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LM51551: not behaving as expected

Part Number: LM51551
Other Parts Discussed in Thread: LM5155

We recently switched to the LM51551DSSR due to the unavailability of our previous boost IC.

I generated a design with the TI WeBench Power Designer, and verified the schematic with the typical application in the datasheet, and everything looked good to me. The 12V_BOOST_EN pin is connected to an MCU GPIO pin that is 3.3V when high.

What I've noticed is that the load voltage is the same as the input voltage, whether I have the 12V_BOOST_EN GPIO set to high or low. I've also tried wiring that pin directly to the BIAS pin, but with the same result.

Are there any obvious mistakes that stand out to you? I wanted to double check before doing a deep dive into the datasheet.


  • Hello Lidong, 

    Thanks for reaching out. First I would suggest to try to power up the LM5155 with the UVLO pin connected to BIAS, like you said. It looks like from the schematic that you have 3 capacitors on the COMP pin, can you please check the compensation? Other than this, it looks like the diode is missing, can you please add a rectifier diode (see page 25 section 10 of the datasheet) . Let me know if I can help you furthermore. 

    Kind regards, 


  • Thanks for the tips. I checked the COMP pin, and it looks like I used the correct resistor, but used the capacitor symbol on the schematic by mistake.

    I soldered a wire to connect UVLO directly to BIAS, but the output voltage is still the same as the input. I can't add the diode without redesigning the board, which will take a couple weeks, but I wouldn't think the diode would cause the issues that I'm seeing.

    I probed the GATE pin with an oscilloscope, and it looks like the pin oscillates for 200us before stopping, and restarts every 80ms.

    Is there anything else you suggest checking?

  • Hello Lidong, 

    Thanks for your response. You cannot have a boost converter without a rectifier diode (please see Typical Application section in the datasheet). Without the diode, the Vin is shorted directly to the Vout and this is why you see the same voltage at input and output. 

    Kind regards,