This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

  • TI Thinks Resolved

LM5113: Disabling the Gate Driver for protection

Part Number: LM5113

Hi,

Can you please suggest a way to disable the Gate Driver LM5113 in case an overcurrent/Over temp is occurring.

Pulling LI and HI Input pulses to ground would mean that i am exceeding the output source current limitation of microcontroller. If using a current limiting circuit for the same would mean i am introducing a more delay while normal operation.

Will pulling the HB Pin to Ground be enough for disabling the gate pulses for HI Switch.

Please suggest a good approach for disabling the LM5113 Gate Driver.

Thanks 

Arjun

  • Hello Arjun,

    Thanks for your question and interest regarding LM5113. Because there is no enable pin on LM5113, there is no other ways other than disabling HI and LI can disable the driver. However, the input HI and LI are both high impedance input, and should need only very small amount of current. The input capacitance of HI and LI are also low, so that even if you have a resistor in series with HI and LI signals, it should not cause too much delay.

    You could pull HB to ground to disable gate pulses for HI switch, since by this way you are losing the supply for high side driver. However, please be noted that this short is also a RC type of discharging, which can take some time, especially when you have boot strap cap to be around 0.1uF.

    Thanks and regards,
    Lixing
  • In reply to Lixing Fu:

    Hi,

    I Had tried simulating the circuit. But it was not successful. I am Attaching a word file with screenshots.

    Kindly suggest me a solution and find out what is the problem.LM5113_simulation.docx

  • In reply to arjun raj:

    Hi Arjun,

    Lixing above was incorrect in stating that you can disable the part by pulling the HB pin to ground. If you successfully do that, then when the low-side switch turns on, the full VDD supply will be applied across the internal bootstrap diode. Lots of current will flow (many amps) and it will quickly destroy the bootstrap diode. In your simulation, you are seeing up to 2.5 amps of current but even that is not enough to hold the HB pin low. Doing this will destroy the LM5113.

    The only method is to use some method to keep LI and/or HI low in the event of a fault. A dual two-input AND gate on both inputs such as the SN74LVC2G08 would do nicely and only adds up to 4.8ns of delay at 5V. A faster one may be found if you only need 3.3V operation. Trying to pull the inputs low and over-drive the microcontroller output is also not recommended as you say because of the excessive current taken from the MCU pin, however the AND gate gets around this problem.

    Regards,

    Nathan

  • In reply to Nathan Schemm:

    Hi Nathan,
    Thank you for the clarification.

    In SN74LVC2G08 there is no enable pin. My protection circuit is active High Output so I cant use it directly as the second input of AND gate to disable the pulses.
    So do you recommend to pull the output of AND gate to ground in event of an over current. Is that acceptable.

    SN74LVC2G125 is a dual bus buffer gate with Enable pin. Do you think using this would be better.

    Regards
    Arjun.
  • In reply to arjun raj:

    Hi Arjun,

    The SN74LVC2G125 has an output-enable. So when the enable input goes high, the output (which will be the LI and HI input) will go to high-impedance. You maybe could use it if you put a pull-down resistor on the output so that it would then be pulled low. However the speed of the disable function may be a problem as it is set by the pull-down resistor value which probably can't be too low.

    If you only need to disable one input (you indicated you seemed to be most interested in disabling the HI input), you could use a device such as the SN74LVC1G98. If you look at figure 3 in the datasheet you can see how to connect it to make a two-input AND gate with one input inverted which is the function you need.

    Otherwise the dual-AND gate with a separate inverter is always an option if you need both inputs disabled or two of the SN74LVC1G98.

    Regards,

    Nathan

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.