Re: Better buy a new multimeter then a new battery!
- From: Jan Panteltje <pNaonStpealmtje@xxxxxxxxx>
- Date: Wed, 24 Dec 2008 11:26:46 GMT
On a sunny day (Mon, 22 Dec 2008 10:19:18 -0800 (PST)) it happened linnix
<me@xxxxxxxxxxxxxxxxxx> wrote in
<ffb0dda1-29d9-4b5c-b7cd-3b2ed490500a@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
I wish to mount it permanently in my cars, just to check the car
battery. So, I was trying to remove the internal 9V battery, or at
least make it rechargeable. But grounding the black lead to the
internal battery does not work. Any idea to make it rechargeable?
Yes, those chips need an isolated supply.
I am facing the same thing just now with a thing I am designing.
This little voltage converter circuit, that uses a very small potcore,
has saved me several times in the past:
ftp://panteltje.com/pub/voltage_converter.jpg
The base loop back winding is half a turn, for a 12 V battery you need 12 turns... etc.
For 9 V secondary 9 turns.
It runs at a few kHZ, makes a sinewave, has little harmonics, causes no RFI.
That one is from some other project (eighties).
The secret is 1 turn per volt, so if you have 6 meters,
then you need to wind 6 x 9 turns, use 6 diodes, and 6 electrolytics, to power
the meters.
I'v run out of potcores (the very small ones), maybe I will now try one with a ringcore.
Wire diameter can be hair thin, hardly any current.
So I tried the ring core solution, took some ring core with about 50 turns from a choke
in some old cellphone charger (switch mode).
Cut the wire in the middle of the ringcore, so that left me with 2 separate
isolated turns.
Connected it as in above referenced diagram, but used an NPN this time:
ftp://panteltje.com/pub/DC_DC_converter.jpg
The grey wire through the ring is the base feedback, if it does not oscillate connect it
the other way around.
And, yes it works:
ftp://panteltje.com/pub/DC_DC_converter_waveform.jpg
It runs at 200kHz with 47nF parallel, a much higher frequency then the potcore versions,
different permeability I think ...
The wavefornm is not such a nice sinewave either.
But, 12mA zero load, so at 10V about 120mW losses, likely the core
(transistor stays cold).
Vin is a little below Vout, depends on where you cut the original wire (transformer ratio).
So, this circuit does it again, it works from less then 3V input to way above 15V.
Cost: zero.
If you test it, test it with say 100 Ohm in series,
in case you got the phase wrong of the feedback,
and it does not oscillate, that will limit DC current.
.
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