> I need 2∗10 positions, and can only get 2∗6 or 1∗12.
> For selecting cell count on a battery charger (no current at all).
> Almost all handheld meters i've ever seen use rotary switches, and
all i
> took
> apart had "custom" printed contacts. Oh, yes, there was one rotten
piece
> of bad engineering
> that had four buttons instead. The damn thing had dozens of design
flaws,
> like blowing up
> each time you pulled the current plug....
>
> I could try to pull some mechanical stunt and stack 2 1∗12 switches.
> I don't know if that would be possibe with the switches i can get.
> But then, one switch for 2.50eur is not exactly cheap, not speaking
of 2...
Would be nice to have a couple more details on the circuit this is
going to be connected to.
Are you changing a resistance? Could you use a potentiometer and then
work out some kind of mechanical detent? Why do you need a dual-pole
switch? Have you looked at analog multiplexer chips? Digital
potentiometer chips? Don't reject them without looking at them, they
are not all expensive. A 16:1 analog mux/demux CD74HCT4067 costs 90
cents in DIP, 48 cents in SOIC. Two of those give you your dual-pole
circuit if you need it. They accept a simple 4-bit input to select
which output goes where. You could provide that input with a
16-position hex switch, or a 4-bit counter (rollover at 12) operated
by pushbuttons. Another 50-cent 74HC154 4:16 decoder could give you an
LED feedback showing which output is selected.
It would be neat to have 12 LEDs surrounding a single central
pushbutton which increments the LED position.
Or maybe you could just go for a rotary switch with fewer positions,
like 6. Then you use a dual-pole switch to choose between high range
and low range. Though even 6-position dual-pole rotaries are all very
expensive unless you get lucky in surplus stores.
As for rolling your own? It'll just have to be trial and error.