees celsius at the hottest point. The BQ24450 has quite a lot of thermal isolation from this (is usually no more than 40 degrees celsius). That gives a lot of headroom for enclosure + ambient temperature.

    Regarding the second point: I have tested the circuit running across the whole voltage range and it appears to perform admirably.

  • 0 in reply to liam liam
    TI__Mastermind 31120 points

    Looks like you did a good job.  Nothing beats good looking waveforms and components that do not get too hot.