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TPS92518HV: Fast LED switching with minimal ripple, output capacitor?

Y Bertrand
Prodigy 60 points
Part Number: TPS92518HV
Other Parts Discussed in Thread: TPS92518, TINA-TI, TPS92641

Hi,

We want to use a fast LED driver for making very short LED flashes for use in a medical device. The current pulses must have low output ripple.

So I thought the TPS92518 would be interesting since it supports shunt PWM dimming.

However, looking at the TINA-TI simulation, using the stock *.tsc file provided, I see a massive ripple in current when running the simulation.

I immediately noticed the recommended design at the output does not figure a parallel *output capacitor* (COUT) at the buck power stage (after the inductor).

Why is that so? The current regulation seems horrible. If the load resistance is relatively low the current will probably cross the 0A point.

Of course, I didn't start trying to optimize it since I was shocked at the ripple with the stock simulation, and prefered asking here.

But still, shouldn't these devices try to reduce the output ripple to acceptable levels? Why is there no COUT? Is there a better product in the TI family that can feature better output ripple (perhaps the TPS92520?)

Thanks

  • 0
    TI__Mastermind 28655 points

    Hello,

    Shunt FET dimming shorts the LED string through a MOSFET to keep the current flowing.  When the Shunt FET opens it can return current back to the load quickly.  Two reasons for low or no output capacitor.  The first reason is because it becomes switching loss, the output capacitor has to get discharged every time the Shunt FET turns on.  At high shunt FET dimming frequencies this can become substantial.  The second reason is the capacitor has to recharge after the shunt FET is opened.  At low dimming duty cycle/on-time it causes a delay due to recharging the capacitor.  This can/will affect light output.

    As a buck converter the LED series load resistance without a capacitor has almost nothing to do with current ripple.  The current ripple is determined by the inductor value, switching frequency, Vin and Vled as well as a few other contributors.  Buck converters have continuous current at the output.  If the inductor value was very large the current ripple can be quite small.

    As far as low dynamic impedance (impedance or resistance in series with the LEDs) an output capacitor would have to be larger to reduce ripple since it is supplying or absorbing current based on I = C * (dv/dt).  If 'dv' gets smaller due to lower dynamic impedance 'C' has to get larger.

    If a 1 uF capacitor was on the output of a 1A LED driver with a 24V LED string it would take 24 us to recharge the capacitor back to 24V.  None of that current goes to the LEDs.  At lower shunt FET off times the LEDs won't even light.

    If shunt FET dimming at 10 KHz, 24V LED string the loss in the shunt MOSFET will be 2.88W.  This power is taken away from the LEDs so if trying to deliver 2W to the LEDs they will not light since it takes 2.88W to charge the 1 uF output capacitor at 10 Khz with a 24V LED string.

    So shunt FET dimming with an output capacitor depends on what the application needs.  Higher inductor value and higher switching frequency will reduce output current ripple.  Low frequency shunt FET dimming without low output on-times can use a larger capacitor across the LEDs.

    Output ripple doesn't have to do with the controller/driver being used if they are all buck converters.  It is determined by the inductor value, switching frequency, Vin and Vled.  If there is a capacitor on the output the dynamic impedance of the LEDs  gets added in to the calculation.

    Best Regards,

  • 0 in reply to Irwin Nederbragt
    Prodigy 60 points

    I appreciate your detailed answer. I didn't really think about the capacitor discharging in fact, somewhat I forgot this was a constant current operation...

    What about a switched capacitor (i.e. a MOSFET in series with the capacitor to block it's discharging, switching it exactly at the same time as the LED shunt)?

    Also, why is the TPS92520 application schematic recommends an output capacitor, contrarily to the datasheet of the TPS92518?

    Same for the TPS92641 : albeit being strongly recommended for shunt dimming, why is this converter using an output capacitor?

    The product I'm working needs a lot precision. The flashes must be short (switching LEDs down to a few microseconds, I'm trying to achieve 1us), yet the output ripple is important as well (the luminance must not vary much within a flash). Hence why I'm asking all these questions. Thank you very much for your time.

  • 0 in reply to Y Bertrand
    TI__Mastermind 28655 points

    Hello,

    The capacitors are recommended to meet stringent automotive EMI requirements.  Output capacitors aren't ideal for matrix or shunt FET dimming however the matrix dimming is at fairly low frequency.

    Yes you could open a capacitor and close it with a switch (MOSFET).  It gets a bit tricky because an N-channel MOSFET has to be opened before the shunt FET shunts and since there is a body diode it cannot be referenced to the circuit common.  A low value capacitor with low ripple current may work for your design.  Let us know your application.

    Best Regards,