Is it good practice to use resistors (parallel to load) on LDO output? The MIC29302WU LDO I've bought has ~7 mA minimum current requirement for stability reasons.

What other/better options I have? (Ignoring it, using Zeners (why?), NTC resistors ,...)

  • \$\begingroup\$ If you are using a 3 A LDO regulator, I would expect that your circuit would draw more than 7 mA and you wouldn't need a resistor. \$\endgroup\$
    – Mattman944
    Commented May 4, 2021 at 21:27
  • 1
    \$\begingroup\$ Regarding the "ignoring it" option: Your regulator's datasheet has a section titled Minimum Load Current that says "If the output current is too small, leakage currents dominate and the output voltage rises". Could the rest of your circuit tolerate an overvoltage when the current drops below minimum? \$\endgroup\$
    – Theodore
    Commented May 4, 2021 at 21:39
  • \$\begingroup\$ this LDO will supply only these ICs : one Atmega328AU (SMD) , one 74LS14 and five other chips supply. can I assume minimum current is maintained all the time? (no sleep mode is used in atmega) \$\endgroup\$ Commented May 4, 2021 at 21:47
  • \$\begingroup\$ You can choose a regulator that has no such minimum requirement. The TI 'cap free' NMOS output types don't require any load to be stable. \$\endgroup\$ Commented May 4, 2021 at 21:49
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    \$\begingroup\$ If you are set on using this regulator, the best thing to do is put a place in the design for a resistor in parallel. If it turns out that the circuit doesn't draw enough current, you can install the resistor. Otherwise you can leave the resistor pads empty, and maybe remove them in a later board revision if there is one. Don't agonize over it, though. Most LDO's do not have a requirement like this. At least the ones I have used. \$\endgroup\$
    – user57037
    Commented May 5, 2021 at 0:36

1 Answer 1


If the LDO is otherwise suitable for your project, and if current consumption during idle time is not critical, then yes - it can be good practice. In fact, if you look at the datasheet from any adjustable linear regulator you will see that the feedback network is often chosen to create the minimum load automatically.

Your IC is not an exception. The reference example is using suspiciously low resistor values for a simple divider. This way, the resistor divider gives a "free" 6-7 mA load at all times. There is a chapter mentioning that you can use this if you don't know how much your load consumes. If, as you say, you know that the CPU is always running at full speed you may be able to ignore this, but it could be better to design your hardware to allow for some future changes.

The alternative is to spend some time searching for an LDO more suitable for your project. If you care about power consumption you want to look for "Low IQ" circuits. I don't know about your input voltage and other requirements, but the first one I found from the same company is the MCP1703A, which doesn't seem to have a minimum load required at all, for stable operations.


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