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TPS2596: About current limit and thermal shutdown..

Ernest Cho
Expert 3171 points
Part Number: TPS2596

Tool/software:

Hello team, 

I'd like to know if the e-fuse output exceed current limit which I set with external resistor, is it always happen thermal shutdown?

For example, TPS2596 has ~100m-ohm Rds(on), I set up the 1A limit, and R_theta_ja has ~50'C/W.

I can assume at most +5'C ambient temperature increase. 

But what if load is try to consume 1.5A or 2A then TPS2596 will try down the 1A limit, what will it be happen?

Can we still calculate it with Rds(on) in the datasheet or is the MOSFET inside of TPS2596 work in the Ohmic region?

Can we calculate when will it be turned to THD(Thermal shutdown) ? (time to shutdown)

Thank you.
Best Regards,
Ernest

  • 0
    TI__Mastermind 38785 points

    Hi Ernest,

    If the device is limiting the current, we can simply find the power dissipation at the device by using (Vin-Vout)*ILIM and then use the time to TSD graph to estimate the time. 

    Best Regards,
    Arush

  • 0 in reply to Arush Gupta
    TI__Expert 3171 points

    Hi Arush,

    I got it. You mean we can utilize the "Thermal Shutdown Limit Plot(graph)" introduced in the Tdvdt paragraph in the datasheet. 

    But how can we know Vout without measuring it?
    With my case, typical application is 12Vin and 1A limit case(same case with me) and there's graph for over-current. 
    Is there a calculation to expect Vout during current limit?

    Bests, 
    Ernest

  • 0 in reply to Ernest Cho
    TI__Mastermind 38785 points

    Hi Ernest,

    When you have the overcurrent event, you will know the load value. You can find the Vout using ILIM*R_load. Based on this you can estimate.

    Fig 26,27 are the time to TSD plot.

    Best Regards,
    Arush

  • 0 in reply to Arush Gupta
    TI__Expert 3171 points

    Hi Arush,

    I would like to confirm if my understanding is correct.

    Arush Gupta said:
    (Vin-Vout)*ILIM
    Arush Gupta said:
    You can find the Vout using ILIM*R_load.

    Based on the power dissipation is calculated with above calculation. 
    But Vout is vary with load resistance. 
    If Vin =12V, I_lim=1A, Rds(on)=100m-ohm, Rds(on) is unchanged until I limit is less than 1A.

    Once Iout became 1.2A, means R_load = 10-ohm and e-fuse MOSFET became ohmic region to cover 1A, means MOSFET act as 2-ohm and Vout becomes 10V. 
    If Iout became 2A, R_load = 6-ohm, to flow 1A out, Vout need to be 6V, MOSFET act as 6-ohm.

    Is my understanding correct?

    Bests, 
    Ernest

  • +1 in reply to Ernest Cho
    TI__Mastermind 38785 points

    Hi Ernest,

    Device during normal operation remains in ohmic region and during current limiting mode, remains in saturation region. 

    Yes, you are correct when saying that FET will change the effective resistance based on the external scenarios. 

    Best Regards,
    Arush