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For a battery-powered application, I'm trying to reduce the current draw of a chip when its voltage rail is removed.

What is the current through the enable pin's pull-down resistor when the 5V USB-supplied rail is removed? In particular, I'm using this TPS62A01 step-down converter as shown here with a 1MΩ pull-down.

1MΩ pulldown

Shutdown current

Enable pin voltage and current characteristics

Can I base the pull-down resistor value on the enable input leakage current from the datasheet as explained here? Chosing a pulldown resistor

Or should I base it off the max shutdown current as suggested here? Pull-Down / Pull-Up Resistor value for minimum current leakage

My thinking for the 1MΩ resistor value is based on the enable input leakage of 100 nA: Ven = 100nA * 1MΩ = 0.1V which is less than the enable voltage threshold.

EarthIsHome
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    But that 1 Mohm resistor has nothing to do with it. When the chip is powered - V_USB_5V0 = 5 V, then the input leakage current comes from the supply, not through the pulldown resistor. The 1 Mohm pulldown carries 5 uA of current from the supply. – SteveSh Dec 20 '21 at 00:46
  • Knowing what you said, is the OP's request impossible, difficult or unnecessary? – GT Electronics Dec 20 '21 at 01:36

2 Answers2

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What is the current through the enable pin's pull-down resistor when the 5V USB-supplied rail is removed?

Well, when you remove that power source, you also remove it from the chips power pin (pin 3) hence, the chip and pull down resistor are totally isolated from any power source and, there is no leakage current argument to make.

My thinking for the 1MΩ resistor value is based on the enable input leakage of 100 nA: Ven = 100nA * 1MΩ = 0.1V which is less than the enable voltage threshold.

When the chip is powered then, the current through the 1 MΩ resistor is 5 μA.

Andy aka
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The 1M resistor has no function because you have VIN and EN shorted. If you are interested in saving power, remove it altogether.

tobalt
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