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LM4120: LM4120 Input voltage range Question.

Part Number: LM4120

Dear Support member,

My customer used LM4120.

I have a question.

Q1
Datasheet (SLVS049F)The circuit of Fig 20 on page 11 contains transistors.

It is assumed that the voltage between the base and the emitter of the transistor added to the IC is about 0.6 V and the input voltage of the IC above + 0.6 V is necessary.

If trying to output the output voltage of 3.3 V, the input voltage is not 3.5 V - 5.5 V but 4.5 V - 5.5 V, is not it?

Best regard.
Bob Lee.

  • Hi Bob Lee,

    I agree that the input voltage should be 4.5 V - 5.5 V if this schematic is going to fully use the FET and output 3.3V at 1 A. I feel the original creator used 3.3V input voltage for no load conditions. I would be more comfortable with a higher voltage such as 3.5 V minimum just for the voltage drop on the resistor.

    -Marcoo
  • Dear Marcoo Z,


    Thank you very much for reply.

    I ask the data sheet to be corrected.

    Best regard.
    Bob Lee.

  • Dear Marcoo Z,

    May I ask add a question?


    Your reply answer that
    『I agree that the input voltage should be 4.5 V - 5.5 V if this schematic is going to fully use the FET and output 3.3V at 1 A.』


    Although the circuit diagram of the data sheet is a PNP transistor,
    is it the same also answer when using a PNP transistor?


    Best regard.
    Bob Lee.

  • Hi Bob Lee,

    I will add this fix into the next datasheet revision.

    I assume you meant NPN instead of PNP. You can use a NPN to replace the PNP but it will not work in that configuration. Typically NPN current boosting applications have the base connected to the output of the VREF instead of the input to make sure that the device turns on when it needs too. If the base is connected to the input then the NPN will turn on as soon as the Vin is greater than 0.6V and not be regulating properly. It will also have no current limit and sink current into the voltage reference before the reference is even turned on. It is more common to have NPN current boosters in applications that have feedback pin to adjust for the voltage drop.

    -Marcoo
  • Dear Marcoo Z,

    Thank you very much for reply.

    There are additional questions from the your reply answers.

    Datasheet D45HB in Figure 20 on page 11 is a PNP transistor.


    Q1.
    Is this circuit not working?


    Q2.
    In order to operate this circuit,
    How can I do it?

    (Ex)
    Vin voltage range change?


    I attached question file.


    Best regard.
    Bob Lee.LM4120_Question.pdf

  • Hi Bob Lee,

    I have added my answers to your questions below. Sorry for any previous confusion.

    Q1.
    Is this circuit not working?
    A1.
    This circuit works because it is a PNP.

    Q2.
    In order to operate this circuit,
    How can I do it?
    A1.
    In this circuit the PNP is being used to supplement the current of the LM4120 without being on at all times. The purpose of the 500 Ohm resistor from the base to the emitter of the PNP is to ensure there is a sufficient voltage difference to turn on the PNP. Typical Vbe values to turn on PNPs are around 0.6V. Using this value we can see that the PNP will turn on and allow current to flow from emitter to collector once there is a minimum of 1.2mA going through the resistor into the LM4120.

    Another way to look at this is that the LM4120 will source all the current up until 1.2mA and the PNP will remain off. Once the LM4120 sources more than 1.2mA the PNP will turn on and supplement the LM4120 up to the desired load condition that is limited by the transistor.

    (Ex)
    Vin voltage range change?

    Vin Min (No Load)
    Vout + Drop out + Iq*500 Ohms = 3.3V + 0.08 V + .138 V = approximately 3.52 V
    This value is higher than the 3.3V in the image.
    When you start adding in a load then the voltage will need to be higher.


    -Marcoo
  • Dear Marcoo Z,

    Thank you very much for your answer.

    Best regard.
    Bob Lee.