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LM5050-1: Selection of series resistor R1 value to avoid failure during startup

Manuel Madlener
Prodigy 50 points
Part Number: LM5050-1

Hello

I am facing the following issue:

My circuit uses a 47Ohm series resistor between IN-pin and VS-pin to supply the VS-pin of the IC and the bypass capacitor C1. (c.f. resistor R1 in fig. 27)

When the power supply (supplying the IN-pin) is enabled (fast voltage rise), the series resistor initially experiences the full supply voltage of 51V, which causes losses of (51V)^2/47Ohm=55.3W- This damages the resistor until it fails to an open-circuit.

I could solve this issue by using a pulse-rated resistor, but I would prefer to keep it a generic 0603 resistor with a 0.1W power rating.

Since the gate driver apparently only requires up to 288uA (table 6.6), I am thinking that the resistance value for R1 can be as high as R=(30V-5V)/(288uA)=86.8k, where 30V is my minimum supply voltage and 5V is the minimum required supply voltage at the VS-pin.

Furthermore, if the resistor is selected such that R>(51V)^2/(0.1W)=26k, it should be impossible to damage the resistor based on the maximum 51V supply voltage.

Therefore, I would assume that a resistor of e.g. 47kOhm should work nicely.

However, the datasheet specifies a typical resistor of 100Ohm, which is orders of magnitude lower.

Do you see any issues with using a resistor in the range of 47k for R1?

What is the recommended way for avoiding over-stressing the resistor R1 when the input voltage to the circuit is enabled?

  • 0
    TI__Guru 57737 points

    Hi Manuel,

    Firstly, this resistor is not required in your case as the minimum voltage in your application is 30V. You can connect the VS pin to IN directly.

    The resistor will have power dissipation only for the duration during which the capacitor is being charged. This happens only once during power up and that too for short interval of time. For a 100Ω resistor and 0.1uF capacitor the time taken to charge the capacitor is 5RC = 50usec. Also the power loss calculation done by you is the peak power loss and as the capacitor starts to charge the power dissipation across the resistor gradually reduces during the 50us time duration. This power loss for a small duration should not be a problem for 0603 resistor. We have used 0603 resistor in our EVM and haven't seen any issue. Did you see the resistor getting damaged and becoming an open-circuit ? 

  • 0 in reply to Praveen GD
    Prodigy 50 points

    Hi Praveen

    Thank you very much for your rapid response!

    I am a bit hesitant to remove the resistor altogether (or use a 0-Ohm resistor) because the resistor is apparently also useful to protect the device. (e.g. " D1, D2 and R1 protect against these
    spikes which might otherwise exceed the LM5050 100-V survival rating." in section 7.1)

    Yes, I am seeing consistent failures of the resistor (0603, 47Ohm, RC0603FR-0747RL) after approximately 20-50 voltage steps (0V->51V) on the input.
    This might be caused by our design having C1=1uF, which is 10x of the typical value from the datasheet. (unclear to me why exactly this was used) This would obviously increase the stress on the resistor.

    Looking at my case of a C1=1uF, the total enegy loss in the resistor is approximately E=0.5*C*U^2=1.3mJ, with a pulse duration in the order of 5tau=235us. This corresponds to a average pulse power of 5.53W. 
    Which (looking at a random datasheet with pulse rating curves e.g. https://www.vishay.com/docs/28705/mcx0x0xpro.pdf, page 6 top figure), exceeds the maximum power of ~4.5W for a ~200us pulse.

    If I were to use a C1=0.1uF instead (but keep R1=47ohm), the energy is 0.13mJ with a duration of 23.5us but unchanged average power of 5.5W, which is still too high.

    For C1=0.1uF, R1=100Ohm, as you say the duration is 50us, which results in a average power of 0.13mJ/50us=2.6W, which is within the specifications of this specific part.
    But I am not convinced, that every generic 0603 resistor would be able to handle this, especially considering that the maximum power could be substantially higher, in the order of (51V)^2/100Ohm=26.1W.

  • 0 in reply to Manuel Madlener
    Prodigy 50 points

    Another aspect/question:
    We experienced some severe failures of the whole circuit, where also the MOSFET and the driver IC failed. May assumption so far was that due to the supply resistor failing, the IC looses supply, cannot turn on the MOSFET anymore and the MOSFET overheats. 
    But my testing shows, that even if the resistor R1 is completely removed and the circuit is only operated with 20V, the voltage at the VS pin is still 4.3V and the ideal diode circuit is still operating (24mV drop across the FET for 8A current).

    What is the expected behavior if R1 is a complete open circuit? Should the ideal diode circuit still operate normally?

  • 0 in reply to Manuel Madlener
    TI__Guru 57737 points

    Hi Manuel,

    Yes you are correct. Instead of just charging the VS capacitor once, if you're constantly charging / discharging (not allowing it to cool) it then you need to bother about your average power. Other wise for a normal application where startup happens only once the resistance shouldn't get damaged. A 0.1uF should be sufficient at the VS pin for the device to operate. Having 1uF will increase the power dissipation in the resistor proportionally.  

    If the resistor to VS pin opens, then the VS pin is not powered anymore. I would expect the device to not power up and hence the Gate would remain low, FET OFF and the FET body diode would conduct. 

  • 0 in reply to Praveen GD
    Prodigy 50 points

    Hi Praveen

    Sorry, I wasn't clear on that aspect:

    I mean after 20-50 power enable cycles over the lifetime of the circuit, the resistor fails. I tested this with a period of 10s, i.e. the IN voltage would step to 51V, stay for 5s high, then go low for 5s etc. This period of 10s means that the average power in the resistor is approximately 1.3mJ/10s=0.13mW for a 100mW rated resistor. Furthermore the duty cycle of the resistor loading is in the order of 235us/10s=0. Therefore, the resistor has more than enough time to cool.

    This means it's not the average power, but the power peak during charging that kills my resistor. 

  • 0 in reply to Manuel Madlener
    TI__Guru 57737 points

    Hi Manuel,

    Thanks for the clarification. let me review and get back to you.

  • +1 in reply to Praveen GD
    TI__Guru 57737 points

    Hi Manuel,

    Thanks for the explanation. Understood your point.

    You can consider using 100 ohms, 0805 resistor with a 0.1uF cap to reduce the peak power dissipation and still not compromising on device performance.