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UCC28019A: Gate Resistor for MOSFET

SJZ
Intellectual 280 points
Part Number: UCC28019A
Other Parts Discussed in Thread: UCC28019

Hello,

I have simulated TINA average model of UCC28019A for 2.5kW classic PFC and have questions related to the MOSFET circuit.

In the UCC28019A EVM design (here), resistor R1, R5 and Capacitor D4 are connected to reduce ringing/ turnoff time.

How will I get the values of these specific components for my design?

Here is the datasheet of MOSFET I am using (without any gate driver, just like the EVM)

Regards,

  • 0
    TI__Mastermind 19435 points

    Hi,

    Thank you for the query on UCC28019.

    As you correctly pointed out R1 is used to reduce ringing / control turn on time, D4 is used to quicken the turn off time.

    R5  pulls the gate voltage down to the source voltage to ensure that FET is off. Without a resistor, external noise coupling, leakage currents, or other unintended biasing will bring the gate voltage above the threshold and turn the FET on. For this resistor it is a standard practice to choose 10k.

    The following app note provides details on the choice of gate drive resistor.

    https://www.infineon.com/dgdl/Infineon-EiceDRIVER-Gate_resistor_for_power_devices-ApplicationNotes-v01_00-EN.pdf?fileId=5546d462518ffd8501523ee694b74f18

    Please let me know if this helps

    Regards,

    Harish

  • 0 in reply to Harish Ramakrishnan5
    Intellectual 280 points

    Hello,

    Thanks for the reply,

    Since the calculated peak current according to the MOSFET parameters(Vgs, Ron(int)) is coming out to be lower than the gate current peak sink of UCC28019A ( 2A as mentioned in the datasheet). I am considering the circuit without GateDriver.

    In that case, what values of Rg(R1, R5) should be used?

    As you mentioned standard value of 10kohm is used for R1, and 3.3ohm for R5 and diode D4 (as used in UCD28019AEVM)

    Will the same values work fine without Gate Driver?

    Regards,

  • +1 in reply to SJZ
    TI__Mastermind 19435 points

    Hi,

    The EVM board was rated for 350W and your application demands 2.5KW, nearly 10 times the power, so it is going to be challenging without a gate driver to replicate the same without simulations/calculations. You can get a rough estimate of the value of gate resistor in the following way. 

    Suppose your application requires 100khz Fsw, in which and you want to achieve switching time of 100 ns. You need to plot a graph of Vgs vs Qg for the FET you are using and try to come up with the point after miller region where Vgs starts to increase for second time. Obtain Vgs and Qg and let's say you have got them as 6.8V and 16nC (These points are subject to change with the operating current)

    The required gate drive current is derived by simply dividing QG of 16 nC by the required switching time of 100 ns, giving a result of 160 mA. The designer can estimate the required gate resistor value. If the drive circuit applies 14 V then a gate resistor of about 50 Ω would be required, according to (14 V – 6.8 V)/160 mA.

    UCC28019 with 2A/1.5A source/sink capability should be able to handle this. But I would recommend simulating as per your switching time requirement as per the above method which I suggested as it would require some trial and error.

    Thank you

    Regards,

    Harish

  • 0 in reply to Harish Ramakrishnan5
    Intellectual 280 points

    This was really helpful, thank you.

  • 0 in reply to SJZ
    TI__Mastermind 19435 points

    You are wecome. 

    Thank you

  • 0 in reply to Harish Ramakrishnan5
    Intellectual 280 points

    Hello Harish,

    I am also checking the Total Power dissipation across the Gate Resistor by using: P = Qg*Vgs*fsw*Rgate

    Taking Qg as 107nC, Vgs as 14V, fsw of 100kHz and Rgate as 10 Ohms. Pdissipated is coming out to be a little higher.(around 1W)

    So do I need to parallel two 20Ohms resistor of 0.5W or am I missing something here.

    Regards, 

  • 0 in reply to SJZ
    TI__Mastermind 19435 points

    Hi,

    Thank you for the query.

    I think you have wrongly multiplied the resistance in this formula. The gate drive losses are only P = Qg*Vgs*fsw. So the losses here are only 0.15W

    The gate resistance need to be suitably sized to handle this.

    Regards,

    Harish

  • 0 in reply to Harish Ramakrishnan5
    Intellectual 280 points

    Hello,

    Thanks for the clarification.

    Regards,