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ISO5852SDWEVM-017: why the source and sink current can reach 20A with 5A buffer transistor?

Howard Zou
Genius 12365 points

Part Number: ISO5852SDWEVM-017

Hi,

In user guide, I see transistor buffer Q1-Q4 are 5A, why the source and sink current can reach 20A?

Is it related to peak pulse current of transistor? But how do we know that the transistor can endure the IGBT turn on or turn off time?

  • 0
    TI__Genius 13445 points
    Hi Howard,

    Thanks for considering our part, My name is Mamadou Diallo and I will help address your concerns.

    Yes the transistors can handle 5A of continuous collector current and 20A of peak pulse current.
    You can estimate the souce/sink current (which determine the turnon and turnoff delays of the IGBT) as follow:
    Isrce = (voltage across the IGBT gate) / (total resistance, on-path)
    Isink = (voltage across the IGBT gate) / (total resistance, off-path)

    where total resistance on/off path is typically: Rtotal = Internal gate resistance of IGBT + Collector resistance in the on/off path +

    Please find attached a reference design that you may find helpful in determining the npn/pnp transistor pair for the external current buffer and source/sink current of the buffer.

    www.ti.com/.../tiduc70a.pdf

    Regards,

    -Mamadou
  • 0 in reply to Mamadou Diallo
    TI__Genius 12365 points
    Mamadou,
    so after calculating Isource & I sink, then we choose the buffer transistor.
    And we should use the "collector current - peak", not "collector current - continuous" to determine if the transistor can be used in the circuit, right?
    As long as "collector current - peak"> both Isource & Isink, it can be used, right?
  • 0 in reply to Howard Zou
    TI__Genius 13445 points
    Howard,

    You're correct. Remember the calculated values are the maximum values that can be sourced/sinked into the IGBT. Reason why you consider the peak collector current.
    In reality though, the actual values will be slightly less than the calculated values due to the effects of turn-on delays and rise/fall times of the BJTs. However, you still want to give the transistors enough margin by selecting higher current capability than the calculated values.

    If this answered your question, please press the green button.

    Regards,

    -Mamadou