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I have a circuit, Arduino Leonardo clone, which I'm attempting to power with a buck regulator. Specifically, I'm using a DSN MINI 360. It's based on the MP2307 buck regulator. If I power my board through the buck regulator, the board works fine. However, if I disconnect the regulator from input power and then try to power my board with USB, the board seems to draw too much current until the PTC kicks in. With the regulator connected to the +5v on my board, there isn't enough power supplied through the USB circuit to enable the circuit to run.

If I disconnect the regulator and power the board through USB, the circuit works normally.

I do not have a schematic for the regulator, but it seems to follow the typical application circuit described in the regulator's data sheet.

Below is the complete schematic for my custom circuit based on the ATMEGA32u4. The circuit, as built, is complete, except U2 (SPX3819 Linear Regulator) has been removed. All other components are present on my circuit board.

My question is why does the buck regulator seem to be drawing so much power through its output, and is there any way to prevent this from happening without having to disconnect the buck regulator when using USB? Would a diode be appropriate in this situation?

Circuit Schematic

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  • \$\begingroup\$ Have you installed the soft start capacitor on the regulator board? \$\endgroup\$ – John Birckhead Aug 1 '17 at 22:42
  • \$\begingroup\$ You say you disconnect the regulator from input: is this true, or maybe you ground the buck regulator input? \$\endgroup\$ – Vladimir Cravero Aug 1 '17 at 22:56
  • \$\begingroup\$ Even if you do not ground the input, you power the regulator via the SW node. It tries to start up, failing, possibly with several low side pulses to charge the bootstrap cap. \$\endgroup\$ – Vladimir Cravero Aug 1 '17 at 22:59
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Your buck regulator is trying to run backwards

Your buck regulator will power up if voltage is applied to its output with its input floating -- this happens through the body diode of the high-side FET inside the buck IC. The overcurrent is coming from the resulting confusion, I bet -- the controller may try to turn the low-side FET on, which shorts the SW node, and thus your 5V supply, straight to ground via the switcher inductor. Oops!

You have the right idea with the diodes, just use the right diode for the job

The easy fix is to stick a pair of suitable Schottky rectifiers (a couple of standard 1N5819s will do) in series with both the USB and buck converter power lines -- the ATmega isn't picky about its power supply, and you want to protect the USB power line this way as well because backpowering USB ports has been known to be a motherboard-frying mistake.

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  • \$\begingroup\$ Thank you. One thing I noticed with the diode was that as the load increased, the voltage drop across the diode also increased. I don't yet fully understand why this is. A diode, if it drops .7v, should drop .7v, right? The regulator should compensate for any load. 6v out of reg, -.7v diode drop, should get 5.3v as long as load is within limits of reg. However, before the diode I see 6v, but after the diode, the voltage drop becomes more pronounced as the load increases. The load never goes over 20mA. \$\endgroup\$ – Steve Groesz Aug 31 '17 at 8:36
  • \$\begingroup\$ @SteveGroesz -- yes, you'll see an increase in drop (Vf) as current (If) increases -- look at the Vf/If curve on any diode's datasheet \$\endgroup\$ – ThreePhaseEel Aug 31 '17 at 11:37

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