Powering Cerberus


I have a question about powering the FEZ Cerberus board without using one of the GHI power modules.

Since we want to power the FEZ Cerberus from the power-line (230V AC), I want to know if the Cerberus will work when we supply 3.3V on pin 1 and 5V on pin 2. Or is it needed to set some voltage on one of the other pins?

I am a newbie and with my little knowledge I can see that in the schematics of the USB Power DP module, PIN 7 have Vcc on it. Also, when I using a multi-meter I have found that pin 7 is 3.3 V when I have attached the USB Power DP module to an adapter.

I have found a PCB mounted power supply from MeanWell. The PM10-5 (https://www.meanwell-web.com/product-info/ac-dc-power-supply/pcb/pm-10/product/PM-10-5) seems suitable to me. Combining the GHI USB Power SP module logic and so using a LM1117-3.3 with an electro cap, the 3.3V can be substracted from the 5V in my opinion. Should it work or am I missing something?

Any help would be appreciated. Thanks in advance!


Yes, you can power through these pins on any socket. As long as the power provided through only one socket.
Yes on your proposed solution as well.

First of all, I’m not sure what you mean when you refer to the “module logic” – there is nothing on the USB SP module other than a USB connector and a voltage regulator. If you don’t need a USB port, then providing power to the Cerberus is sufficient to get it to start up. Don’t worry about power sequencing, since the Cerberus itself only uses the 3.3V rail.

That actually brings up another thing to think about: you usually need a 5V rail. Unless one of your modules explicitly requires it, you only need to provide the 3.3V rail to the board, so check the schematics of your modules first.

If you don’t need a 5V rail, try to find a switching converter that outputs 3.3V directly. Your same supplier has a 3.3V version:

If you do need a 5V rail, I would recommend you avoid using a 3.3V linear regulator, as it burns a lot of power (roughly 0.2 Watts!). A switching regulator is usually around 80% efficient in these sorts of applications; some require nothing more than an additional inductor and capacitor.

One more thing to think about…
Cerberus by itself uses less than 100 mA of current on the 3.3V rail, so you can probably get away with a cheaper power supply if you’re designing a cost-sensitive application. I generally like using non-isolated buck converters designed around one of those Power Integrations ICs, since they’re easy to work with, they’re cheap, and they don’t take up a lot of PCB space.

Food for thought. It all depends on how much time you want to spend messing with power.

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First of all, thanks to all for your effort.

@ Jay, with “module logic” I mean the so-called ‘voltage regulator’ but since I am not familiar with electronic schema’s I did not know how to call it.

I need the 5V rail since both the Cellular Radio Module and Character Display module need it.

This is what I need to power:
3.3V 100mA

Cellular Radio Module
5V 100mA
3.3V <1 mA

Character Dispaly module
5V 30mA

Flash Module
Unknown since the status is still ‘TBD’

Ultrasonic Sensor
5V 3.1mA nominal

5V 20mA (unknown, but I currently use the Relay Module (http://www.ghielectronics.com/catalog/product/327), but I only need 1 relay).

I currently power this all using the GHI USB SP Power module with a USB charger which at least provide 1000mA. In-field tests with USB chargers <1000mA did periodically result in unexpected behaviour of the Cellular Radio Module. Since depending on the signal quality the needed power will increase I think the displayed 100mA might sometimes become a lot more.

The cellular radio is the beast in your setup. We’ve heard of people whose voltages sag at the peak draw times so if you could get more than 1A supply that would be beneficial.

Ultimately, what you’re doing is building your own Gadgeteer power module, so you could always look at the Gadgeteer specs to get more clarity too - but ultimately, pins 1, 2 and 10 on a Gadgeteer socket are what you need to have connected.

A follow-on question here: is the 3.3v pin1 required for the Cerb to work? Or is supplying 5v on pin 2 alone sufficient? I know this is a noob question, but I assume it IS required since the Cerb itself (without the DP or SP modules) will not convert 5 to 3.3 on its own. Right?