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I was looking at a breadboard power module (link). This board has two inputs: a DC plug (7-12V advised) and a USB mini (5V). Using either of those two inputs, the board provides two voltages (3.3V and 5V) using fixed AMS1117 voltage regulators. So far, so good.

However, looking at the module, and after some path testing with a multimeter, it seems that something strange is happening. When using the USB as input to power the module, the 3V3 regulator gets its input through the 5V regulator. But in a strange reversed way. (The voltage at the input pins of both AMS1117's is around 4.5V.)

The schematic of the board is this (excluding the output headers and switches): Schematic of the power supply module

Using either DC or USB as input power, both 5V and 3V3 rails are working. But I suspect that the 3V3 rail using USB as input power, might not be a proper supply.

Am I correct in assuming that this reversed use of the 5V regulator to power the 3V3 regulator is less than desirable? Or am I missing something?

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    \$\begingroup\$ Seems normal. You are experiencing the body diode or protection diode built into the linear regulator. \$\endgroup\$
    – winny
    Commented Feb 24, 2020 at 20:27
  • \$\begingroup\$ Yes that would explain the slight voltage drop (5v -> 4.5V), but all power drawn from the 3V3 rail, should now pass in reverse through the 5V regulator. Is that ok? \$\endgroup\$
    – Philip
    Commented Feb 24, 2020 at 20:29
  • \$\begingroup\$ Yes that is OK, see the datasheet, page 4 section "Protection diodes". Diodes between the input and output are not usually needed. Microsecond surge currents of 50A to 100A can be handled by the internal diode between the input and output pins of the device That sounds like an ESD protection diode which can also handle the currents you're using. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 24, 2020 at 20:30
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    \$\begingroup\$ "The best way" depends on what is best. For a cheap module like this it is all you need. A proper switching circuit with MOSFETs and whatnot would be overkill and make the module more expensive. Would I design like this for a critical system in an aircraft? Nope. Is it good enough for a hobby and experiments module: Yes. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 24, 2020 at 20:36
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    \$\begingroup\$ You might want to put a Schottky diode in there, since some regulators are stressed by too much reverse current. A Schottky will also have a lower voltage drop. Also, I think you have your regulators switched above. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Feb 24, 2020 at 21:09

1 Answer 1

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Using an internal "feature" of an IC for an unintended purpose may always work, or sometimes work, or stop working in un-understood circumstances.

In this case, adding a Schottky diode across the regulator from Vout to Vin to take the 'reverse' current would be an effective, cheap and sensible solution.

TI & Nat Semi datasheets contain references to an internal protection diode.
A sample Asian datasheet did not. They still MAY exist.

TI datasheet
Section 7.4 page 12. Figure 13

Natsemi data sheet page 10 fig 4

Both say:

Under normal operation, the LM1117 regulators do not need any protection diode. With the adjustable device, the internal resistance between the adjust and output terminals limits the current. No diode is needed to divert the current around the regulator even with capacitor on the adjust terminal. The adjust pin can take a transient signal of±25V with respect to the output voltage without damaging the device.

When a output capacitor is connected to a regulator and the input is shorted to ground, the output capacitor will discharge into the output of the regulator. The discharge current de-pends on the value of the capacitor, the output voltage of the regulator, and rate of decrease of VIN. In the LM1117 regulators, the internal diode between the output and input pins can withstand microsecond surge currents of 10A to 20A.

With an extremely large output capacitor (≥1000 μF), and with input instantaneously shorted to ground, the regulator could be damaged.In this case, an external diode is recommended between the output and input pins to protect the regulator, as shown in Fig 4 (13)

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