9 + 9 = 18 ?

[Virginia Belle]

I Know that this is a dumb question but, I am curious anyway.

Lets just say that I need an 18 volt DC power source and that the source
has to come from one of those transformers? that plug into the wall and
change AC to DC.

What if I only have two of those transformer devices that are 9 Volts
DC each.

Now, If I cut the insulation off of the ends of both of those
transformers, take the two wire ends from each and twist the 4 ends
together to make two again, plug them in, will THAT make me the 18 volts
that I need?

9 + 9 makes 18 doesn't it?

PLEASE tell me that I have made SOME sense here?

Thank You
Brenda

 

[Jay]

Hello Brenda,

In article <29303-3FBBCB52-39@storefull-2237.public.lawson.webtv.net>,
BRWVABell@webtv.net says...

I Know that this is a dumb question but, I am curious anyway.

Lets just say that I need an 18 volt DC power source and that the source
has to come from one of those transformers? that plug into the wall and
change AC to DC.

What if I only have two of those transformer devices that are 9 Volts
DC each.

Now, If I cut the insulation off of the ends of both of those
transformers, take the two wire ends from each and twist the 4 ends
together to make two again, plug them in, will THAT make me the 18 volts
that I need?

9 + 9 makes 18 doesn't it?

PLEASE tell me that I have made SOME sense here?

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.

 

[Robert Baer]

Jay wrote:

Hello Brenda,

In article <29303-3FBBCB52-39@storefull-2237.public.lawson.webtv.net>,
BRWVABell@webtv.net says...
I Know that this is a dumb question but, I am curious anyway.

Lets just say that I need an 18 volt DC power source and that the source
has to come from one of those transformers? that plug into the wall and
change AC to DC.

What if I only have two of those transformer devices that are 9 Volts
DC each.

Now, If I cut the insulation off of the ends of both of those
transformers, take the two wire ends from each and twist the 4 ends
together to make two again, plug them in, will THAT make me the 18 volts
that I need?

9 + 9 makes 18 doesn't it?

PLEASE tell me that I have made SOME sense here?

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.