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Original question:

TPS92515: Buck mode LED driver support camera flash/strobe

TPS92515HV-Q1: flash control through the PWM pin - with a GPIO signal

Tzachi Bear97
Intellectual 320 points
Part Number: TPS92515HV-Q1
Other Parts Discussed in Thread: TPS92515

hi

in my design i require to sync the flash signal of the camera with the LEDs,

and  i have a 5%-10% duty cycle for the LEDS, 

can i use the PWM signal as a GPIO to toggle the signal up/down to turn on the leds, or do i need to add an external FET after the inductor to control the on/off of the leds?

 

i have two rows of 3 leds each, each row is ~9.6@1A

thanks for your reply

Tzachi

  • 0
    TI__Mastermind 28655 points
    Hello Tzachi,

    It depends on how fast you need it to respond. Using the PWM pin disables and enables the TPS92515 so there will be a delay as the inductor current rises and falls. The rise and fall will also depend on the design, input voltage and output voltage. The MOSFET across the LEDs is SHUNT FET dimming. This is much faster but it will also be less efficient because the TPS92515 is always running. There is output current bypassed by the SHUNT FET that is not going to the LEDs. This would be at 90%-95% in your design if the LED duty cycle is 5%-10%.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points

    Thank you for that answer

    just to make sure that i understand - how much of a delay are we talking about? what parameter in the spec is referring to that?

    also when i am using the shunt FET option - does the TPS92515 consumes the same amount of power even when not connected to the LEDS?

    Best regards,

    Tzachi

  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points

    Hello Tzachi,

    The delay is actually related to the inductor value, the input voltage, the output voltage and the DC current level.  If PWM has been low the current will be zero.  Once PWM goes high the TPS92515 MOSFET turns on and starts charging the inductor (V = L * (di/dt)).  If Vin is twice Vout the current ramp up and ramp down will be similar since the voltage across the inductor will be about the same, just opposite polarity.  During on time the 'V' will be Vin - Vout, during the off time the V will be Vout plus the freewheel diode Vf.

    The delay is the charge and discharge of the inductor.  If that becomes a significant portion of the on PWM pulse it may affect the light output.

    Best Regards,

  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points

    hi Irwin

    and assuming i go with a SW in the form of FET on the output that passes or blocks the current to the load with a GPIO control

    you have mentioned that this will result in power consumption, since the load is not connected how much power is being consumed?

    thank you very much for your detailed answer

    Tzachi

  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points
    Hello Tzachi,

    Yes, there will be power consumption. You will also have to add the off-time circuit for shunt FET dimming. It will be switching but at a much lower switching frequency. There will be conduction loss and the freewheel diode will be near 100% duty cycle. If Vout is near Vin it may actually have less loss then when shunt FET dimming because of the near 100% duty cycle of the freewheel diode. This can all be calculated once you know your parts and operating parameters.

    Vf of freewheel diode times output current plus conduction loss of the inductor. Switching loss should be down quite a bit due to low frequency operating.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points
    Hi Irwin,
    if a simple 3.3v IO for an on/off SW will do the trick at ~200nSec i think it is more power efficient and BOM efficient,
    is there an option to verify with you the design? should this be handled by TIs rep?
    Thanks
    Tzachi
  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points
    Hello Tzachi,

    Are you saying your PWM on-time is 200 ns? Check V = L * (di/dt). You'll probably find that the inductor current cannot slew that fast. I can calculate what will happen if I have the following information.

    Vin
    Vout(Vled) and if there is any output capacitance.
    Inductor value
    LED current when on
    PWM on pulse duration (time)
    PWM off pulse duration (time)
    is it a repeating signal?

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points
    i didnt mean that the PWM would be on for 200nS only that this is the delay from the command till it stabilizes

    Vin =12v
    Vout(Vled) and if there is any output capacitance. 3X3.2@1A (9.6v)
    Inductor value 3.3uH
    LED current when on 1A
    PWM on pulse duration (time) 2msec
    PWM off pulse duration (time) 38msec
    is it a repeating signal? no - it is triggered from the camera

    thanks
  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points
    Hello Tzachi,

    This should work fine. The rise time to get to 1A using 12V and 9.6V for Vin and Vout and 'L' being 3.3 uH is about 1.375 uS.
    The fall time is about 344 nS.
    With the inductor value being so low I assume you have this design set to fairly high switching frequency? What is your switching frequency?
    Have you designed the circuit yet?

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points
    hi Irwin and thank you for all your help :)
    yes, i have the scheme, i'd be happy to send it by mail
  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points
    Hello,

    Actually just wanted to know your Roff and Coff values, Rsns (current sense resistor(s)) and if you are using Iadj what it is set to.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points
    Coff - 470pF
    Roff - 4.53k
    Rsns - 0.1
    Iadj - connected via a 100k to VCC
  • 0 in reply to Tzachi Bear97
    TI__Mastermind 28655 points
    Hello Tzachi,

    What I get from these values is approximately:
    854 KHz
    0.681A peak to peak inductor ripple
    2.06A average current

    To get 1A average you would need to change Rsns to 0.179 ohm which gives 1.34A peak current sense (0.24V/0.179 ohm). Otherwise this should work fine.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 320 points
    Hi Irwin,
    i have two parallel 1A LED strings, so 2A is what i am aiming for
    Thanks you very much for all of your assistance
    :)
    have a great weekend
    Tzachi