The MCP1702 used in JeeNodes and other circuits here at JeeLabs is a nifty little voltage regulator. It’s not particularly good for high voltages (limited to about 13V), but it has a very low idle current. That’s very useful for low-power circuits, especially when trying to get months or even years of service out of one set of batteries.
Let’s find out how it’s performing. I’ve set up this little test circuit:
Well, it’s not really a test circuit at all – it’s in fact exactly what you need to use one of these things.
Putting 5V on the input (Vin), and leaving it a few minutes to settle, gives me a reading of 2.2 µA. Great, just what the specs say. It doesn’t really go up much with input voltage: 2.8 µA @ 12V. This is at no load.
With a 10 kΩ load, current draw rises to 338.1 µA, i.e. 330 µA for the resistor and about 8.1 µA for the regulator (assuming perfect calibration). That’s only 6 µA more, i.e. about 2% of the load.
Now let’s take the input voltage down, and see how this “low-drop” regulator behaves:
- 5.0 V -> 2.2 µA
- 4.0 V -> 2.2 µA
- 3.5 V -> 2.1 µA
- 3.3 V -> 2.1 µA
- 3.2 V -> 90.4 µA
- 3.1 V -> 84.1 µA
- 3.0 V -> 78.2 µA
Whoa… it sure doesn’t like being fed less voltage than it needs to supply a regulated output of 3.3V! The change is quite pronounced between 3.30 and 3.28V (I couldn’t get my power supply knob to adjust any more accurately).
For yesterday’s AA power board discussion that’s actually bad news, because it means that when the supply drops a mere 20 mV, it could cause the regulator to start wasting a fair amount of current (relatively speaking).
Let’s try something different: no power on Vin, but power on Vout instead.
This simulates connecting a battery to the +3V pin, and bypassing the MCP1702 voltage regulator altogether. Evidently, it’s not a good idea putting much more than 3.3V on there, since now we’re reverse feeding the regulator, something not all regulators are happy with:
- 3.40 V -> 1000 µA
- 3.35 V -> 500 µA
- 3.30 V -> 90 µA
- 3.25 V -> 90 µA
- 3.20 V -> 90 µA
- 3.10 V -> 84 µA
The conclusion seems to be: when using the AA Power Board in combination with a JeeNode, it might be better in certain cases to omit the regulator altogether if you’re aiming for minimal power consumption!
It’s not enough to short the regulator’s Vin and Vout, it really has to be taken out of the circuit. Fortunately, you don’t have to disconnect all three pins – just disconnect the ground pin (on the JeeNode v4..v6, that’s the one closest to the FTDI connector).
But if all you want to do is power the whole thing from some power source over 3.3V, such as a 3x AA battery pack or a LiPo battery, then all is fine… you’ll see that ultra-low 2.2 µA figure – as intended in this design.
Just beware of low voltages: weird things do happen!
This must be the start of a Jeenode v7-design. Why not add a fet with source connected tot the output of the regulator and drain on the ATmega-side. The gate then connected to the input of the power regulator via a voltage devider. That way the regulator only participates when external voltage is applied. Maybe even a fet parallel with the voltage regulator to bypass it when the input comes too close to the 3.2 volt. But then the ATMega needs to track the voltage with an analog port. Connect ADC4/ADC5 via a solder jumper perhaps?
Now that the diode is not used anymore since v6. You have some room to put at least one fet in that space. ;-)
Good tips – there is always room for improvement, but please don’t hold your breath for this one. I don’t expect to get to implementing a board redesign for quite some time. There are simply way too many other items on my list!
It turns out I’m not crazy after all, they do pull power when connected to the AA board! I’ve just removed mine from the board and slip one back in if I need to power it from the FTDI connector. Note than you can also just short Power to 3V and run the ATMega at 5V long enough to flash it. The RFM12B datasheet says it is ok up to 6.0V “not at working mode” whatever that encompasses.