Electronics > Beginners

Some noob questions

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Brumby:
The first thing to understand is that an 'ideal' multimeter measuring current will have zero resistance across its terminals which means it will allow as much current to pass as the circuit it is in can provide.  Such an uncontrolled flow can be a source of danger.  In a real multimeter, however, this resistance will not actually be zero but it will be as close to zero as the designer can get - so it is always wise to think of your multimeter (in current mode) as being a simple piece of wire between between one probe and the other.

Now imagine what would happen if you stuck a piece of wire into the active and neutral of a mains power point that was switched on!!!  Some fireworks, a blown fuse or both - as well as a risk of injury to yourself.  This is the sort of problem that can happen when you don't use measuring equipment properly.

In the case of your wall wart, this is exactly the sort of thing you were doing - but you were fortunate in that things were on a smaller scale and the meter was designed to be able to carry that current.  I would also guarantee you were overloading the wall wart and if you had left it connected it would have heated up very quickly and very likely have been destroyed.

To understand why you kept getting 2A, I will offer you the following very simplified diagram.  This represents your wall wart - but only in a very basic sense.  It is missing a lot of detail that you would find in a real device, but it shows why you would get the same current reading.  To make the maths simpler, I will use the 'ideal' current measuring meter that has zero ohms between the probes.



The story goes like this...
As shown, there are four sections of the transformer that generate 3V each - but the wire that is used in each section has a resistance of 1.5 ohms.
 * When you select 3V, there is only 1 section used and it has a resistance of 1.5 Ω.  Using Ohm's Law V=IR, you get a current of 2A.
 * When you select 6V, there are now two sections used and while this gives you 6V, the resistance across the two sections is 3 Ω.  Again, using Ohm's Law, you get a current of 2A.
 * When you select 9V, there are now three sections used and while this gives you 9V, the resistance across the three sections is 4.5 Ω.  Once more using Ohm's Law, you get a current of 2A.
 * When you select 12V, there are now four sections used and while this gives you 12V, the resistance across the four sections is 6 Ω.  Yet again using Ohm's Law, you get a current of 2A.

This resistance is buried inside the wall wart and something you can't really do anything about it, other than include it in your calculations.  This sort of internal resistance has a name: "Equivalent Series Resistance" or ESR and it is found in every electrical and electronic device - unless it happens to be a superconductor.

It is this ESR and how it is distributed through the wall wart that gives rise to the "constant 2A" phenomenon you observed.


For those wanting to challenge the above, I would like to add that there are several ways of providing a multi-voltage wall wart - and the above is only showing one of those in a very basic conceptual form, purely for the purposes of education, not as a construction project.

ArthurDent:
The wall-wart that you have should have a rating label on it that will tell you what the voltage and safe current rating of the supply is. It sounds like this is a supply that no matter what voltage selection you choose, the short circuit current is limited to 2 amps. You are lucky that it is limited because you never should try to measure the output of an unknown supply by setting a DMM to current and measuring directly across the supply output. A lot of wall-warts would be destroyed by what amounts to a direct short across the output.

Imagine what would happen if you tried this by setting the DMM to current and plugging the leads into a wall outlet. I believe your mains are 230 VAC and with enough current to run appliances. If you plugged the DMM leads into a wall outlet it would blow the DMM’s internal 11A fuse, trip a breaker, or create an arc and burn the probe ends; or some combination of the above. 

As to the confusion about why 2 amps isn’t adequate for everything, what does the work is power, not current or voltage alone. A hair dryer has far different power requirements than a LED nightlight but both may be designed to operate from the same mains voltage.

An automotive bulb may be designed to operate at 12V/1A and an incandescent bulb for your house may be designed to operate at 230V/1A. Both bulbs are designed to load the proper supply at 1A but if you try to run the 230V/1A bulb on 12V it will have too high a resistance at 12V to even glow and will draw very little current while the 12V/1A auto bulb will brightly flash once and be destroyed on 230V because its resistance is far too low for the mains voltage and it will try to draw a very large current and burn out.

To measure current a DMM is generally placed in series with a load like a 12V/1A auto bulb to measure the actual current the bulb draws. A DMM set to current has a very low resistance between the leads so it doesn’t have a noticeable effect on the voltage being delivered to the load. To accomplish this, internally a DMM measures the voltage drop across a short heavy calibrated wire called a ‘shunt’ and the voltage drop across this shunt is displayed as current, and the total series resistance of the DMM leads, the shunt, and fuse is probably a fraction of an ohm. If you are measuring a 1A load, the voltage drop across the DMM is probably a couple of tenths of a volt. For instance IxR=E (the voltage drop across the DMM) might be 1Ax.2ohm=0.2V drop so if you have a 12V supply connected to a 1A load with the DMM reading current placed in series with the load, the load will see 12V-0.2V or 11.8V and the DMM will display about 1A or whatever the actual current at 11.8V is.

Jwillis:
With the analogy, flow is defined as an amount over time. Speed is not a factor in the "flow" of electrons .We'll use yet another analogy to explain .Think of a Newtons cradle and that best represents how electrons move down a wire.Although in reality its much more chaotic with many other factors . Analogies are used to to best explain a process in the simplest terms.Its much easier to picture in the mind something that may be familiar.
It's not an attempt to insult your intelligence in any way.Just a simple way to explain the relationship between voltage , current and resistance .

KL27x:
The water analogy isn't sorta kinda a half-ass analogy. It is pretty close to perfect.

A high pressure washer and your kitchen faucet might produce the same volume of water per time, but the water coming out of the high pressure washer has more velocity. Thus, it has more kinetic energy. Thus, it can do more work.

In the water analogy, it is most common to refer to the height of the water as a measure of its potential energy. But if you were to pour it to "ground," that turns into kinetic energy. In absence of "resistance," 1 gallon of water poured from 100 feet high would be moving faster by the time it reaches the ground than 1 gallon of water poured from 3 feet.

So, this is why hydroelectric generators are positioned at the bottom of tall water falls, not just stuck in the middle of a big, slow moving river that passes 1000x the volume of water per unit time.

Your adjustable PSU (most likely) has a linear regulator. All of the water it can pour starts out higher than 12V. When you set it under this level, the regulator absorbs the excess "velocity," and turns that energy into heat. But when you pass it through your 0.1ohm current meter DMM, the actual voltage you would record across your DMM leads (if you put another DMM on there on voltage setting) would be close to zero (maybe 1V or 2V tops)*, no matter what voltage you set it to. Almost all the energy is just turned into heat in your PSU. There are other sources of resistance in the PSU that are higher in sum than the 0.1R shunt in your DMM which are busy turning this energy into heat.

*ALL real PSU have internal resistance called impedance in this specific instance. 12V PSU with lower impedance (stronger PSU, essentially) will get that reading higher than one with higher output impedance. This is not necessarily proportional to the max current output, but there's usually a correlation. This stronger PSU would make your DMM do more of the job of turning this energy to heat.

Mr D:
Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

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