• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 1 answer
  • Subscribers 60 subscribers
  • Views 257 views
  • Users 0 members are here
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

LM5180-Q1: LM5180 Voltage level problem

Max Av
Intellectual 470 points
Part Number: LM5180-Q1
Other Parts Discussed in Thread: LM25184, LM5180

Hi,

According to the datasheet, I determined the feedback resistor for the 3.3 output voltage.
However, I get more than 30 volts at the output, and there is no reaction to changing the FB resistor. Transformer 1:1

  • 0
    TI__Mastermind 35482 points

    Hi Max,

    You need a zener across the output to clamp the no-load voltage , e.g. 13V for a nominal output of 12V. If you apply some load, it should regulate as expected.

    If the turns ratio is 1 : 1, the feedback resistor should be 121k for a 12V output (see the LM25184 12V EVM for example). Also, the primary-side clamp zener should be set at 150% of the reflected voltage, something like 18-24V in this case. You can use the LM5180 quickstart calculator to assist with component selection.

    Regards,

    Tim

  • 0 in reply to Timothy Hegarty
    Intellectual 470 points

    Hi,
    Thanks for the quick response!
    As I understand it, if I put a TVS diode, then a current will always flow through it, and when the load is connected, the voltage will drop. Do you need to ensure the minimum current in the circuit so that the microcircuit receives feedback? What is the minimum current that must flow for correct regulation?

  • 0 in reply to Max Av
    TI__Mastermind 35482 points

    Hi Max,

    A Zener on the output is preferable as it provides an accurate clamping voltage (whereas a TVS clamp level varies with current). The Zener should be set at 110-120% of nominal Vout.

    There is a min switching frequency of 12 kHz, which dictates a certain amount of power delivered to the output = 0.5*Lmag*Imin^2*Fsw-min. The load current should be high enough to overcome this for regulation to occur (~0.5% full load).

    Regards,

    Tim