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Voltage dropout at power up.

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New Member
I drive my PCB with a 5V USB that comes from the PC.

My PCB consumes 50mA, and I noticed that when I connect the PCB to the USB, then the RST pin of the MCU is getting lower and then getting HIGH, causing an external reset to the MCU right after power-up.

I know that this happens due to voltage dropout, since when I used instead of the USB, a "strong" DC power supply, there was no external reset, I assume that its because the DC poewr supply can sustain the 50mA current drawing.

I have always thought that current drawing causing a consistent voltage dropout since the wires have resistance.

Now I experience that the voltage dropout is just for a brief moment at start-up.

My question is:
Why does drawing current cause a breif voltage dropout and then the voltage gets back to its normal level?

Thank you very much.
Last edited:


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USB is very dependent on what mother board you are using.
Often there is a reset-able fuse (PTC) on the USB ports. It is a fuse that has low resistance when the current is low and when faced with high current it heats up and has high resistance, maybe 1k ohm. With low current it will cool off and have low resistance again. Some mother boards have a solid state version of a resetable fuse.

Your PCB probably has too much capacitance on its +5 volts line. When the USB is trying to charge the input capacitor, current limit happens. There is a spec for ‘max capacitance on supply’ for a USB device.


Well-Known Member
Most Helpful Member
Question: even though your circuit only draws 50mA, how much capacitance is hanging across the 5V going into it?


Mike's probably got it, if you're using a decent amount of decoupling caps the current draw when the circuit is first connected is going to be large very briefly.
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