0
\$\begingroup\$

Below is a circuit I'm making. A 24 VDC PSU (VCC) goes through this switching regulator and outputs 5V to power the load. However, the load can also be powered directly by the USB 5V (USBVCC).

Now if the USB is connected but the 24 VDC PSU isn't, I'm getting a ~3.3V reading on the input line of the regulator. Is there a specific name for this? I've been calling this a "voltage back-drive" for now. Also, is this ok for the regulator or will it be damaging in the long run?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
5
\$\begingroup\$

You've got reverse current flowing in the regulator, probably through protection diodes in one or more of its chips.

It's a module, and it doesn't specify that situation, so you don't know if it'll be damaged or not.

If it were me, and I were determined to use a module there, I'd start by shopping around for one that does specify that sort of a back-drive situation.

If my back were up against the wall and I just had to ship product with that module, and I couldn't make changes, I'd measure the current flowing from the +5V line into the regulator -- if it's less than a mA, I'd reluctantly say the product is OK to ship -- then I'd go home and make sure my resume is current.

Ideally, if you use that module, you'd put a switch between that regulator and the rest of the circuit that cuts off the output of the regulator when the 24V supply drops below 8V, or when USB power is applied.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for the info! Now I'm thinking of putting a P-MOSFET between the output of the regulator and 5V line. Use the 24V line to let the Gate either open or close the MOSFET. \$\endgroup\$ – Agriculex Aug 25 at 15:36
2
\$\begingroup\$

The switching regulator is a buck and depending on the internals, there is a high chance this is due to the body diode of the switching FET

schematic

simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I see. If it's flowing through the body diode, then it shouldn't be too bad then? \$\endgroup\$ – Agriculex Aug 25 at 15:51
  • \$\begingroup\$ 5V - 3.3V seems like a lot of drop, though, especially at low current -- I'd expect more like 0.6V to 0.7V for a body diode (or am I cracked? I've never measured that at low current). \$\endgroup\$ – TimWescott Aug 25 at 16:05
1
\$\begingroup\$

"reverse leakage" might be the normal term.

It's probably in the microamps and 10K or 100K from IN to GND will probably just about eliminate it.

Worth trying as an experiment to measure the leakage current : I doubt it could cause any damage whatsoever.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for the info! \$\endgroup\$ – Agriculex Aug 25 at 15:52
1
\$\begingroup\$

You could add schottky diodes to isolate the two power supplies and prevent any reverse current being fed into them (which often is dangerous). In your design, if only one power source will be on at a time, then the attached schematic will work fine. Just make sure that the current rating and power dissipation of the schottky diodes exceeds your required supply current.

P.S - there could be a voltage drop around 0.4v for schottky diodes. So the Rload may see only 4.6V input. This usually isn't a problem as most 5 v sources like usb actually provide 5.1 volt, and after the diode drop it will be 4.7V which is mostly okay for a load which expects 5v. A simple solution is to use a higher current rated schottky diode which will drop less voltage at currents lower than its rated max capacity, but high current schottky diodes can have a higher reverse leakage current (sometimes more than 20mA) which you need to keep in mind. read the datasheet carefully.

SIMPLE POWER SUPPLY ISOLATING

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.