I've got a question about the Vee pin on a 4051B mux.

I'm breadboarding a very simple little 8 step sequencer out of found parts and I'd like it to output + and - voltages.

I'm using a 4051B multiplexer to switch the voltages coming from 8 voltage dividers in sequence controlled by a 4029B up/down counter.
This is NOT a serious instrument, I'm doing this just for fun and for learning about 4000 series stuff.

QUESTION:
I'd like it to be able to send out positive and negative voltages (-9V to +9V) so does this mean that I should connect the Vss pin to 0V and the Vee pin to -9V? Or should Vss and Vee just both be 0V?

I ask this because...
the Toshiba TC4051B datasheet says
"The digital signal to the control terminal turns "ON" the corresponding switch of each channel, with large amplitude (Vdd-Vee) can be switched by the control signal with small logical amplitude (Vdd-Vss). For example, in the case of Vdd=5V, Vss=0V and Vee=-5V, signals between -5V and +5V can be switched from the logical circuit with single power supply of 5V. As the ON resistance of each switch is low, these can be connected to the circuits with low input impedance."

The common output of the mux will got to an Opamp buffer connected to +/- 9V so too much current won't ever be drawn through the 4051B.

 

Interesting question. I have a National Semiconductor data book. For the 4051 is says maximum Vdd =18, but recommends Vdd of 15V.
Going further it lists analog inputs of +7.5V and -7.5V that is with pin 8 grounded, pin 7 -7.5V and pin 16 + 7.5 V.
You could try it with + and - 9V to see what happens, or you could use a high resistance voltage divider on each input to divide the voltage in half.
Then us an op amp on the output to double the voltage. When doing this you could use + and - 5V or +9V on Vdd and -5V on pin 7.

 

Connect Vdd = +9V and Vee = -9V.

 

Interesting question. I have a National Semiconductor data book. For the 4051 is says maximum Vdd =18, but recommends Vdd of 15V.
Going further it lists analog inputs of +7.5V and -7.5V that is with pin 8 grounded, pin 7 -7.5V and pin 16 + 7.5 V.
You could try it with + and - 9V to see what happens, or you could use a high resistance voltage divider on each input to divide the voltage in half.
Then us an op amp on the output to double the voltage. When doing this you could use + and - 5V or +9V on Vdd and -5V on pin 7.

Hmm, I was thinking of maybe using less than +/- 9V and then amplifying it with the opamp instead of just using it as a buffer like I was planning to do.
The Toshiba datasheet says
DC Supply Voltage Vdd - Vss -0.5~20 and Vdd - Vee -0.5~20
but that sounds high compared to every other 4051B datasheet I've found.

How high of a resistance would I need for dividers at the inputs?

OR maybe I'll just create a +5V and a -5V with a 7805 and a L7905CV I just found in one of my bins.

 

In case of doubt inline a resistor, and see if the voltage goes down (IC consumes abnormal current).

Normally, if you approach a spec. limit, at first, current consumption will increase.

If you go beyond spec. there is also some risk the components may latch suddenly one day (in general).

 

I just found a more modern datasheet from Toshiba because I noticed the old one from 1997 labled Pin #2 incorrectly.
My first self discovered datasheet error

The 2012 datasheet fixes the problem.