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TPS61022: Under Performance

Andrew Charnley
Intellectual 640 points
Part Number: TPS61022

Hi,

I've posted a few times regarding this chip and continue to have issues with it. I'm now at the point of looking at a viable alternative.

Previously the chip would self-destruct due to the phase issue. I've since found a combination which works across the input voltage range of 1V to 5V whilst giving reasonable current output without the chip self destructing. The issue now is one of regulation.

For the 61022 I have 22u 10v close by on the input side and 44u 10v for the output side. The input side also has 1500uF of solid polymer close by, and the output has a 220uF of solid polymer close by. The input source is two large super-capacitors which can sink many times the current the 61022 is able to draw.



The boost is set for 5V output. A 120p capacitor from the calc.xls is used and indicates the chip is stable at all current draws.

If the output current is too great there is a UVLO set for 4.5V. This will turn the output draw off until it recovers to 4.9V.

Here is the output circuit.



The output from the buck is a peak of 2A at 4V. So 8W, and factoring a 10% efficiency drop is 8.8W. Divided by 5V = 1.76A.

When this is requested with a 4V input the 61022 doesn't regulate to 5V and the UVLO fires at 4.5V, so the 1.76A is too great an ask.

My understanding is one can have wide operating input voltage with stability if the output current is quite low compared to the headline figure of 8A.

If there is any other understanding around this it would be good to know.

Regards

Alex

  • 0
    TI__Genius 12465 points

    Hi Alex,

    Thanks for reaching out.

    Andrew Charnley said:
    When this is requested with a 4V input the 61022 doesn't regulate to 5V and the UVLO fires at 4.5V, so the 1.76A is too great an ask.

    My understanding is one can have wide operating input voltage with stability if the output current is quite low compared to the headline figure of 8A.

    Sorry I am a little confused about your questions here. Do you mean that you find the TPS61022 can't work normally in the condition when VIN=4V, but can work normally in other conditions?

    Is the inductor you are using 1uH? I don't find the inductance in the schematic. I have checked the stability when VIN=4V with your setup (L=1uH), there is no stability risk in this condition. Do you have any waveform, like VIN, VOUT, SW, ILOAD, so that I can check what happens when VIN=4V?

    Or is there a load transient? I mean maybe the VOUT drops below 4.5V because the there is a undershoot when load transient happens. If that, we recommend to increase cross-over frequency to decrease the voltage drop by changing feed forward capacitance.

    Look forward your feedback.

    Regards,

    Nathan

  • 0 in reply to Nathan Ding
    Intellectual 640 points

    Hi Nathan,

    That is the issue with this chip, you can't have your cake and eat it.

    To gain stability over a wide input voltage range have to add feed forward, which reduces the voltage regulation.

    In my case I restrict the output current when the input voltage drops to 1.2V to protect the chip from self destructing. Although the calc.xls shows for the capacitance I have it should be stable and on my IR it doesn't get that hot, I don't trust it.

    However when the input voltage is above this, let's say 4.2V, when 2A/4V from the output buck is drawn, which computes to about 1.76A from the 61022, the voltage regulation falls from 5V to 4.6V with transients around the 4.4V range which trigger the UVLO. I have had to set the UVLO to 4.3V to get it to pass. Basically more than 2A is impossible. The 8A headline figure is a fallacy - the only time it'll be doing something remotely close to that is in pass-through mode.

    L1 is 1uH, large and iirc about 12A saturation at 30% drop.

    I don't have waveforms to hand but previously the PCB footprint area was looked at by TI and signed off as good, there are none of the usual switch-mode mistakes.

    Regards,
    Andrew

  • 0 in reply to Andrew Charnley
    TI__Genius 12465 points

    Hi Andrew,

    The 8A is inductor current limit, not output current, the output current can't be as high as 8A.

    And more, It is a stable value, when there is a load transient, there will be undershoot and overshoot at output voltage. So that is why I want to check based on the waveform, if the 5V drops to 4.6V only for a short time or for a long time. Because TPS61022 can support 4V to 5V/2A condition theoretically. You can see this from our datasheet.

    Regards,

    Nathan

  • 0 in reply to Nathan Ding
    Intellectual 640 points

    Indeed your data-sheet says that but it's on the assumption the feed forward is adjusted for the input voltage level in question. Since this is fixed in real-life it means the voltage under or over-shoots at some point depending on the input voltage/output current combination. Anyone using the chip with super-capacitors like I am is going to find the chip is 'over-sold'.

    I have a buck driving an LED on the output of the 61022 so it won't be a nice resistive load. I'm in production with it on two products so I did the best I could with it, which is a 120p capacitor and a hard restriction on the output current when the input voltage is low. The max current at higher input voltage level is around the 2A, due to voltage drop as stated above.

    Given the capacitance is so low perhaps TI should think about 'switching in' feed forward capacitance depending on the input voltage level, which would give a wider mode of operating range with expected regulation performance.

    I have since found I can remove the 220uF polymer, so my combination now is 5x 22uF 0805 10V ceramics and 120p feed forward. If I didn't use a fast 5MHz UVLO on the output and a max current restriction at low input voltage level the 61022 would burn.

    A.

  • 0 in reply to Andrew Charnley
    TI__Genius 12465 points

    Hi Andrew,

    The maximum output current  is actually related to the input voltage, because we limit the input current, the lower input voltage, the lower maximum output current.

    Regards,

    Nathan