Switched battery Supply


Senior Member
Idle wonderings rather than anything else, but how would one control a battery supply so it is on when a switch is open and off when closed, drawing uA when off and taking little when on ?<code><pre><font size=2 face='Courier'> .-----.
.--------| ? |----&gt; +4V5
__|__ + `--.--'
-.- |
3 x AA | O|
__|__ + ===|
-.- O|
| |
`-----------^-------&gt; 0V </font></pre></code> The thought was that power could be controlled by the N/C contacts in the 3.5mm socket of a PICAXE download cable.

Edited by - hippy on 20/08/2007 18:00:59


Why not put a NPN transistor where you have the ?. Tie its base to 4.5 v with say a 10k.
Pushing the switch will pull its base to 0v and turn it off.
A darlington pair would be better, depends on what your current draw is.


I'd go with jaka. It's called an &quot;emitter follower&quot;. An old fashioned form of voltage regulator. Essentially, the output will be at whatever the voltage on the base is at. Hence clamping the base to 0v will give 0v out. Replace the switch with a zener and you have a (not perfect) regulator. Leave it open and the output will be a diode drop or so below the input.


Senior Member
So, it's a simple as that - Thanks ! You can tell who round here 'does software' not hardware :)

A 10K pull-up would itself suck 500uA, not entirely excessive but not incredibly low either. Could it be made lower ? Would increasing the pull-up value start to have adverse effects on the transistor behaviour ?

This was my main thought when considering there probably would need to be a pull-up somewhere and to get down to 1uA ( not necessarily achievable ) it would need around 4M pull-up. Would 470K and ~10uA be achievable ?


New Member
i would use a darlington, either as a single package or two discrete transistors. the current gain is effectively equal to the two individual transistors multiplied together...a current gain of 10000 should be achievable


Indeed, the higher the transistor gain (hfe) the higher value resistor that you can get away with. Darlingtons certainly have the highest gains but you pay the price of a diode drop in output voltage.
If the resistor value is too high then the output voltage will collapse. Max load current is roughly base current*hfe.

Edited by - beaniebots on 21/08/2007 08:07:14


Senior Member
Many more thanks. As it seems to be so simple, I'm sure I can find some transistors to try, and I'm betting on having a &quot;wow&quot; moment :)


Senior Member
You really want to use a mosfet. Gives you infinite gain so you can use say a 1Mohm pullup resistor. This gives you zero OFF current and minimal ON current, and it also switches rail-to-rail.

The down side is it's also static sensitive to the extreme! There is no built-in static protection in a plain mosfet aside from it's natural capacitance. The Gate absolutely must have good protection at all times!

Making use of that higher inbuilt capacitance of high current mosfets I recommend buying an extra heavy one for your switch. On top of that I'd still add a 100nF ceramic capacitor directly across the Gate-Source pins. And adding say a 6 volt zenor also across the Gate-Source pins is normal, striped end to the Source.

And finally putting the pull-up resistor also across the Source-Gate pins completes the switch construction.

Using a P-channel, the connections are:
Source to the battery positive
Drain to load
Gate to pushbutton

You could also add a series resistor between the Gate and the pushbutton.

Edited by - evanh on 22/08/2007 03:27:33