Voltage 5V pin difference Argon vs Boron


We are using the VUSB (Boron) and VIN (Argon) to power a 5V device.

Measuring with a multimeter we found that all Borons have a voltage drop (4.76V) while this does not happen with the Argons (5.06V) using the same adapter rated for 3W (5V 0.6A 3W), which causes our external sensor to give incorrect values.

Is this a known issue? I could not find anything on the forum regarding this.

Kind regards,


I don’t quite understand the situation. Which pin are you using on the Argon? VIN isn’t a labeled pin on the Argon.

Also are you using VUSB as an input or an output? If an output, what is the power input? Micro USB?

If you are using micro USB as the power input and VUSB as an output on the Boron, the voltage drop is likely caused by the Schottky diode that prevents voltage from flowing from VUSB to the USB connector. This is necessary because the VUSB pin can be used as a power input up to 17 volts on the Boron because of its bq24195 PMIC, and that voltage would damage a USB device plugged into the micro USB.

On the Boron the best option if you need exactly 5V is to use a 5V power adapter to power your sensor and VUSB as an input instead of using Micro USB.

Hello @rickkas7 ,

I’m using a micro b USB as the power input for the Boron and Argon and the VUSB (on the argon) and VIN (on the Boron) as an output for 5V input for my sensor.

If I understand correctly, it’s best to use a 5.2V adapter to prevent this.

Is there something we could do on the software side to fix this voltage drop?

I’m not sure you could find a 5.2V micro USB adapter, since that would violate the USB specification.

The problem is that there is a Schottky diode between VBUS_NRF (the micro USB connector) and the schematic net VIN, which is connected to the VUSB Feather pin.

This is necessary because on the Boron only, VUSB/VIN can be used as a power input or an output. You are using it as an output, and the diode is causing the 0.2V voltage drop.

The reason the diode is there is because VUSB/VIN can also be used as a power input. In this case, the diode is necessary to prevent voltage from flowing backwards from VUSB/VIN to the micro USB connector (VBUS_NRF) and nRF52 VBUS pin. These can only take 5V, but VUSB/VIN can be up to 17V; the diode prevents the higher voltage from going to these pins.

What I’m saying is use a 5V (non-USB) power adapter and connect to to VUSB/VIN instead.

Thanks for your explanation! This is all I need for a proper solution.

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