This voltage divider confuses me

[ayayay!]

 

More specifically, I don't see why the IC1 & IC2 are listed on there from +9v to ground. 

[sean k]

I think if the author had put + and - symbols either side of the IC, like   0V____ - IC1 + _____ 9V, you would have got it. 

[slacker]

Yeah I think the schematic is trying to show 9 volts going to the +supply pin and ground going to the - supply pin of what ever IC1 and 2 are.

[ayayay!]

Sorry, I should have posted the full schem.  Here it is.  I get that it's going to the +supply pin.  But when I look at the schem all the supply pins are getting +4.5v.  Do I just send +9v to the IC on another supply pin anyway???

[R.G.]

Those are not the supply pins getting 4.5V, they are input pins, using +4.5V as a reference voltage. The supply pin is not shown separately that I see.

[slacker]

The schematic doesn't show the supply pins. The + and - pins shown are the inverting (-) and non-inverting (+) inputs. On a standard op amp that is pins 2 and 6 for the - input and 3 and 5 for the + inputs.
The supply pins are 8 + and 4 - so they go to 9 volts and ground.

Hope that makes sense

[ubersam]

Looks like the supply pins are represented by the texts "IC1" and "IC2" between +9V and GND.

[R.G.]

I know of two and only two dedicated-function ICs (as opposed to a bunch of separate transistors in a DIP package) that have no power supply pins, either +V, -V or ground.

Anyone want to guess? 

[snap]

I know of two and only two dedicated-function ICs (as opposed to a bunch of separate transistors in a DIP package) that have no power supply pins, either +V, -V or ground.

Anyone want to guess? 

could it be some kinda smart voltage converter?

[ayayay!]

Got it!  Thanks guys.  I figured it was something simple.