> Hi all. I'm a little bit confused about the CD4052 4 
> channel
> multiplexer. I'm just unsure about the Vee pin. I'm 
> considering using
> it to switch some audio signals (consumer line level) and 
> I'm really
> not quite sure what voltage Vee wants to see. There is a 
> datasheet
> here if anyone wants to have a look:
> http://focus.ti.com/lit/ds/symlink/cd4051b.pdf
>
> I guess what I'm trying to figure out is if I can run this 
> all off a
> single ended 12V supply or if I need both positive and 
> negative
> supplies.

It has a digital control and logic voltage range and an 
analog range.

The digital signals must lie between Vdd and Vss as usual.

The analog side shares the same +ve rail = Vdd but has its 
own -ve rail = Vee, which can lie below the Vss ground.

Analog signals can lie anywhere in the range up to Vdd 
(above ground)  and down to Vee (can be below ground).

If you don't need the analog signal range to extend below 
ground then you can tie Vee to Vss. The digital and analog 
range then both lie between Vdd (+ve max) and Vss (= Vee 
= -ve max) .

In your case tying Vee to Vss gives you a single ended / 
single supply switch with 12/0 switching signals and a 
0=-12V analog range.
Clear as mud :-).

There are limitations on the range Vdd-Vee depending on the 
size of Vdd but that doesn't affect you here.


        Russell McMahon







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