Specific output of current?

I'm getting a little more knowledge about current measured in mA, but it's still all confusing. From my understanding, your project draws the current needed out of your battery. If my LED uses 25 mA, then the LED will be sucking 25 mA from the battery. If I have two LEDs, then it would be sucking 50mA from the battery, thus reducing its life. So, based on that, how can wall warts have a specific output of amps? How can a wall wart put out 1000 mA? If you use a project that only takes 700mA what happens to the other 300 mA? What if your project use 1,500 mA?

And. . .

How do you measure the current being pulled through a basic circuit and your average 555 circuit?

Thanks,

 

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First of all LEDs are a bad example since they will pull as much current as possible till something fails. Typically the LED fails. This is why you have to have a current limiting resistor in line with LEDs.

But if a basic circuit, in general, needs only 25mA that's all it will pull from the power source.

Wall warts have a max amperage rating. That 1000mA is the most it is capable of supply without something in the wall wart failing. It's also the rating for best performance. Many wall warts are not regulated so if your load is below 50% of the wall warts ability (pulling 500mA or less from one rated for a max output of 1000mA) then the voltage output may be very high. I have 9V wall warts that put out nearly 15V when it doesn't have a sufficient load on it. Keep in mind the wall wart isn't pushing out 1000mA it just has that power available. Just like a well may have several hundred gallons of water in it and your bucket is only pulling 2 out at a time. You don't want to run it at a full 1000mA. Its most efficient if your using about 80% to 90% of the rated output of the wall wart. You don't loose anything if you use less than that but your also not getting the most out of it.

If you try pulling more than 1000mA from the wall wart it will fail on you. If it goes really badly it could catch fire.

Finally there are a couple of ways to measure it. If your meter has the ability to measure current then have the positive lead of the power source connect to the positive lead of the meter. Take the negative lead of your meter and connect it to the positive input of your circuit and power it on. Your meter will now measure the current. If you have a basic voltage meter that you will want to get place a current shunt (very low resistance and accurate resistor) in line with the positive power lead like you would have with the example above. Use your meter to measure the voltage across the resistor. 1000mA. Now you do a bit of math. Take the voltage you read across the resistor and divide that by the resistance of the resistor. This will give you how much current is being drawn by your circuit.
HavocRC (author)  mpilchfamily4 years ago
Hey thanks guys, that make total sense. It just a max output. Skinnerz, what I'm asking is how to measure it, though it may not be necessary, I still would like to measure it.

After looking at wikipedia's article on ohms law, I cam up with this. . .
Does this look good?

9v across resistor, 1k resistor = 0.009A, 9 mA
12 across resistor, 5.2k resistor = 0.002 A, 2 mA ?
Yes your math is correct.

HavocRC (author)  mpilchfamily4 years ago
Haha thats the best thing I heard all day! Ok, one more question. I understand how to measure the current over one resistor(thanks!), but what about the circuit I attached? (assuming we're using a 9v battery)
Also, what about things like a servo, assuming it's working very hard and using 5v?
Screen Shot 2013-02-01 at 4.59.51 PM.png
Just like i said. Place your meter between the power source and where the power source would connect to the circuit.
HavocRC (author)  mpilchfamily4 years ago
Well, too bad my multimeter doesn't do that.
So it can only do voltage and resistance? Well then you need to invest a basic $40+ meter that can measure current up to at least 10A. That is if you want to continue this electronic hobby. Or spend a few dollars on a good shunt resistor and measure the voltage across the shunt to calculate the current draw.

But the above circuit will operate well within 500mA so there is no need to worry there. As for a Servo they will say on them how much peak current they draw. All motors will have a peak current rating on them or should. This is the max current they draw when they first get power and it only lasts a fraction of a second then they settle to a lower current draw.
HavocRC (author)  mpilchfamily4 years ago
Oops, actually it does, I was looking at the wrong manual :P
So, you probably shouldn't listen to me :P. Oh, and by the way, with 6 of my servos running with no load, they pull 80 mA. Thanks!
lemonie4 years ago
Why did you choose the name "electricloser"?

L
HavocRC (author)  lemonie4 years ago
Lolz lemonie- because at the time I first made my account, I didn't know ANYTHING about electronics! Now that I know a whole ton more, I'm thinking of a new name. Instructables offers me the chance to change it once, but I can't think of any better names :P Any suggestions???? Why did you choose lemonie? XD
lemonie HavocRC4 years ago
There is/was a change your name page. Lemonie was in a song a friend made up, they were doing RatherGood.com-type-of-stuff before the internet happened.

L
HavocRC (author)  lemonie4 years ago
Well since it concerns you so much, I'll change my name soon :P
The current rating on the power supply is the most it can continuously deliver. Below that limit, whatever you plug into it will only draw the current it needs, but if you try to draw more current than the supply is rated for, the voltage may start to drop out, it may overheat, or it could break entirely.

The 555 IC itself only needs a small amount of power, so it is not normally necessary to worry about it. The supply will need to be able to deliver enough current to whatever the 555 is switching on/off when it is on (the maximum load). However, at higher frequencies, you only need to supply the average current, calculated by multiplying the duty cycle by the current drawn by the load when in is on.