I have a circuit with an Exar XRP7659 5v 1.5A regulator and a Torex XC6210B 3.3V LDO Regulator. I am using this circuit to charge USB devices at full current and have noticed that when the device is charging I get negative voltage from any two ground points on the breadboard.

It appears that this negative voltage is coming from the GND pin on the USB port, as I read the highest negative voltage there, but it's seen throughout the circuit. The result is that the LDO goes slightly under voltage to 2.8 - 3.1 volts. The LDO returns to 3.3v when the USB device is unplugged. The issue is not apparent when using a 5V 1.5A power source but this results in ~4.4V out of the 5V regulator, which is unacceptable. I am using (and have tried multiple) a 12V 1.25A DC input source.

What could cause this negative voltage and how do I resolve it?

Here's a schematic for my circuit Power Schematic

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    \$\begingroup\$ You say you are seeing negative voltage on ground. Question: measured against what? In other words, where is your other meter probe connected? If it is on another ground point, this indicates resistance between different points on your ground. \$\endgroup\$ – DoxyLover Apr 15 '15 at 23:56
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    \$\begingroup\$ Additional question: why are you feeding 3.3V into a USB device? USB is always 5V. Feeding 3.3V in is not likely to work well. \$\endgroup\$ – DoxyLover Apr 15 '15 at 23:57
  • \$\begingroup\$ Aha! You're right - I measured against the ground point of another DC wall wart and measured no voltage across the two grounds. There is however 50-60k ohms of resistance when the device is charging and no resistance at all when the USB device is unplugged. As per your USB question, sorry if I was unclear but I am charging off the 5V - The 3.3V has no load on it and is for an unintegrated portion of this circuit. \$\endgroup\$ – t3ddftw Apr 16 '15 at 0:05
  • \$\begingroup\$ Thangs are still unclear. How are you measuring resistance? You cannot measure resistance on a powered circuit with a standard meter. Turn off the power and measure resistance between the ground points. Most likely you'll see a few ohms or tens of ohms in your ground path. That is dropping the voltage when under load (Ohm's law: V=IR). \$\endgroup\$ – DoxyLover Apr 16 '15 at 1:06
  • \$\begingroup\$ I measured resistance with the circuit powered on initially. You're correct, there's about .5 Ohms of resistance from ground to ground on the unpowered circuit. When I power the circuit, resistance stays about the same (1 Ohm) but when I hook up the USB device to charge, resistance jumps anywhere between 50k-110k Ohms. \$\endgroup\$ – t3ddftw Apr 16 '15 at 1:21

As discussed in the comments, you have about a 0.5 Ω resistance in your ground wiring. You are losing about 0.6 volts when under load. Ohm's Law (V=I*R ==> I=V/R) says I=~0.6/~0.5 = about 1.2 amps flowing through that ground. It all adds up.

Fix the resistance in the ground wiring and you'll fix your voltage drop.


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