Hero999
Banned
The internal resistance can easily be calculated using Norton's theorem
Norton's theorem - Wikipedia, the free encyclopedia
Norton's theorem - Wikipedia, the free encyclopedia
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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.
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.
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.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.
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.
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?
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?