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TPS54260 "Dangerous" Current

Antonio Stenio
Prodigy 70 points
Other Parts Discussed in Thread: TPS54260

What does it mean the controllable time of 135ns for the IC TPS54260?

If this "blanking time" means that NOTHING protects the IC, even the internal switch current limiting, considering that the internal FET has only 0.2ohm resistance, a quite dangerous peak current can flow through the IC when I connect the power supply. In my application, I am applying 21V at the input, have Cout of 0.01ohm ESD and inductor of 0.05ohm resitance, which could give me approx. 70A current for 135s, possibly destructive to the IC.

Best Regards,

Antonio Stenio

 

  • 0
    TI__Guru**** 191445 points

    The current limit is not active during the blanking time.  But there are still physical limits. The change rate of the switch current is limited by the inductor.  During the blanking time, the current can change by di = ((Vout-Vin) x 135 nsec) / L.  So for example Vin = 21 V, Vout = 5 V and L = 10 uH, at start up Vout = 0 V and I out = 0 A, so at the first initial start up pulse the switch current can only reach 284 mA during the blanking time.

  • 0 in reply to JohnTucker
    Prodigy 70 points

    Thank you John.

    Then, could you please give an example where I can have a peak current equal or bigger than my IC current limit of 6.1A?

    In my application, I am using an inductor L=27µH with 3.5A(max.), and If I have any situation where the current jumps to this limit of 6.1A or more, my inductor will saturate and loose inductance, which makes the current even bigger and possibly causing damaging to IC. Right?

    My point is that I found a 10A output peak current that destroys my IC during start-up, but I don't really understand where it comes from.

    Best Regards. 

  • 0 in reply to Antonio Stenio
    TI__Guru**** 191445 points

    If you are concerned about fault conditions, then you will probably want to size your inductor saturation current rating appropriately.  Even under fault conditions, the TPS54260 will source the maximum current but at very low (near 0 V) output voltage.  this protects the IC, but the inductor can saturate.

    The blanking time is required because the IC does not actually measure "current"  but rather a voltage proportional to that current  There has to be some settling time after teh switch transition to get an accurate reading of the voltage.  there are other safeguards as well including frequency foldback:

    During short-circuit events (particularly with high input voltage applications), the control loop has a finite minimum controllable on time and the output has a low voltage. During the switch on time, the inductor current ramps to the peak current limit because of the high input voltage and minimum on time. During the switch off time, the inductor would normally not have enough off time and output voltage for the inductor to ramp down by the ramp up amount. The frequency shift effectively increases the off time allowing the current to ramp down.

    During steady state operation, the switch current is balanced.  It falls the same amount during the off time that it rises during the on time.  So to get a switch current above the internal current limit, it will have to occur during a fault condition or load step transient.  During a load step,  the on time will increase towards maximum duty cycle to allow the inductor current to "unbalance" and ratchet up to the demanded current (which may be greater than the max current limit).  In that case, when one switching cycle ends with switch current just below the limit, it may exceed the limit on the next cycle before the min on time is completed.  But the on time will be terminated immediately at the end of the blanking time and there will be a long off time to allow the switch current to fall further.


    At any rate, these ICs are rather resistant to brief over current events.  You can usually run them into a dead short without damage.  So I rather doubt that the IC is damaged by a 10 A transient.  More likely, that is causing an over voltage condition somewhere.  The IC is not very tolerant at all of over voltage.


  • 0 in reply to JohnTucker
    Prodigy 70 points

    Dear John,

    Thank you very much for your nice explanation.

    Actually, I am driving a inductive load (motor) with this application, and this maybe the cause for the "load step transient" you mentioned above, right? Also, I noticed that when I plug my circuit to the charger (48V), I see some chattering(many current pulses) coming in.

    Anyway, now that I know this 10A pulse is unlikely to destroy my IC (actually, it is burned!!), I am very interested now in what you call "OVER VOLTAGE" situation. In this case, can we say that some voltage instability from the charger could be the reason for that? And what is the maximum tolerance of this IC against over voltage peak?

    Best Regards,

    Antonio Stenio 

  • 0 in reply to Antonio Stenio
    TI__Guru**** 191445 points

    Driving an inductive load could affect the operation of the circuit.  The inductive load must also follow the laws of physics (inductor current cannot be discontinuous).  For the dc/dc converter that is why there is a catch diode from PH to GND (otherwise PH voltage would go to - infinity when the high side switch is turned off).  So that could be the source of over voltages.

    The absolute maximum voltage ratings for each pin is given in a table in the datasheet.  I would not let any voltage exceed those levels.

  • 0 in reply to JohnTucker
    Prodigy 70 points

    Dear John,

    Thank you so much for your Great help.

    Best Regards,

    Antonio Stenio