Computing stuff tied to the physical world

JeeNode power pins

In Hardware on Dec 7, 2009 at 00:01

The JeeNode has a number of connections for external power. This is the place before the 3.3V LDO regulator. Keep in mind that all these pins are tied together, electrically:

JeeNode power pins.png

This has two important consequences:

You can power a JeeNode through any of these pins. It does not matter how you connect power to a JeeNode, all of them will have the same effect. You don’t have to connect a battery to the battery connector or the FTDI connector, you can also tie it to a PWR pin on any of the port headers (as I tend to do with Projects On Foam).

You should avoid connecting multiple power sources. If you connect a 9V battery, for example, the JN will work fine. But if you leave it connected when plugging in an FTDI interface such as a USB-BUB, then the 9V will find its way to the FTDI chip, and through the internal ESR diodes probably also to some I/O pins. Chances are high that you’ll damage at least the FTDI and the ATmega chips at this point (thanks to Paul Badger for pointing this out). A 3- or 4-cell Alkaline / NiMh battery pack can probably take some abuse without damaging anything.

There are no diodes in these various connections because the JeeNode is designed to be usable under very low power conditions. Throwing away 0.6V (or even just half that with a Schottky diode) is not always an option. The MCP1702 voltage regulator on the JeeNode was specifically selected to have an extremely low drop-out voltage (under 0.1V at low power levels) and a minute quiescent current draw (under 2 µA). This means that you could power a JeeNode from just 3.4V, perfect for 1 LiPo or 3 NiMh batteries, for example. The price is reduced protection against incorrect use.

The JeeNode USB is slightly different. It is intended to always be powered from USB, i.e. with PWR fixed at 5V. There are no FTDI and battery connectors, and all the other pins are connected to the 5V line via a 350 mA polyfuse (PTC) as protection against over-current on the USB power line. You can probably power a JeeNode USB through any of the port or PSI connectors, as long as the voltage is around 5V. The current draw will be several mA higher than with a plain JeeNode due to the on-board FTDI chip.

  1. Have you considered the MIC2940A? The disadvantage to it appears to be the higher quiescent current but its advantage is much lower voltage dropout and better availability.

    • Thanks, hadn’t seen that one (it’s a bit larger, but it’ll go up to 1A). The 240 µA quiescent current would be a show stopper for the ultra-low power uses I’d like to explore. The 0.04V dropout is indeed even less than the MCP1702. FWIW, the MCP1702 is available from Farnell and DigiKey – I haven’t looked for other sources, though.

      Needless to say, you can always hack any regulator onto the board with a couple of short wires.

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