9 + 9 = 18 ?

If you use two of the DC "wall warts" (small AC to DC adaptors) that each put out 9V, you can get 18V by putting them "in series", assuming the wall adaptors are 2-pin devices and not 3-pin (i.e. they don't use the Earth grounding pin).

Assuming a setup like this:

Adaptor one has plugs: (-) and (+), let's call them A1(-) and A1(+) Adaptor two has plugs: (-) and (+), let's call these A2(-) and A2(+) .

Connect A1(+) to A2(-). Use A1(-) as the negative terminal and A2(+) as the positive to whatever needs 18V.

If you measure the voltage across A1(-) and A2(+) with a voltmeter you should read something close to 18V, perhaps slightly higher (these wall adaptors usually put out a bit more than what they state unless they are under a load).

Now, you don't state what kind of kind of current the 18V load will draw nor do you mention the current rating of the 9V adaptors you have. Make sure that your 18V load doesn't draw more current than the rating on either of the 9V adaptors(i.e. if one 9V adaptor is rated for 1000mA and one for 500mA, your load shouldn't draw more than 500 mA).

Let us know how it goes, Jay.

120 percent *or more* is a hell a lot more than "a bit" !!! A 4,5V wallbump puts out about 6V with nominal load; that is 130+ percent.