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LM4132-Q1: loss of power supply

Lutz Naumann
Expert 6140 points
Part Number: LM4132-Q1
Other Parts Discussed in Thread: LM4132

Dear Team,

I have a specific question regarding LM4132CQ1MFR3.0. In a design we can not always guarantee the LM4132 is supplied correctly or supplied at all, whereas the rest of the circuit might operate. The following two conditions can happen:

1) Complete loss of power supply. Voltage present at the output but not higher than the max ratings. The question is how does the part behave? We will see a leakage current through the feedback divider, What is the max current we have to assume? Also does the transistor block or will there be some leakage path through the bipolar transistor as well and what current / behavior do we expect here? Is there an ESD Structure which does start conducting?

2) Vin can be as low as 2,9V. I understand that the REF can not regulate anymore. It is assumed the Vout tracks Vin with a Vdrop. Firstly at 2,9V does the part still operate as described? Secondly what will be the voltage drop from Vin to Vout be at 2,9V supply?

many thanks

Lutz

  • 0
    TI__Expert 3835 points

    Hi Lutz,

    In the design, do you expect the LM4132 to supply load current when supply is not above the 3V PLUS dropout voltage?

    1.) Expect max current to be the maximum supply current(Iq) spec on the datasheet. I believe at 2.9V, the transistor will not block and there will be leakage current. Looking at the graph below from the datasheet, When the input voltage is near the output, there is a spike in supply current, before it goes into the linear dropout region. Although this is a graph for Vref = 2.048 expect similar behaviors for the +3V part. 

    2.) At 2.9V the REF does not regulate but will have an output voltage tracking the input voltage like you said. The Vdrop will depend on the load current. The minimum input voltage for regulation is 3.175V at 10mA load.  Expect Vdrop to be at minimum the dropout voltage plus some effective resistance from the BJT since it will be in the linear region. I can't give you a definite answer since we do not characterize it under minimum input voltage. We also can't guarantee the voltage drop will be the same across since load and line regulation is out of spec.

    Hope this helped,

    Ethan

  • 0 in reply to Ethan T
    TI__Expert 6140 points

    Thanks Ethan,

    The REF is supposed to set the voltage for an ADC of an MCU, this will be the load condition. We wanted to understand if in case the voltage is down to 2,9V the output is still kind of predictable say within a 20% error condition, so that the ADC could detect the loss of supply voltage by measuring Vsupply of the REF with the output of said "bad" REF.

    Regarding the unpowered state I understand we have to assume the complete Iq as leakage. This does imply on the other side that the part will survive this condition, correct?

    What about the max current thought the resistive divider? What is the lowest value of those resistors?

    thanks

    Lutz

  • 0 in reply to Lutz Naumann
    TI__Expert 3835 points
    Yes it is safe to assume that during unpowered state, the supply current would be considered iq leakage current.

    Regarding the current and values of the resistor dividers, I will get back to you as soon as possible.

    Ethan
  • 0 in reply to Ethan T
    TI__Expert 6140 points

    Ethan,

    how are we doing on the resistor diver front? Can we have the information please? This is kind of urgent now as we started this conversation a week ago.

    thanks

    Lutz

  • 0 in reply to Lutz Naumann
    TI__Expert 3835 points

    Hi Lutz,

    I have followed up with our design team, but they have yet to come back to me. I will forward them the urgency and get back to you when I have an answer.

    Ethan

  • 0 in reply to Ethan T
    TI__Expert 3835 points
    Hi Lutz,

    The design team cannot give an exact value of these resistors by looking at the schematic/layout of the device. In previous testing, we put a source across the output and estimate the range of values it can be.

    Seems like the part would fail around 20uA of continuous source current. R1 is around 150K to 300K and R2 is around 100K to 200K. These resistor values are trimmed to match the bandgap reference voltage that varies across devices.