TPS92510: Problems with BOOT UVLO

Hello David,

If the input dips to around 3.5V to 4V and the output current is continuous the freewheel diode will conduct when the internal MOSFET turns off and it should recharge the boot capacitor. If not this note applies:

"Attention must be taken in maximum duty cycle applications which experience extended time periods with little or
no load current. When the voltage across the BOOT capacitor falls below the 2.1 V UVLO threshold, the highside
MOSFET is turned off, but there may not be enough inductor current to pull the PH pin down to recharge the
BOOT capacitor. The high-side MOSFET of the regulator stops switching because the voltage across the BOOT
capacitor is less than 2.1 V. The output capacitor then decays until the difference between the input voltage and
output voltage is greater than 2.1 V, at which point the BOOT UVLO threshold is exceeded, and the device starts
switching again until the desired output current is reached. This operating condition persists until the input
voltage and/or the load current increases. It is recommended to adjust the VIN stop voltage greater than the
BOOT UVLO trigger condition at the minimum load of the application using the adjustable VIN UVLO feature with
resistors on the EN pin."

Do you know if it is hanging up at Vout/Vled or allowing the freewheel diode to conduct?

Best Regards,

Yes, that is the paragraph that I was referring to on page 15 of the datasheet.

I believe it is getting stuck where Vout is too low to put sufficient current through the LED, which leads the freewheeling diode to stop conducting as well. How can I best test that? I don't want to give you wrong information and send us in the wrong direction.

Hello David,

If you look at the switch node, and maybe the boot voltage, with an oscilloscope you would see that the voltage on the switch node is not dropping due to freewheel conduction.

I'm not sure if you have access to other voltages in your design but if you do you could add a separate boot charge diode if the voltage is high enough.

Best Regards,

I do have a 5V rail available. So, a diode from 5V to boot should help?

Is there anything that can be done to help pull down SW faster, and thus increase Vboot from the other direction? My quick calculation tells me that a 1K resistor would discharge a 2uF output capacitor from 3V to 1V in about 2mS, and would only cost me 3mA of static current draw. Would this actually work - to pull down the SW node faster so that the boot cap charges without waiting for Vin to rise?

Hello David,

Yes, I would try that. It might not be enough but it would be worth trying. Try using a fairly low Vf diode. If it doesn't quite get there adding the load resistor with that diode would make the refresh quicker.

Best Regards,

I've looked at the SW node on the oscilloscope, and I can see that it does stop switching, and is no longer pulled low, once VIN droops low enough, and then resumes switching once Vboot rises high enough. Without any external circuitry that takes about 10mS for VIN to rise. With a 1K pulldown I can see Vsw pulled down in a few mS, and then the part resumes switching. While the 2mS delay is better it still causes some distracting flickering. Increasing the resistor current shrinks that delay but at a certain point I'll be burning too much power there.

I have run into similar issues in the past with voltage regulators and found what appears to be a good solution in your LMR16010. From that part's description of the bootstrap functionality:

"When operating with a low voltage difference from input to output, the high-side MOSFET of the LMR16010 will
operate at approximate 97% duty cycle. When the high-side MOSFET is continuously on for 5 or 6 switching
cycles (5 or 6 switching cycles for frequency lower than 1 MHz, and 10 or 11 switching cycles for frequency
higher than 1 MHz) and the voltage from BOOT to SW drops below 3.2 V, the high-side MOSFET is turned off
and an integrated low side MOSFET pulls SW low to recharge the BOOT capacitor."

So that part has a small MOSFET that pulls down the SW node when necessary, while the TPS92510 relies on SW being pulled down externally, or VIN pulled up. Is that a correct analysis of the difference between these parts? Does TI have any LED drivers that operate like the LM16010, that can refresh its own bootstrap capacitor even when Vin is too low? Or, would an LED driver with a maximum duty cycle less than 100% also fix this?

Hello,

There are parts that have the internal pull-down, TPS92515 is one of them, just checked, 5 mA pull-down when boot hits UVLO. It won't operate low enough for your design. When the switching stops what is Vout voltage at that point?

Best Regards,

Switching stops when Vout is about 3.5V, and then resumes when Vin rises to 5.5V. This is without the pulldown resistor, so the SW node is stuck at about 2.5V the whole time.

The TPS92515 is interesting but I do think the voltage won't go low enough. I think a min operating voltage of 4.0 is probably the highest I can go.

Is there some particular parameter that I can look at to search through TI parts for that internal pulldown? Or is it just insider knowledge of the various parts available?