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

[Hero999]

The internal resistance can easily be calculated using Norton's theorem
Norton's theorem - Wikipedia, the free encyclopedia

 

[audioguru]

If you load the adapter with 350mA which is its rating then the adapter cannot provide any more current to do something useful.
Subtract how much current you need for a load from the 350mA then load the adapter with a resistor that uses the remaining amount of current.

 

[bychon]

Or just sit there and smile because you proved the adapter was not broken, plugged in your device, got the right voltage and everything is working correctly.

 

[Hero999]

Most circuits designed to work from 9V will use 16V capacitors and similarly rated components so unless the current draw is really low then it won't matter. You could stick a 15V 1.5W zener in parallel with the adaptor and it'll be fine.

 

[Birdman Adam]

Great explaination of wall-warts:

Unregulated Power Supply Tutorial

 

[redepoch7]

WOW!

Tons of information to take in. Thanks so much for responding. I can't believe how many concepts I have been introduced to by asking a silly question. This is awesome.

Hero999:

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.

So would that be in parallel with the entire circuit? Should I place that right across the +/- coming from the adapter?


audioguru:

Subtract how much current you need for a load from the 350mA then load the adapter with a resistor that uses the remaining amount of current.

How do I find out what the current is for the entire circuit? Can i use my multimeter and just measure at the power source? (you have to measure current when it's plugged in and operating... correct?) Or...
Do I need to draw up a schematic and calculate all the values?

Again... thanks SO MUCH for all the information. I'm really thankful that this forum exists.

 

[Hero999]

Yes, the capacitor should be in parallel with the whole circuit, the +/1 DC coming from the adaptor.

 

[audioguru]

How do I find out what the current is for the entire circuit? Can i use my multimeter and just measure at the power source? (you have to measure current when it's plugged in and operating... correct?) Or...
Do I need to draw up a schematic and calculate all the values?

You should know how much current the load uses. Or the cheap simple unregulated wall adapter will have an output voltage that is much too high because it is not loaded enough, or the wall adapter will be overloaded.

 

[Hero999]

Why not just get a regulated 9V mains adaptor?

It's not like they're that much more expensive.

RadioShack should sell them.

 

[audioguru]

An unregulated 9VDC/500mA wall adapter costs $6.00 at an electronic parts store.
I don't know what RadioShack calls it so I couldn't find one.
They sell a regulated 5V USB adapter for $40.00 that is worth $10.00.

 

[redepoch7]

You should know how much current the load uses. Or the cheap simple unregulated wall adapter will have an output voltage that is much too high because it is not loaded enough, or the wall adapter will be overloaded.

So, when you say I should know how much current the load is using... how do I figure that out?
Can I just measure the current with a multimeter when the circuit is on?

Thanks!

 

[bychon]

yes.......

 

[redepoch7]

Ok that worked.

Is it normal for the probes to spark when they touch the circuit? They felt like they were sticking to the metal a little bit.

But it gave me a reading.

On another note, I think I fried the circuit. I was experimenting with a voltage divider to power the toy keyboard. 9v went to the square wave generator and a voltage divider was branching off to give 3v to the keyboard.

It worked for awhile. I was trying different values of 3 equal resisters. 3 100 ohm resistors worked but the resistors would heat up quite a bit.
I tried other values but I couldn't seem to get the keyboard to power. I assume that is due to lack of current?

Anyway, it's all fried now. Both the square wave and the keyboard. Not sure what I did.

Oh well. Break n' learn I guess. On to the next one.

 

[Hero999]

You didn't connect the current meter across the power supply rails did you?

You should never do that, the current meter should always go in series with the load you're measuring.

Connecting the meter across the power supply will just measure the short circuit current which will be very high and will probably blow the fuse inside the meter.

Using a potential divider is a bad idea because, apart from being inefficient, the output voltage will change as the current drawn by the load varies.

 

[carmusic]

just get a switching adapter at 9V, they are often cheaper than transfromer adapters and they always have 9V whatever is the load

 

[chopchip]

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).

if you take a 9 volt battery and put needles in the terminals and insert one needle into your left hand finger and another into your right hand finger the low resistance conductor of your electrolyte solutions in your body will allow a large current to flow. I=E/R. I=9/low resistance of your body fluids. this will pass through your heart and cause defibrillation and you could get dead if someone doesnt treat you.

 

[redepoch7]

if you take a 9 volt battery and put needles in the terminals and insert one needle into your left hand finger and another into your right hand finger the low resistance conductor of your electrolyte solutions in your body will allow a large current to flow. I=E/R. I=9/low resistance of your body fluids. this will pass through your heart and cause defibrillation and you could get dead if someone doesnt treat you.

I'm not really sure what that has to do with the conversation. If put my head in a sharks mouth I might find that it will bite my head off... but I'm not going to do that.

I was just asking whether I needed to be worried about fiddling with electricity of that voltage. Thanks for the comment though.

 

[redepoch7]

You didn't connect the current meter across the power supply rails did you?

You should never do that, the current meter should always go in series with the load you're measuring.

Connecting the meter across the power supply will just measure the short circuit current which will be very high and will probably blow the fuse inside the meter.

Using a potential divider is a bad idea because, apart from being inefficient, the output voltage will change as the current drawn by the load varies.

First, I did connect it across the power rails. I did this because I asked if I should and someone told me (in this post) that that was how to do it.

I will always measure in series from now on... thanks for the heads up.

Second, instead of a potential divider, what should I use? How do I get 3 volts from 9 to power the little keyboard?

 

[bychon]

You should post a new thread. Your question has changed from "why cheap wall warts change voltage" to "how to design a 3 volt regulator".

 

[Reloadron]

First, I did connect it across the power rails. I did this because I asked if I should and someone told me (in this post) that that was how to do it.

I will always measure in series from now on... thanks for the heads up.

Second, instead of a potential divider, what should I use? How do I get 3 volts from 9 to power the little keyboard?

OK back to the basics, you did ask:

So, when you say I should know how much current the load is using... how do I figure that out?
Can I just measure the current with a multimeter when the circuit is on?

bychon answered your question with a yes...

Bychon assumed that when you asked that you were aware of how to correctly measure current with what was your meter. He could not have known that you planned to place the meter in a current mode and place the leads directly across the source. Bychon fails at mind reading 101 and you failed to mention you were unaware of how to measure current with your meter.

You have a 9 volt simple everyday wall wart designed for 120 VAC to 9 VDC @ 350 mA. Early on in this thread someone (maybe Birdman) suggested a very good link for you to read, did you read it? Additionally a Google of Wall Wart(s) will bring up a wealth of knowledge on what they are and how they work.

A wall wart is not a regulated precision device. Matter of fact if your mains voltage is not 120 VAC that will even vary the output. Following rectification in the wall wart there is a big filter capacitor. When that wall wart is unloaded it is normal to see a higher voltage than on the label. All of that and why was explained in this thread as well as in detail in the link provided. You need to start understanding how things work with a focus on the wall wart.

Now if you want to sort of "Wing It" you can measure the input resistance of the keyboard and that means measure the input resistance using correct polarity of your meter. Then using Ohms Law calculate what the current should be. All of this will be approximate! The current actually drawn by the keyboard will likely vary with volume and other factors. So lets say the keyboard draws an estimated 200 mA (.2 Amp). OK you need to shunt maybe 100 to 150 mA. ( Volts / .1 = 90 Ohms so place about a 100 Ohm resistor in parallel with the power input to the keyboard. That is winging it but would likely be close enough for this application.

Nobody here has your keyboard or wall wart or even your meter in front of them. You have gotten some very good suggestions.

Ron