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Original question:

TPS54202H: Schematic check for 5V to -17V inverting buck converter

TPS54202H: Inverting design using TPS54202H not starting

Derek
Mastermind 9792 points
Part Number: TPS54202H
Other Parts Discussed in Thread: TPS54202

Hello,

  Built boards with the circuit in the linked forum post.

https://e2e.ti.com/support/power-management-group/power-management/f/power-management-forum/1320239/tps54202h-schematic-check-for-5v-to--17v-inverting-buck-converter

However, I'm not seeing any output voltage. It's supposed to be -14.6V. We disconnected any output circuitry to isolate the issue. EN pin (pin 5) is 1.300mV, and VIN (pin 3) is 5V. I probed the SW signal (pin 2) and am not seeing any oscillations.

The EN pin is going through a voltage divider of 22k over 7.63k with the top attached to 5V and bottom attached to the -17V rail. So it should be 5.66V above the -17V rail, well within the 7Vmax,EN.

Inductor is 56uH, and feedback network is 47k over 2k with 100pF (compensation as per Webench) from Vfb to the output of the inductor (GND).

Thanks,

Derek

  • 0
    TI__Mastermind 27276 points

    Hi Derek

    Can you please attach your updated schematic?

    Before the device start up, the '-17V' hasn't start to build up, the IC_GND is 0V same as system GND. So when 5V Vin is applied, the EN to IC_GND voltage is 5V*(7.63k)/(22k+7.63k) =1.28V, which is quite near to the V(EN_RISING) specified in the datasheet. Please configure the voltage divider so that the EN to IC_GND is higher than 1.35V to make sure the device can normally start up.

    After the '-17V' build up, the the EN to IC_GND voltage is (5+17)V*(7.63k)/(22k+7.63k), and you are right we need to keep this value <7V.

    I am not sure how comes the EN pin (pin 5) is 1.300mV, can you please check about this? 

    Thanks and best regards.

  • 0 in reply to Gui He
    Mastermind 9792 points

    Schematic is below. I originally tried it with the EN resistor divider of 100k over 7.87k but that didn't work either. What's the best way to ensure that the EN pin is > 1.35V above IC_GND (in this case our -17V rail) but doesn't exceed the VEN,max of 7V? If the resistor divider is 22k over 10k then:

    startup: VEN = (10k/32k * 5V) = 1.56V above IC_GND

    running: VEN = 10k/32k * (5V + 17V) = 6.875 above IC_GND

    That would work, but without very much margin on either side. I guess another way to do it (although sloppy) would be to hit it harder (e.g. a resistor divider of 10k over 10k) so that while running the output of the resistor divider is higher, but then use a 5.6V zener from EN pin to IC_GND to restrict that voltage to 5.6V above IC_GND.

    I re-measured VEN. With the current resistor divider (22k over 7.87k) it is measuring 1.27V which makes sense because (7.87 / (22 + 7.87)) * 5V = 1.3V.

    I also overnighted a replacement TPS54202 in case this one was damaged. I also have other boards I can test.

    Thanks,

    Derek

  • +1 in reply to Derek
    TI__Mastermind 27276 points

    Hi Derek

    I  the 22k over 10k  solution can work. If you want to leave more margin and the power dissipation on the Zener wouldn't be a problem, you can add a Zener diode between EN and IC_GND.

    Does the circuit operate normally now? 

    Thanks and best regards.

  • 0 in reply to Gui He
    Mastermind 9792 points

    Hello, I changed the resistor divider to 22k over 10k and also replaced the IC and it is working now. Thank you.