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Wall adapter - says 9v dc but measures at 17v

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redepoch7

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I have a wall adapter.

Panasonic AC adapter.

Input:120v ~ 60Hz 6w

Output: 9v - 350mA


However, pluged it in and measured it with a multimeter. The meter said that the voltage was 17v.

Is this normal? I was hoping to add a wall adapter socket to a DIY-synth-keyboard-thing. I was going to base a lot of the design on it being 9 volts.

Any help would be much appreciated. Thanks!
 
Try putting a load on it and then measure. 50 to 200 ohms should work. Transformers always load down when they are working at a job. Another way to put it is, the power supply was not designed to not be connected to anything. Measuring it while it is doing something it was not designed for will give false readings.
 
Thanks so much for the reply.

Pretty new to electronics.

When you say "put a load on it"...
Can I just connect a 50 - 200 ohm resistor in series with the multi meter or do I need more than that?
 
Ok,
I just tested the resistor in series... of course it didn't work (read your reply right after).

So, to test it... can I solder a 100 ohm resistor between the the two wires coming from the adapter plug (adapter goes into a socket with two wires coming out of the socket)...

Then measures at the solder points?

Also, can i get shocked/hurt from 17v (or 9v wall adapter?... what ever it is).
 
You can solder or use alligator clips or hold the resistor on with clothes pins. I think 17 volts would hurt if you put both wires in your mouth, so don't do that. Use your fingers.
 
So, in electornics...

I have seen two wires (in a house... 110v or 120v) touch and blow a fuse/pop a breaker.

If I have enough resistance in between it won't pop? Is that the idea?

In my experiment with the adapter... i have only the resistor (100 ohm) in between. That won't pop anything?
 
HA HA!!

Ok... i plugged it in anyway.

Success!!! except, with a 100 ohm resistor acting as the "load" i am down to 12.5 ish volts.

I guess, I add more resistance to get it down to 9 volts?

And... if I have a circuit designed for 9v... will it be ok to just add the adapter or do I still need the parallel resistance to allow the adapter to work as it was designed?


Thanks again for all the help.
 
When there is enough resistance, electricity does not pop. That is the principle behind every electric device that works and does not pop. As you can see, there are very many things that run on electricity and do not pop.

If you knew Ohm's Law, you could calculate that the resistor would have to be size 3 watts to not over heat.
 
Well, I know ohms law in formula. But I don't really know how to use it in practice.

E/IR ... Right?



9v/350mA = 9/3.5 = 2.571Amps (i got the milliamp from the adapter specs... right?)

So, does that mean that when you calculate that using ohms law that you would need more resistance than what the equation equals to be safe?

Am I in the right ball park? If I am, I still don't really get it... If not...
 
You also need Watt's Law. P = IxE. That's how you find the power that will happen in the resistor. If too much power happens, the resistor will get very hot and threaten to pop. To fix that you use a big enough resistor to hold the heat.
You also need to know that 350ma is not 3.5 amps. It is .35 amps.

If the 17 volts would not go down,

I = E/R
17 v/100 ohms = .17 amp
P=IE
Power = .17 x 17
Power = 2.89 watts
That is how big the resistor has to be to not get too hot.

The voltage did go down. It will probably be 9 volts when 350ma of current is used. That is normal. The wall adapter is not broken.
 
Ok, i went online and found a converter because I don't trust my math skills (or more, my decimal moving skills).

350mA = .350A ... correct?

So,
9v/.350 = 25.714 ohms

I don't think I am doing this right at all. And I definitely don't know what it all means.
 
25.714 ohms is the resistance that will use up the whole 350ma when connected to the adapter. I don't think you have that resistor, so I told you to pick one in a range that would show if the 17 volts is going to go down to 9 volts if you used 25 ohms. It worked. You proved that the voltage would go down to its proper place.
 
Last question.

I did the math as you advised. I found that a 1k resistor "load" would equal .017 amps and I would need a .289 watts resistor... assuming I did the math right... (i think i now understand why there are different wattage ratings on resistors... i knew why, I just didnt KNOW why.)

My question is this... will I be able to hook the adapter strait up to the circuit that would normally take a 9v battery or do I need the 1k (or whatever the correct value is) in parallel with the circuit???

I swear that is the last question.

Thanks again.
 
May I suggest that you may find it easier to use the E²/R power formula? That way you don’t have to calculate the current through the load resister.
 
You have proved the 9 volt adapter is not broken. It will power things that need nine volts and it will be very close to the right voltage if they use 350ma of current. If you connect to a hundred dollar electric piano, it will probably work. If you connect to a $4 transistor radio that uses 10ma, you might smoke the capacitors inside it.

Even after you prove the adapter is not broken, you still have to consider if the machine you want to power will use most of the current, and if the machine is good quality or very cheap and has no ability to survive a few extra volts.

ps, your math is getting better.
 
Well, you were definitely right.

Plugged it in and nothing melted or "popped". I checked the voltage and it was in fact 9v. I still don't quite understand why it reads almost 17v with a 1M, 15v with a 10k, and 12v with a 100. I understand the math but it seems backwards to me.

I think I may have a better understanding of the practical applications of "ohms law" though. Just need to practice now.

Thanks again for all the help.
 
The best way to lower the voltage is to connect a voltage regulator to the output.

Unfortunately most voltage regulators need the input voltage to be higher than the output in order to regulate properly, so you can't have 9V@ 350mA

You can reduce the voltage to 8V using a low drop out regulator such as the LM2940LD-8.0 to get a steady 8V supply which will be enough to power any device designed to work from 9V.

The circuit is on the datasheet.
https://www.electro-tech-online.com/custompdfs/2010/04/LM2940.pdf

By the way, it's probably a good idea to add another large capacitor (1000µF to 2200µF, 25V) across the input to the regulator because the DC from the wall adaptor won't be very steady.
 
I think your only problem with Ohm’s law is that you forgot (or was never told) to apply it to the internal impedance of the power supply. That is why the output is dropping; you are actually measuring it in the middle of a voltage divider. With no load, you see the whole voltage (17 volts). With an external load equal to the internal impedance you measure 8.5 volts.

Do you now see what is going on?
 
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