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My board will be mostly powered with an already regulated 5V input from USB (or wall charger), but I want it protected for higher supply voltage in case somebody plugs in the wrong voltage.

So I'm putting a voltage regulator on the board after Vin.

Initially I was planning to use the LP2985 but realised I may need to draw higher current, probably around 1A, so I'm considering the LM1117.

I see it is used on the Arduino Nano, which accepts 5V as an input, but after reading the datasheet, I understand Vin needs to be higher than Vout.

Can I use the LM1117 to convert 5V to 5V ?

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    \$\begingroup\$ This is an X-Y problem. You need to detect over-voltage and do something with that, maybe crowbar a fuse, not add extra regulation. \$\endgroup\$
    – Trevor_G
    Commented Feb 11, 2018 at 14:25
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    \$\begingroup\$ convert 5V to 5V ? I would just use a piece of wire. Anyway an LM1117 needs an input voltage that is slightly higher than the output voltage, how much higher: read the datasheet. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 11, 2018 at 14:34
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    \$\begingroup\$ From memory, the Arduino Uno has USB in and DC jack in. You might check its schematic to how they managed it. \$\endgroup\$
    – Transistor
    Commented Feb 11, 2018 at 14:37
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    \$\begingroup\$ Reverse voltage protection is simply a series diode. But you must account for the voltage drop of the diode. \$\endgroup\$
    – bill
    Commented Feb 11, 2018 at 19:12
  • \$\begingroup\$ @bill This doesn't account for over-voltage. \$\endgroup\$
    – Daniel
    Commented Feb 11, 2018 at 19:29

2 Answers 2

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Adding another regulator is a bad idea since it complicates the power line.

A better solution is to protect your circuit by shutting off the power when the input voltage exceeds some limit. The circuit below will disconnect the source if the voltage goes over 5.5V. You can make it closer to 5V by reducing the Zener voltage from the shown 5.1V device.

schematic

simulate this circuit – Schematic created using CircuitLab

Description:

When Vin is under 5.5V the zener is off as is Q1. R3 holds the gate of M1 low turning it on as a low resistance. When Vin rises above 5.5V D1 holds the base of Q1 down turning it on which pulls up the voltage on R3 to close to Vin, which turns off M1.

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    \$\begingroup\$ Hi Trevor would you please give some background on how this circuit works? \$\endgroup\$
    – Dumbo
    Commented Feb 11, 2018 at 16:29
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    \$\begingroup\$ @Sean87 added.. \$\endgroup\$
    – Trevor_G
    Commented Feb 11, 2018 at 16:32
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The LM1117, although very cheap, particularly in clone versions, is not really an LDO regulator, more of a medium drop-out regulator.

You can use a CMOS LDO regulator as you suggest, it will just appear as a resistor when your input voltage is 5V, however you would also have to account for the power dissipation at maximum input voltage (and many of them are not rated for much input voltage to begin with). Trevor's circuit cuts off the device for overvoltage so it does not suffer from that particular ailment.

Another issue is the body diode of the internal pass MOSFET which will not provide reverse voltage protection, so you would need to add another external MOSFET and zener diode + resistor to get reverse voltage protection (if needed).

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