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LM5025C: Inquiry about out (pin 9) short.

David Kim2
Genius 3415 points
Part Number: LM5025C
Other Parts Discussed in Thread: LM5025

Hi

There is a problem that the out (pin 9) is short-circuited with the GND during the input on/off test.
I want to improve it, so which part should I review?

Thanks

  • 0
    TI__Mastermind 33115 points

    David,

    Can you please be more specific to better describe the issue you are seeing? what is input on/off test - is this just the removal of input power one time, is it periodic, is the PWM expected to operate for some amount of time when input power is removed? What do you mean the OUT (pin 9) is shorted to GND - is this just saying there is no PWM output when it should be expected? Please help to explain and feel free to attach waveforms, images or timing diagrams that will help to better communicate the issue?

    Regards,

    Steve

  • 0 in reply to Steven Mappus1
    Genius 3415 points

    Hi Steven

    Thanks for your support.

    1. The ON/OFF test is a test that repeats the input power ON/OFF to see if there is a problem with the set.
    2. Short circuit to GND means the inside is dead and there is no output
    3. We also confirmed that the current in OUTB is -1.4 A when it is peak. Can this dead the IC?


    4. If I change C24 from 104 to 472, I think there will be no problem because the current will be reduced.


    5. Is there any problem with using C24 as 472?

    Thanks

  • 0 in reply to David Kim2
    TI__Mastermind 33115 points

    Driver stage is rated for 3A, so Ipk=1.4Apk is not problem. For sizing the C24, refer to section 6 here.

    Steve

  • 0 in reply to Steven Mappus1
    Genius 3415 points

    Hi Steven

    OUT_B(Pin 9) is dead and OUT_B has a peak of 1.25A as shown below. If it comes out to 1.45A, can IC die?


    And the other pin is fine, but only OUT_B(pin9) keeps dying. What's the condition that only OUT_B(pin9) can die?

    Thanks

  • 0 in reply to David Kim2
    TI__Mastermind 33115 points

    Pay attention to the time constant between C24 and R38. When you used C24=0.1μF, R38=10kΩ, you get RC=1ms but when you change C24 from 0.1μF to 4.7nF, you get RC=47μs. You reduced the RC time constant by 21x. Depending what is your switching frequency, equations 47, 48 below will give guidance for sizing C24, R38. Determine the RC time constant (tau) to approximately 100/Fsw as shown in equation 47 below.

    Use the desired tau of ~100/Fsw to determine the value of Cc (C24) according to equation 39 below (I sent this to you previously). It could be that when you cycle the input power ON/OFF/ON, the clamp capacitor is not fully discharged which changes the initial conditions during the second startup, causes the PWM controller OUTB duty cycle to suddenly change and this result in a charge imblance across Cc. You should monitor VIN as well as the clamp capacitor voltage and the OUTB gate drive when the ON/OF test is introduced. Compare the result to a "normal" turn off where you wait a longer time for the circuit to return to steady initial conditions and then start. Also, consider to change your 1N4148 to a BAT54 or equivalent Schottky diode type.

    Regards,

    Steve

  • 0 in reply to Steven Mappus1
    Genius 3415 points

    Hi Steven

    Thanks for your support.

    Then, as you said, if the gate current increases during the on/off/on test and exceeds the range, the IC may die, right?

    I have to find the cause, but the gate current is the most suspicious, so I'm inquiring.

    Thanks

  • 0 in reply to David Kim2
    TI__Mastermind 33115 points

    Follow the example I gave for calculating the RGS, Cc and tau. The gate driver currents shown in the LM5025 data sheet are the maximum peak drive current capabilities of the IC.

    Steve