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bq24400 example circuit

Laura 56109
Prodigy 170 points
Other Parts Discussed in Thread: BQ24400

I intend to use the example circuit in  the datasheet (Figure 8) of the bq24400. My question is, what should i change in the circuit to use ist with higher voltage, like 40 V. In the example circuit are 16V allowed at the DC+. Some application notes about this circuit would also be very helpfull.

Thanks

  • 0
    TI__Mastermind 31120 points

    The IC is a 7V process so the input votlage needs to be reduced to 5V

    The power stage components need to be rated for greater than 40V so 60V would probably be good.


    replace all components that have the high voltage on them with 60V components.

  • 0 in reply to Charles Mauney
    Prodigy 170 points

    Thanks. I would also like to understand the circuit. What function exactly have the transistors Q1, Q2 and Q3 ?

  • 0 in reply to Laura 56109
    TI__Mastermind 31120 points

    Q1 is the switching FET that chopps the input DC into a square wave where the output filter, LC, can average or filter the waveform.

    Q2, D5, Cp and R12 is a turn off circuit where when the MOD pin goes low, Q2 is pulsed on and drives Q1 off.

    Q3 and R11 create a current source to drive Q1 and also "translates" the voltage at the switching transistor (input voltage) down to the 5V level where the IC is biased.

     

    these 3 transistor need to be rated for at least 40V.

  • 0 in reply to Charles Mauney
    Prodigy 170 points

    Hi,

     

    I changed the transistors and tried a 25V Voltage. Something burned then in the circuit. I tried with a new board to increase the voltage and the current was also increasing.  I thought that the IC is keeping the current somehow in some limits through switching. Now i am not sure. Could it be that the current is increasing with the input voltage and the IC is not able to regulate it?

    Thanks again,

    Laura

  • 0 in reply to Laura 56109
    TI__Mastermind 31120 points

    this is a hysteretic switcher, which means the top FET stays on until the current sense threshold is reached (> |+/- 50mV|).

    Use a scope and look at the switching waveform (phase node).  This should be switching between Vcc and ground, unless the current is not high enough and then it probably should not be burning up.  IS the drive to the FET enough to fully turn on the FET?

    Send some waveforms of the IC if you need further help.