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TPS2394, Timing

Vitali Lach
Expert 1130 points
Other Parts Discussed in Thread: TPS2394, LM5067, LM5068

Hello,

 

I was testing TPS2394 on evaluation board und have several questions about that hot swap controller.

Conditions: Vin=-24V, Cflt=4.7µF, Cramp=1µF, Rsense=4mOhm, Cload=47000µF, Rload=25Ohm, di/dt=25A/s

 

1.Equation 4 (page 12 in datasheet) VLSS is 134V in this case.

tss=1260 x 1µF = 1260ms (Equation 3)

Vlss=(3,18µA/(2 x 0,047 x 0,000001 x 100 x 0,004)) x 1,260s²=134V (equation 4)

That means that the load can be expected to charge completely during soft start event. If I take a shot from my scope, I see that the voltage just reaches 14V after soft start period. Please see attached picture: CH1-Vin, CH2-Vout, CH4: I-Inrush power supply (-24V)

 

2.I want to reduce the start up time to about 400ms (equation 2 and 3).

In this case I need di/dt about 80A/s, I don't? Is there any another way to reduce the start up time? With di/dt 80A/s I probably would destroy my MOSFET transistor.

 

Best Regards

 

Vitali

  • 0
    Expert 1130 points

    Hello,

    It's related to question nr. 2:

    I changed now Cramp to 330nF and got very high top current, please see attached picture

    That current would destroy my MOSFET transistor, in this case it limited to 10A by power supply.

    Best Regards

  • 0 in reply to Vitali Lach
    TI__Mastermind 42795 points

    From your second plot, you have the information needed to evaluate the stress on your FET. Use Vds, Id, and turn on time along with the FET SOA curves to help determine if the FET is robust enough. Maybe use the digital scope to compute a RMS for Vds and Id and then use these as voltages and currents in the SOA curves. The main idea is to ensure that the junction temperature of the FET does not exceed the ABS max during startup (with design margin).

    Use the PG pin of the device to keep the dc load off during start up too (so that all the current is used only for charging the output cap).

  • 0 in reply to Eric Wright
    Expert 1130 points

    Hello,

    thank you for reply.

    So I can assume that the figure of the current ramp is Ok, I can't?

    I still cannot understand the calculation issue, equation 4 in datasheet. Do you have any idea?

    Best Regards

    Vitali

  • 0 in reply to Vitali Lach
    TI__Intellectual 1005 points

    Vitali,

    Based on requirments you are trying to charge a very large output capacitor. Remember that the energy dissipated by a linear hotswap at turn on equals the final energy stored in COUT.

    In your case E = 0.5*47mF * 24V^2 = 13.5J.

    You can see that you'll need to be very carefull about MOSFET SOA. TPS2394 is a lower cost hotswap, which is good for lower power applications.

    I would recommend that you switch to LM5067, which is a more sophisticated part with power limiting. They have an excel calculator to help you with your design. The part costs more, but it will save you a lot of headache.

    Good luck,

    Artem  

  • 0 in reply to Artem Rogachev
    Expert 1130 points

    Hello Artem,

    thank you for reply.

    This this actually our issue: we have to replace LM5068 in our design with TPS2394 (LM5068 is similar to LM5067) due to very long insertion time. That delay is the result of appropriate timer capacitor adjusted for 47000µF load.The device should act like a switch without long delays. I'm worried about that current peak of current ramp and wanted to know whether it's the correct function of TPS2394. I'm used to see another current ramp without peaks from other hot swap controllers.

    Best Regards

    Vitali

  • 0 in reply to Vitali Lach
    TI__Intellectual 1005 points

    2703.CAP_BANK_Seperated.pdfVitali,

    Appologies for the slow reply.  The LM5067 calculator is giving you a long insertion time b/c you are trying to charge so much capacitance. Regardless of the hotswap that you use, you will be limited by the MOSFET SOA.  

    Could I ask, why you need so much output capacitance? Are you trying to meet a hold over requirment?

    One potential way to achieve a fast start time and to have a large holt time is to have a 100uF cap that is charged quickly and then to have the remaining capacitance in a capacitor bank that can be charged slowly through a resistor. Just make sure that your resistor and diode can handle the power levels. A quick diagram is attached.

    Regards,

    Artem

     

     

     

     

     

  • 0 in reply to Artem Rogachev
    Expert 1130 points

    Hello Artem,

    thank you for your reply.

    I agree, the long insertion time based on very high capacitive load 47000µF.

    That capacitor we need for current buffering for a galvo motor. Galvo motor is very similar to a loudspeaker, it has a movement angle of only 15°, it moves a laser morror. If that mirror will be moved very fast we have current spikes upt to 30A. My requirement is to make power supply up in the time below 1 second, so I have to charge 47000µF quickly. Thank you for you suggestion: I think we would have some power loss through series resistor and shottky diode.

    My question is, is the current behavior that I measered in previous reply at TPS2394 correct?

    Have a nive weekend.

    Vitali