# 9 + 9 =18?

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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

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I won't go into the right or wrong of it, but to answer your question...

Strip the insulation from each of the four wires. Twist together the negative of one adapter to the positive of the other adapter. This leaves you two open wires, one positive, one negative, and you now have 18V.

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It would make sense only if both Wall Wart outputs are isolated from ground or common. If not, you'll be short-circuiting the output from one. Sadly, this is the case with some (but not all) of these devices.

What you want are Wall Warts with DC outputs isolated from ground and common.

If their outputs are isolated from ground, then by connected them in SERIES you'll either obtain 18V or 0V, depending on the polarity of how you connect them.

e.g., 9V + 9V = 18V ...OR... 9V + (-9V) = 0V.

Hope this helps.

Harry C.

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I have yet to see any wall wart with a ground connection.

If your "wall warts" are designed to replace 9 volt bateries, you can cut the part that go to the appliance in half and connect ONLY the male of one to the female of the other. The unused male and female will have your 18 volts.

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It is also a good idea to use identical wallwarts. If you use different ones, imbalances can occur, especially if they are poorly regulated (like most are...)

-Z

Harry C> It would make sense only if both Wall Wart outputs are isolated from

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It's not a dumb question at all. 9 volt wall warts deliver 9 volts *at their rated current*. If you run them at a different current, the voltage could be considerably higher or lower.

If the two wall warts have identical current specs, you'll get 18 volts at the load for which the transformers are rated. The problem is that MOST wall warts are not regulated. That means that a wall wart rated 9 volts,

100 mA may put out say 12 volts if it is connected to a load that draws 10 mA instead of 100 mA.

The bottom line is this: when you connect the two wall warts in series and then connect the two remaining wires to the load, you may not get

18 volts. You could get well above or well below 18.
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On Thu, 20 Nov 2003 2:18:11 -0800, JeffM wrote (in message ):

Um... Jeff, she posted to *one* n.g.

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In that case, you've lived a sheltered existence. How many have you found without a NEUTRAL connection? Ever buzzed one out with an Ohmmeter?

Harry C.

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Dave She posted seperately to several newsgroups that I look at and as well as the usual many similar answers to the same question she got them in all sorts of variations in EACH newsgroup. What a waste of band width.

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You are starting to sound like someone who looks under his bed before going to sleep.

Actually most don't even have polarized plugs.

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That comes from being a professional engineer whose career depends entirely on the viability and successful operation of anything he/she designs. You learn to worry about the details...the small issues that most non-pros ignore...the stuff that later comes back to bite you on your ass when neglected! :-) Harry C.

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Sure. Putting two wall warts in series will come back to "bite you on your ass."

Get real!

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I don't think I've seen a US/Canadian one that's a true "wall wart", though the "brick" type, both linear and switching types, often do have the output grounded. If the AC plug has only two pins, there's nothing to worry about.

Best regards, Spehro Pefhany

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As I pointed out in another newsgroup, what happens to the filter capacitor when one of the two accidentally gets unplugged and you have not added diodes across the outputs?

Best regards, Spehro Pefhany

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Here is a hint about posting Google results. This URL:

Gives the same result as the longer one above.