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Use of the UCC39002 without sense+ input

Leonardo Muricy
Intellectual 515 points
Other Parts Discussed in Thread: UCC39002, UC3907, PMP10099

Hello,

I'm developing supply system that must have current equalization when paralleling units.

Searching for Texas chips, I read the data sheet and the app note for the UCC39002 (also for the UC3907, but this one seemed to have more than I needed) and it looks like it can do what I need, but it uses a Sense+ input from the power supply, which is not present on my project currently.

The supply's project has a digital control, using the Texas tm4c129x chip, in a way that I can actually adjust it to work with the chip, but for that I need to know exactly what kind of signal the ADJ pin generates as output to the system.

Is it simply a voltage signal to be added (or subtracted) to read output voltage when comparing to the reference? Does this voltage need to be multiplied by any factor?

Thank you for your help

Just to complete the information, the supply has a half bridge converter output configuration, with 45V - 59V and up to 50A.

  • 0
    TI__Mastermind 34465 points

    0820.sch.pdf

    I am attaching a schematic page from the TI reference design PMP10099 and this shows how the UCC39002 is wired up.

    You can see that the ADJ pin adds an error signal to the feedback path of the power supply and this forces all of the modules to share equal currents.

    In your application , you can control the load sharing of the 45V or the -59V rails but you need to be able to access the feedback path of your power supply.

    The product page located at

    http://www.ti.com/product/UCC39002/toolssoftware

    includes reference designs and details as to choosing circuit component values.

    Regards,

    John Griffin

  • 0 in reply to John Griffin1
    Intellectual 515 points

    Thank you for the answer, John.

    I actually have 1 output only, with its value adjustable in the 45V to 59V range continuously. 

    The current feedback path of the supply is just an Analog to Digital converter on the output voltage (with a voltage divider). 

    In order to get the signal I would just have to read the adj signal and use it to add on the feedback variable of the algorithm. What I need to know is the range of values the pin Adj can have. For example, can it be negative? 

  • 0 in reply to Leonardo Muricy
    TI__Mastermind 34465 points

    Hi Leonardo,

    The ADJ pin needs to be connected to a positive voltage and

    VEAO + 1V  < VADJ < VDD

    The max value for VDD is 15V

    ADJ is a current sink and will sink a max of 3.95mA (typical) at the upper limit of the current adjusted range.

    (See page 4 of data sheet)

    John

  • 0 in reply to John Griffin1
    Intellectual 515 points
    This is exactly what I wanted to know, John. Thank you very much.
    Just one more doubt then, on the calculation of the feedback signal, does the adj pin add with the output voltage to make the real offset or does it subtract?

    In other words:
    If I had 50V of reference on the control, 50V of output voltage and 2V on the Adj pin would the next error value be:
    1) 50 - (50 + 2) = -2V (which would make the control try to make the output voltage 48V).
    or
    2) 50 - (50 - 2) = 2V (which would make the control try to make the output voltage 52V).

    My ideia is to use a pull-up resistor from Adj to Vdd and read the voltage using the ADC.
  • 0 in reply to Leonardo Muricy
    TI__Mastermind 34465 points
    ADJ is an error signal.
    As ADJ increases it should force your power supply to turn down its output.
    Its a negative feedback system.
  • 0 in reply to John Griffin1
    Intellectual 515 points
    Thank you very much. That's all the information I needed.
    I'll use the schematic from figure 4 of datasheet.