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UCC28700: Cable Compensation with TPS2511

Hagan McBride
Intellectual 260 points
Part Number: UCC28700
Other Parts Discussed in Thread: TPS2511, UCC28710, PMP8363, UCC28704

I have a power supply circuit (5V, 2.1A) and the power supply has a cable connected to it for the application. I am using the ucc28700 for the power supply circuit and I have a TPS2511 on the secondary circuit for "usb" power control. At the output of the TPS2511 I am getting a good constant regulation around 5V from 0A(no load) to 2.1A (full load); however, when I test the output at the cable end the voltage drops down to 4.75V @ 1A and 4.6V at 2.1A . I believe this has to do with the cable not being compensated on the TPS2511. The cable compensation pin on the UCC28700 is shorted to ground for the max cable compensation. Looking at the data sheet for the TPS2511 the ~CS pin is used for cable compensation for the TPS2511. The schematic shown uses a FB pin coming from the main power supply. How can I achieve cable compensation for the TPS2511 using the UCC28700 since there is no FB? Any advice would be appreciated.

Thanks.

  • 0
    TI__Intellectual 880 points
    Hey Hagan,

    The UCC28700 already takes care of cable compensation. That additional feedback network shown in the TPS2511 dedicated USB charger is for power supplies that regulate using a feedback network. UCC28700 uses primary side regulation by using information from the primary power switch and the auxiliary flyback winding to provide precise control over output voltage and current.

    Taking a look at the TPS2511 datasheet, Figure 13 shows what the output at the end of the cable can look like at various load currents for a charger. Your results are pretty close to what is depicted in the datasheet.

    I suggest taking a look at the PMP4351 reference design, as it is a 5V@2A universal adapter design and may fit your design specifications. From there, calculate the resistor value required for compensation and use that for your cable compensation instead of using the TPS2511.

    Another reference design I suggest taking a look at is the PMP8363, which uses the UCC28710 to generate a 5V@2A output and does not actually use the TPS2511 because it already takes care of cable compensation.

    I hope this helps,
    Davit
  • 0 in reply to DavitKhudaverdyan
    Intellectual 260 points
    Thanks,
    I currently have the CBC pin shorted to ground for the max cable compensation. My circuit seems to work if i take out the TPS2511 however, it does not seem to have great short circuit protection when the TPS2511 is missing. When there is a short I am getting 0V and 2.4A, with the tps2511 at 0V I get 0.00A. Is there another way to add more short circuit protection?
  • 0 in reply to Hagan McBride
    TI__Intellectual 880 points
    Hey Hagan,

    The UCC28700 datasheet Figure 17 shows the typical target output I-V characteristics. When the output current hits Iocc, the UCC28700 will go into CC (Constant Current) control mode while dropping the output voltage until it hits the UVLO turnoff, then your device will shut down.

    Take a look at the UCC28704, which has a 300mV cable compensated output (and replaces the CBC pin for an NTC control pin) and has a unique feature called CCUV (section 7.3.8 of the datasheet). This feature will reduce the output voltage until a certain voltage, then reduce the current as well.

    How are you doing your short circuit testing? Try monitoring the output voltage while doing the test and see if your waveforms show any indication of the current staying the same while the voltage level drops until hitting the UVLO turn off voltage.

    Regards,
    Davit
  • 0 in reply to DavitKhudaverdyan
    Intellectual 260 points
    Davit,
    I have my power supply hooked up to a DC load. The DC load has a short circuit feature. When I turn it on The voltage is cut off but the current ramps to 2.4A. There is obviously some SC protection with the UCC28700 or the circuit would no longer work. Is it bad that the current says 2.4A or should the current go down near 0A at a short circuit?
  • 0 in reply to Hagan McBride
    TI__Intellectual 880 points
    Hi Hagan,

    What should be happening is that once the device detects an over current situation it will keep the current level the same, but start lowering the output voltage until the lower turn off voltage, just like Figure 17 shows in the UCC28700 datasheet (page 15, link here: www.ti.com/.../slusb41a.pdf ). From there the device will attempt to restart.

    There may be something in the e-load you are using that is still content with its output and is not allowing the restart cycle to happen. It can also be possible that the device is clearing the short circuit fault, restarting, then going through the fault again. What you can do is to simulate the short using a hardwire short or using a power MOSFET switch between the output terminal and the ground to simulate that scenario as well.

    Have you tried to get a waveform of the output voltage during the short circuit test? If so, what is the voltage doing during this short circuit?

    Davit
  • 0 in reply to DavitKhudaverdyan
    Intellectual 260 points

    I have simulated the short with a hardwired short and I still get the high amperage. Looking at the waveforms the Voltage drops down below 1V close to 0V when there is a short. The CS pin on the UCC28700 should cause the Ucc28700 to shutdown once the CS pin's voltage goes above the voltage threshold there (0.75V). When I put a short at the output the CS pin never reaches the threshold voltage to shutdown the controller. Here is a model of my SC test. 

  • 0 in reply to Hagan McBride
    TI__Intellectual 880 points
    Hagan,

    Thanks for the additional information!

    Taking the new info into consideration, the solution is to use a -slower- diode for the auxiliary winding diode. On every reset cycle, the auxiliary diode would receive energy and use that to keep the IC powered on rather than shutting down. Using a slower diode, the slower forward recovery means that it will take a little longer for the diode to start conducting, such that the energy spike will effectively not be taken into consideration and the IC will shut down accordingly.

    Regards,
    Davit