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I need a linear regulator for a project, for current sense reasons I need to select a regulator that passes its quiescent current into the load.

Most regulators like this have a minimum load current of a few mA. I cant seem to find anything on what happens if this condition is not met though.

Is it simply a case of regulation quality will degrade? Can I assume that since it is a parameter not listed in the absolute maximums that the chip will not be damaged.

We are currently looking at using the LD1086

I.e. if I don't care about tight regulation in no load condition can I safely use the part.

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If you don't draw the minimum load current, the output voltage will likely rise out of regulation, which could damage the circuitry that is powered. The regulator itself won't care.

For this reason you should use the worst-case value which appears to be 10mA.

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  • \$\begingroup\$ In this case the reason current would drop below the minimum value is because the device being tested is disconnected, when connected the device should be robust enough to deal with the transient while the supply goes back into regulation. \$\endgroup\$
    – Hugoagogo
    Aug 23, 2019 at 2:53
  • \$\begingroup\$ I would guess worst case it won’t be much worse than the output cap charged to the full input voltage. \$\endgroup\$ Aug 23, 2019 at 3:18
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You are misinterpreting the datasheet.

The fixed output voltage versions of the LD1086 do not have a "minimum load current". Only the adjustable voltage versions have. You need this value to calculate the maximum resistor values for the ADJ pin voltage divider.

If you choose too high resistor values the internal circuit cannot draw enough current for its own function and the regulation becomes unstable. Most likely, the output current and voltage drop to zero then.

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  • \$\begingroup\$ I didnt realise the distinction between the Adjustable and Fixed, we are using the adjustible version. As I read it on Table 8 Io(min) is 3mA typ 10mA max. The adjust pin current, which defines the required impedance of the circuit driving it is 40uA typical \$\endgroup\$
    – Hugoagogo
    Aug 23, 2019 at 2:04
  • \$\begingroup\$ Yes, but you still need the voltage divider having a 10mA current run through it from output to ground. That's why it says minimum load current, not minimum adjust pin current. Take a look at the inner circuit schematic and you see how the output is the common node for the whole inner circuit. \$\endgroup\$
    – Janka
    Aug 23, 2019 at 4:34
  • \$\begingroup\$ I understand to get the rated spec, 10mA load is required (and the load could come from anything, not just the divider used to set the current). My question was more on what will happen if I do not apply 10mA load. \$\endgroup\$
    – Hugoagogo
    Aug 23, 2019 at 6:04
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    \$\begingroup\$ The load could come from anything, but you need the divider anyways. \$\endgroup\$
    – Janka
    Aug 23, 2019 at 6:08

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