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Original question:

BATTERY BOOST CONVERTER GSM/GPRS APPLICATION

BATTERY BOOST CONVERTER GSM/GPRS APPLICATION TPS61230A

Blindu Marcel
Intellectual 325 points
Other Parts Discussed in Thread: TPS61230A, TPS61022

There is some way to reduce the inrush current of the converter TPS61230A?

We need 150mA of inrush current. 

  • 0
    TI__Mastermind 34231 points
    Hi Blindu,

    When the device is enabled, the high-side FET turns on to charge the output capacitor linearly by a DC current which is called the pre-charge phase. The pre-charge startup phase terminates untils the output voltage being close the input voltage. The pre-charge mode current is 0.06A(min), 0.6A(max) when Vout =0 V. And it is 1.02A(min) when Vout is 2.5V. It is designed inside the IC so we have no way to change this value.

    What's the application working condition, Vin range, Vout, Iout? I'll see if there is another device that meets your demand if you can tell me these information. Another way is adding a load switch IC which has current limit feature.
  • 0 in reply to Zack Liu
    Intellectual 325 points

    Application working condition:

    • Vin:LiSOCl2 battery (2-3.67V)
    • Vout: 2G Module: 3.8V @ 2A

    By the way:

    • Increase the input capacitor of the converter reduce the inrush current?
    • Decrease the output capacitor of the converter reduce the inrush current?

  • 0 in reply to Blindu Marcel
    TI__Mastermind 34231 points
    Hi Blindu,

    1. TPS61230A input voltage range is 2.5-4.5V. So it's not suitable for the LiSOCI2 battery input. During the startup, Is the load 2G Module working? Could I say the boost converter starts up in no load condition?
    2. Why do you hope the inrush current is not higher than 150mA? When the converter is working at 3.8V@2A, the average input current is much higher than 150mA.
    3. Increase the input capacitor will cause higher inrush current because the input needs to absorb more energy from battery. Decrease output cap could help. But we need to consider the output voltage ripple and load transient voltage drop.
  • 0 in reply to Zack Liu
    Intellectual 325 points

    I explain you the application:

    The microcontroller enable and disable the boost converter and the inrush current appear when the enable signal goes from 0 to 1. I think the inrush current is because there is a large capacitors at the output of the boost converter. Maybe increase the input capacitor C9 decrease the inrush current when we switch the enable signal.

    1. The boost converter starts up in no load condition but exist a large capacitors of the output of boost converter.

    2. What I don`t want is that when we enable the converter the battery voltage goes down a lot and reset the system.

    3. Increase the input capacitor will cause higher inrush current because the input needs to absorb more energy from battery but only when connect the battery not when the enable signal of the boost converter goes from 0 to 1.

    I attach the boost converter schematic and output capacitors of the boost converter.

  • 0 in reply to Blindu Marcel
    TI__Mastermind 34231 points
    Hi Blindu,

    Thanks for explaining it.
    1. Yes, if using signal to control EN pin, the input capacitor C9 voltage is battery and increase C9 could reduce the inrush current. and reduce output cap could reduce inrush current. Use more 22uF in parallel with C9 to see if it will help.
    2. The 150mA inrush current is a peak value or average value? Have you used current probe to see how inrush current pull down the battery voltage?
    3. My concern is TPS61230A input voltage range is 2.5-4.5V. it could not work in Vbattery = 2-2.5V condition. TPS61022 vin range fits this application. You can take a look at this part.
  • 0 in reply to Zack Liu
    TI__Mastermind 34231 points
    Hi Blindu,

    Do you have any updated news? I'll close this thread now since there is no replies. You can post a new thread if you have further findings.