Hello, I need to power a chip that requires +/- 8–18V. Is this possible to do with a single 9 V? (I'm guessing not.)

If not, could I connect two 9V batteries in series, as shown below?

Vs+------G--------Vs-
-|-------- | ---------|
|+ 9V –| --- |+ 9V –|

In the diagram, I'm pulling the positive supply from the positive end of the two connected batteries, the negative supply from the other end, and the ground from between the two.

Thanks for reading.

 

You're 100% right, wiring the batteries in series and using the middle terminal as a ground will give you both +9 (from the positive terminal) and -9 (from the negative terminal). It cannot be done with a single battery.

 

OK, thanks for verifying!

 

In an effort to be thorough, I will add:

Yes, conventionally, vpoko is correct. There is no wire orientation that will allow you the >9 potential difference you require.

However, for the sake of "is this possible", consider if you were to use that 9v battery with a boost converter such as:

http://www.maxim-ic.com/datasheet/index.mvp/id/2040

This would now provide you with a potential difference greater than 9V, let's say it is set up to give you a 20V output. Now all you need to do is setup a virtual ground (which is what you did with your two battery scheme) which would then provide a +/- 10V supply. This website should give you all the details you want:

http://tangentsoft.net/elec/vgrounds.html

I'm not saying this is easy or economical, but to answer your question, yes, it is possible.

 

That's good to know, thanks Stuntman!

 

I have used this chip with good success to generate a negative voltage.

 

Thanks, Jim. I ended up picking up a ICL7660S for a few dollars (http://www.intersil.com/data/fn/FN3179.pdf) which appears very similar to the ICL7662 you recommend.

 

How much current do you need? If it's very low, you can use a MAX232; just tap off the V- and V+ charge pumps. I've used that in the past to power an opamp.
/mike