Batteries for project including AXE-033

[free1000]

Thanks to this forum I now have my AXE-033 up and running.

As someone pointed out on another thread it needs a 'good' 5V, it seems to die under about 4.6V.

As the picaxe project needs 4.5 V I'm wondering about how to power this project up for real use.

I am building a camera trigger that I'd like to be able to run on AA cells and that has a reasonably good battery life. Realistically, is the AXE-033 a power hog?

Could I get away with using a 6V supply (4 cells) or would that be too much for the Picaxe circuit? I suspect its OK, but thought it was worth checking before blasting a chip.

My project is quite a simple one, it involves a push button switch which is used to trigger the camera at a delay stored in a Picaxe 28X. A keypad is used to enter the delay time, and the AXE-033 is used with a simple menu system to allow viewing the delay time entered via the keypad.

 

[papaof2]

Possible long-term power sources:

A. 4 D cells with low dropout regulator to provide 5 volts to the PICAXE.

B. 5 NiCd or NiMH cells (6 volts nominal) with a low dropout regulator for 5 volts and a solar panel to keep the cells charged (there's probably a solar charge controller somewhere in the Forum)

John

 

[eclectic]

Possibilities

Free1000. A few comments/ideas/ramblings.

I've just tested a 28X and Axe033 with 4 x NiMH (~5.05v.)
Measured current through the Axe033 ~ 2.3 mA.

Questions.
What is the longest “session” time your set-up is likely to run? (hours/days)
What are the time intervals? (seconds/minutes/hours)

If the longest time is hours or a few days, then before each “session”,
start with fresh NiMh.
Or,
Use “Sleep” commands to reduce Picaxe current instead of “pause”.

Alternatively, power the AXE033 with 6v, but use two diodes for
the Picaxe supply.

I'm sure there'll be lots more ideas.

e.

 

[free1000]

Alternatively, power the AXE033 with 6v, but use two diodes for
the Picaxe supply.
e.

Thanks for the thoughts. I think the idea of running with the 4 AA battery pack seems the best one as I can always rustle up some AA cells whereever I go in my travels. Alternatively I could switch to D cells, but they are a bit bulky and less easy to get hold of in some more remote places.

Interesting point about the pause commands, I'm using a busy loop continuously so anything that helps reduce power consumption seems useful.

I'd be grateful for a reference to a circuit for connecting up the diodes... is there something in the picaxe docs?

 

[BeanieBots]

The diode idea suggested by eclectic would be the simplest and cheapest.
Simply put a diode (or two) in series between the 6v and the PICAXE. Pointy end towards the PICAXE. Each diode will drop about 0.6v.

 

[free1000]

The diode idea suggested by eclectic would be the simplest and cheapest.
Simply put a diode (or two) in series between the 6v and the PICAXE. Pointy end towards the PICAXE. Each diode will drop about 0.6v.

Thanks Beanie, so basically using the diode like a resistor, but removing any reverse current.

Cheers

 

[eclectic]

Any more info?

Free.
Could you post your circuit and code?

I'm sure that you could get many more ideas for current reduction.

Another question: Is your project for a delayed action trigger,
or an intervalometer?

e.

 

[Dippy]

What is the total current requirement of your complete circuit? And as suggested, please post it complete with component types/values.

Is it possible/convenient to have power switching for the AXE033 board? (This is the LCD board isnt it?). You don't need it switched on all the time do you?

 

[BeanieBots]

so basically using the diode like a resistor, but removing any reverse current.
Cheers

Um.. no.
A resistor will 'drop' volts due to current passing through a resistance in accordance with ohm's law. V=I*R.
A diode is a semiconductor device and does not have a fixed resistance like a resistor does. It has P and N junctions which require a certain amount of voltage before ANY current will flow. (about 0.6v for a normal silicon diode).

If you used a resistor, the amount of voltage 'dropped' would depend on (and vary with) the amount of current your circuit uses. Unless, you measure it, it is an unknown amount and therefore you would not know what value resistor to use. Also, anything switched on/off would change that current and hence the resistor would need to change. This is actually how a linear regulator works, it continually changes the resistance between the supply and the load in order to keep the voltage constant. Using a 'diode drop' is the poor mans version of that. A constant voltage drop from an already 'constant' supply.
Stopping reverse current does not come into it with this application.

By the way. A diode is not perfect. Think of it as a battery a resistor and a perfect diode all in series. The battery is about 0.5v and the resistor about 10ohms. Thus, the other end of the diode is 0.5v less than the supply less whatever is also dropped by the 10 ohms.

 

[Peter M]

@Free1000,
depending on whether you need the display on all the time, perhaps You could use the 28x to turn on/off the supply to it, if it is only ~2.3mA as eclectic suggests (or via a transistor, mosfet or relay if its above 20mA).
You could program it to turn on for a certain time, to allow the display to be read when you are updating the times, and/or a button to bring the display when needed... along with combinations of the suggestions above.

@BeanieBots,
by the way, love your decription of a diode.