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LMR62014: Reduce Input Current ripple when driving 2 Wire FAN Load

Deniel Go
Intellectual 395 points
Part Number: LMR62014


Hello,

We want to know if the input current ripple we are seeing from the boost converter is the result of having a 2 wire brushless dc fan connected at the output/load. 

Below is the circuit we have for the boost converter. It has an output set to 12V. The input voltage range is from 5.5 to 8.4V. The average current of the fan is around 300mA at 12V. 

The output of the boost converter is connected to the FAN and LED driver circuit that consumes around 30mA. We are using a current probe to see the waveform of the FAN and boost converter input currents. 

Below is the photo of the FAN current waveform coming from a 13V Power Supply. From datasheets of brushless fans, this is the expected waveform of a 2 wire brushless DC fan.

RMS current is around 376mA, peak to peak current of 456mA and a peak ripple current of 496mA. The values will be a bit lower when the input voltage is 12V.

 

The PCBA we have has two of those boost converter circuits shown above. The way it connects to the FAN and LED circuit is via a cable of 6FT in length. Output of both boost converters is around 12.5V and the voltage when it reaches the fan/LED circuit is around 12V. Both Boost converter share the same input voltage. When powering ON our board with just the FANs and LEDs and the microcontroller (consumes only 150mA), below is the photo of the joined input current waveform both boost converters. The input current waveform is similar to the waveform of the FAN with higher current values because of the conversion ratio of a boost converter. 

The RMS Current is 1.88A, Peak to Peak current is 2.44A and the Peak Current 2.6A. There is about 0.8A upper peak current in the input of the waveform. This is our main concern.

What we want to know is, Is this the typical behavior of a boost converter? Does the input current (before the input capacitor) follow the shape of the output current of load attached to it?

Is there a way to a have filtered version of the input current where it does not follow the shape of the fan current? Is there an additional circuit to add to improve the input current? 

This current waveform (fan current shape) bleeds into the DC line. Is there a way to avoid it? 

Let me know if you need more data or information regarding our design

Input waveform of both boost converters joined. 

Best Regards,

Deniel

  • 0
    TI__Mastermind 19620 points

    Hi Deniel,

    While the boost working properly, the input current will follow the shape of the output current. For the converter ratio of Vout/Vin, the input current can be regarded as an "enlarged version" of the output current.

    As in the application you mentioned that has a rapid change in Iout, usually placing a large electrolytic capacitor around 150uF between the input power source and Vin will work.

    Just let me know if you have further questions.

    Regards

    Lei

  • 0 in reply to Lei Zhong
    Intellectual 395 points

    Hello Lei,

    Thanks for the response and clarification regarding the boost converter's operation. I want to verify the location of the 150uF capacitor you mentioned. 

    What do you mean by 150uF between the input power source and Vin? Do we place the 150uF from the positive pin of our power source to the positive pin of the Boost converter Vin? Or is one pin of the capacitor connected to both VIN and VBAT and the other pin connected to Ground?

    Which of the two schematic diagrams below are you referring to? 

    Thanks,

    Deniel

  • 0 in reply to Deniel Go
    TI__Mastermind 19620 points

    Hi Deniel,

    Sorry for the confusion. I refer to the right schematic diagrams in your picture.

    Regards

    Lei