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TPS92641EVM: Shunt FET recommended from data sheet for SDIM is to hot at PWM of about 50kHz~100kHz

Takeshi Hojo
Prodigy 20 points
Part Number: TPS92641EVM
Other Parts Discussed in Thread: TPS92641, TPS922052, TPS922054, TPS922055, TPS922053

The TPS92641EVM(32V,1A) is inputted by PWM about 50kHz~100kHz and blinks the LED. However, after several minutes of operation, the shunt FET (NTD3055-150T4G) became hot enough to melt the solder.

Is this the default? 

And I would like to know how to select an appropriate FET.

Thanks

Takeshi 

  • 0
    TI__Mastermind 22800 points

    Hi Takeshi,

    I do not think this is a normal situation. May I know your input voltage, output voltage, output current, and the components you have changed on the TPS92641EVM?

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    The parameter is input Voltage:48V、Output voltage:32V and Output corrent:1A. The components I have changed on the TPS92641EVM is shunt FET(NTD3055-150T4G) of Q3 for SDIM. This is an in accordance with TPS92641EVM.
    And Please refer to the attached

    Thanks

    Takeshi

  • 0 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    I do not see any signal for SDIM in above picture. How do you drive the shunt FET? 

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    Hi Steven,Thank you for youre reply

    The oscilloscope waveform above is what I measured here(from Fig.).
    Also, at least the light is blinking when PWM is applied to TP3 (SDIM) ("Signal of Photo from LED" is slow, so I only checked if it is blinking or not). Therefore, I think that the shunt FET is driven.
    But like this quietion, the shuntFET gets very hot after a few minutes of driving.

    Where is the signal for SDIM that you pointed out?
    Which TP should I measure?

    Best regards

    Takeshi

  • 0 in reply to Takeshi Hojo
    Prodigy 20 points

    Also, I think that the faster the PWM input frequency (50kHz～), the hotter the Shunt FET gets will be.
    For example, if the input PWM frequency is 1kHz, it will not get hot after a few minutes of operation, but with PWM (50kHz), it will become so hot that you can't touch it in a few minutes.

  • 0 in reply to Takeshi Hojo
    Prodigy 20 points

    By the way, what is the upper limit of the input PWM frequency to the SDIM?

    What is the recommended range of input PWM frequency to SDIM?

    Please tell me 

    Best regards

    Takeshi

  • 0 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    Sorry. I overlooked the SDIM label by mistake.

    Takeshi Hojo said:
    Also, I think that the faster the PWM input frequency (50kHz～), the hotter the Shunt FET gets will be.
    For example, if the input PWM frequency is 1kHz, it will not get hot after a few minutes of operation, but with PWM (50kHz), it will become so hot that you can't touch it in a few minutes.

    Your understanding is correct. Normally the higher the SDIM input frequency, the more energy will be consumed by the shunt FET, which will make the FET hotter.

    Takeshi Hojo said:

    By the way, what is the upper limit of the input PWM frequency to the SDIM?

    What is the recommended range of input PWM frequency to SDIM?

    Please tell me 

    It is hard to give an exact limit on the PWM frequency you can apply to the SDIM as the application conditions differ from case to case. May I understand why you need to drive the shunt FET with such a high-frequency PWM? We can talk through e-mail if needed.

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    Hi Steven,Thank you for youre reply.

    ・What I want to do is make the LED blink at high speed from 50kHz up to 1MHz. If you have an LED driver for such high-speed flashing, please let me know.

    ・Also, I would like to know in detail why "Normally the higher the SDIM input frequency, the more energy will be consumed by the shunt FET, which will make the FET hotter." (Is it switching loss?) For example, a buck regulator(Q1,Q2) switches on/off at about 500kHz, so could you please tell me why this buck regulator doesn't get hotter than a shuntFET? Please also tell me how to calculate the thermal loss of this TPS92641EVM?

    ・Either by e-mail or this community is fine.

    Best regards

    Takeshi

  • +1 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    Takeshi Hojo said:
    ・What I want to do is make the LED blink at high speed from 50kHz up to 1MHz. If you have an LED driver for such high-speed flashing, please let me know.

    May I understand what your input voltage, LED (output) voltage, and LED (output) current requirement are? What is the minimum current pulse width you would like to drive the LED?

    Takeshi Hojo said:
    Also, I would like to know in detail why "Normally the higher the SDIM input frequency, the more energy will be consumed by the shunt FET, which will make the FET hotter." (Is it switching loss?) For example, a buck regulator(Q1,Q2) switches on/off at about 500kHz, so could you please tell me why this buck regulator doesn't get hotter than a shuntFET? Please also tell me how to calculate the thermal loss of this TPS92641EVM?

    The reason why the shunt FET consumes more energy than the switching FET is due to the output capacitor because the shunt FET needs to consume the stored energy in the output capacitor each time when it is being turned on (LED being turned off).

    Let's do a rough estimation. Suppose the output capacitor Cout = 0.1uF, output voltage Vout = 32V, and you are driving the shunt FET in frequency f_SDIM = 1kHz. If we assume Cout is being discharged to nearly 0V every time the shunt FET is turning on, then the power (stored by the output capacitor) consumed by the shunt FET will be equal to P_shuntFET = 0.5 * Cout * Vout^2 * f_SDIM = 0.5 * 0.1u * 32^2 * 1k =  51.2mW.

    However, if we are going to drive the shunt FET at f_SDIM = 50kHz. Then the above consumed power will be P_shuntFET =  0.5 * Cout * Vout^2 * f_SDIM = 0.5 * 0.1u * 32^2 * 50k = 2.56W, which may exceed what the shunt FET can withstand. Here we do not take into consideration the conduction loss of the shunt FET. If the conduction loss is also considered, then the overall loss on the shunt FET will be larger.

    Here I view the loss from the perspective of output capacitor discharging. You can also view it in the form of switching loss but that will make the estimation much difficult.

    You can also estimate the loss (switching loss + conduction loss) of the switching FET (f_sw = 500kHz) and compare it with the loss on the shunt FET (f_SDIM = 50kHz) under the same circuit (Vin = 48V, Vout = 32V, I_LED = 1A, all MOSFETs are the same). You will see there is a large gap between them.

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    Hi Steven,Thank you for youre reply.

    Thanks to you, I now understand the mechanism by which Shunt FET generates heat. Regarding heat, it seems that this problem can be solved by improving heat dissipation mechanisms such as heat sinks.
    It was very helpful!

    The input voltage is 32V, the input current is 1A, and the minimum current pulse width is about 3uS (the minimum current pulse width is the desired value).
    We plan to evaluate blinking using a high-speed response photodetector.
    Also, is there any problem with the fact that it takes about 2.5uS for the I_LED to rise in the figure below?

    For future reference, is there a series of LED drivers with faster control speed than TPS92641?

    Best Regards,
    Takeshi

  • 0 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    Takeshi Hojo said:
    Thanks to you, I now understand the mechanism by which Shunt FET generates heat. Regarding heat, it seems that this problem can be solved by improving heat dissipation mechanisms such as heat sinks.

    Yes. Improving heat dissipation may help.

    Takeshi Hojo said:
    The input voltage is 32V, the input current is 1A, and the minimum current pulse width is about 3uS (the minimum current pulse width is the desired value).

    May I understand what you LED voltage and current is?

    Takeshi Hojo said:
    Also, is there any problem with the fact that it takes about 2.5uS for the I_LED to rise in the figure below?

    Could you also measure the inductor current and put them in the same oscilloscope capture to help me judge what's happening here?

    Takeshi Hojo said:
    For future reference, is there a series of LED drivers with faster control speed than TPS92641?

    If extra fast current rising and falling time is needed (for example if you need several hundred nanoseconds of current pulse), the shunt FET with an LED driver like TPS92641 is the best choice. Otherwise, you can take a look at TPS922052 / TPS922053 / TPS922054 / TPS922055.

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    Hi Steven,Thank you for youre reply.

    ・Maximum drive current of LED:1200mA,LED voltage:35.6V
    ・As shown in the figure, the inductor is surface mounted, so it seems difficult to measure the inductor current. Is there somewhere I can measure it instead?
    Do you expect the delay to be caused by C_out's charging or L_cout's electromagnetic induction?

    Best Regards,

    Takeshi

                                                  FIG

  • 0 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    Takeshi Hojo said:
    Maximum drive current of LED:1200mA,LED voltage:35.6V
    Takeshi Hojo said:
    The input voltage is 32V,

    Please double check your input voltage.

    Takeshi Hojo said:
    As shown in the figure, the inductor is surface mounted, so it seems difficult to measure the inductor current. Is there somewhere I can measure it instead?

    Sorry I forgot you are using SDIM to do the PWM dimming instead of UDIM. There is no need to measure the inductor current.

    Takeshi Hojo said:
    Do you expect the delay to be caused by C_out's charging or L_cout's electromagnetic induction?

    I think this is due to the charging of C_out. When the shunt FET is turned off (LED being turned on), it needs some time to charge the output capacitor. For example, let's assume that the LED will conduct little current before Vf > 30V. Then it will take t = C_out * Vout / Iout = 0.1uF * 30V / 1A = 3us before the LED current rises. This estimation is close to what you have observed in the above picture.

    Best Regards,

    Steven

  • 0 in reply to Steven Li
    Prodigy 20 points

    Hi Steven,Thank you for youre reply.

    I'm sorry for lack of explanation and my misunderstanding.

    The input voltage(32V) is Vout to the LED of the TPS92641EVM (i.e. Vout in the TPS92641EVM data sheet)
    LED voltage(35.6V) is the specifications of the LED used (from the LED's data sheet,Min:32.7V,Typ:35.6V,Max:38.5V)


    Best Regards,

    Takeshi

  • +1 in reply to Takeshi Hojo
    TI__Mastermind 22800 points

    Hi Takeshi,

    So you are going to drive a 35.6V, 1.2A LED with minimum 3us current pulse. Is my understanding correct? If yes, then TPS92641 with shunt FET or TPS9222052/3/4/5 with shunt FET are good choices. The shunt FET is necessary.

    Best Regards,

    Steven