Megaphat Information Networks



Electricity 101

Voltage, Amperage and Power

So often I have to explain some basic things to people. Whether my kids, wife, mother, sister, customer, a tech (I prolly shouldn't say that but it's true). here's the thing. 12VDC @ 3.5A, wtf does that mean? Ok well, 12 Volts Direct Current at 3.5 Amps. First what is misleading is that I said Direct Current. No I didn't mean to say Alternating Current, I meant Direct Current. We'll get into that whole AC/DC thing in a bit.

So Direct Current actually describes the way current flows, but it's worth noting the importance of the fact that I am still talking about Voltage. So when discussing Voltage, why am I using the term Current? Are they the same thing? No. Are they similar? No. So then, wtf?

Okay so Voltage is how many Electrons travel in a path at the same time. So for example, 2 trains are traveling through a mountain tunnel. The tunnel can fit 1 train at a time. Trying to squeeze 2 trains into this tunnel at the same time will cause severe damage to the tunnel, the tracks (rail system) and of course the trains. So what we need to focus on here is the "rail system" as that will be the example of the electrical circuit. A circuit is designed to operate with a specific number of electrons to the components. Exceeding this number of electrons will simply fry the components. However the current is the direction that the electrons are flowing. Direct current means that the electrons are flowing inwards on one path and outwards on another. These paths are what we commonly call positive and negative. When the positive and negative flow of electrons are on dedicated paths the current flows direct. As most micro-electronics operate on DC voltage we see that there are so many power adapters used to plug them in. Some people call them power bricks, others wall worts, and some simply call them adapters. These adapters are rated at specific voltage and amperage.

We can see that the trains were traveling on 2 paths, inbound (positive) and negative (outbound). As the path of the electrons is direct, we call this Direct Current. What if the trains could zig-zag back and forth from positive to negative and back? But why would we want a zig-zag? While electricity never loses power, there is a possibility that there is something in the path that will slow down the electrons (or the train). We call this Resistance. With electricity, even the metal wiring can have Resistance. In order to minimize the effect of Resistance, we can Alternate the current from positive to negative and back. This will help avoid the depleting effect Resistance plays on current traveling down a path. So to visualize this, imagine riding on a bike path which is parralel to another bike path. As you encounter a pedestrian you hop from path 1 to path 2, which helps you avoid the need to slow down (but of course there is the possibility of just running the pedestrian over, but that's not part of this story). When you see a pedestrian on path 2, you hop back to path 1.

Alternating Current (AC) is when the positive and negative flow of electrons alternate paths. This is especially useful over long distances in order to retain a constant flow of electricity. The inherent Resistance of the wiring has a lesser effect in AC current than DC current. While most electrical components are DC driven, AC is delivered from the street because it can easily travel further distances. Last note about AC current is that it is rapidly changing paths. Unlike DC, AC creates an resonating frequency (some call it a hum) as it Alternates. So when you read AC Voltage ratings, you may see 110-120VAC @ 60Hz where Hz represents Hertz which is a measure of frequency, or how fast the current Alternates from one path to another. Since it is important I will simply say that Hertz represents a cycle or Alternation in a single second. So 60Hz means it Alternates 60 times per second. I will cover frequencies another time (believe me there is a lot to cover there).

Earlier I mentioned that the path of the rail system (circuit) is designed to handle a certain capacity. How many trains is not only dependent on inbound and outbound, but the size of the rails as well determines the size of the train. So in other words, if the circuit is designed for a train of 12 Volts, you cannot put a 48 Volt train on those rails. By doing so will damage the rails because the train is 4 times larger than the rail system can handle. This not only will damage the rails (circuit) but the Resistance of the path will probably create a large amount of heat and the train catches fire. So the train must be at the right size in order to enter the rail system. So what if the train is smaller than the rated Voltage of the rail system? Well the train isn't just traveling for no reason. The train has a purpose. For that the train is carrying cargo and the cargo are electrons. The circuit requires a specific number of electrons to be delivered to the components. Too little and the circuit cannot function. To great and the circuit will fry. This is the function of Voltage.

Now the trains need to deliver electrons at a specific (more like a minimum) speed. While the circuit is capable of limiting the rate at which the electrons can be delivered by the Voltage by slowing down the flow rate (or going back to our train analogy, the station inside the tunnel has traffic control mechanisms which can slow down the rate of the trains traveling within the rail system), the trains still need to deliver the electrons (yes you got it, electrons are the passengers, duh). Circuits are designs to control the speed of delivery for electrons just the same. However if the number electrons delivered are too slow, the circuit cannot function and the circuit cannot increase the speed of delivery.

Ultimately the circuits need Power in order to operate. Power is the number of electrons delivered in each train is the actual power. Typically this is measured in an hourly rate. So let's put this altogether.

Power = Voltage * Amperage (or Current)

Now let's use the mathematical equivelant to say this.

P = V * I

So if you look at a DC power adapter (brick, wort, supply or whatever stupid term you wanna use) it will say something like:

INPUT: 100-120VAC @ 60HZ
OUTPUT: 12VDC @ 3.5A
MAX POWER: 42W

So MAX POWER is P = V * I or 42 = 12 * 3.5

Did you catch that? I said 42W, what does W mean? WATTS BABY WATTS! Wattage is the measure of Power delivered using electricity.

So what if the circuit (or device) I have is rated for 60 Watts and not 42? What if I use my 12VDC @ 3.5A power supply (42W) for a 12VDC/60W circuit? Will it work? Probably yes BUT I am sure you can feel the supply adapter heating up as the circuit is demanding that the supply speed up the delivery of electrons at a rate the supply was not designed for. Now remember at this point we noted that the Voltage of the circuit is 12VDC. What if we increase our power supply to a 20VDC @ 3A? Well since the circuit was not designed to handle the number of electrons being delivered at once, the circuit will simply begin to fry because it was not designed to handle that large of input. So a 12VDC 60 Watt circuit needs 12VDC at 5 Amps (12 * 5 = 60). But 20 * 3 = 60 as well, but 20V is way too many electrons at once so be sure you're ready to watch your circuit burn. Oh but wait, 3 Amps is slower than 5 Amps so the electrons are not coming in as quick, so that should be okay. No, it's not okay. It doesn't matter how fast the electrons come in, it matters that too many are coming in at once and the circuit cannot handle that capacity.

So on one hand, too little current will cause the supply to heat up because of the demand on the circuit. Most supplies will brave through it and continue to work but some poorly made supplies will simply burn out. Too great of voltage to the circuit will certainly burn out your circuit because the number of electrons at once is too great.

Let's take another gander at the formula P = V * I for a moment. If we have a circuit requiring 42W from 12VDC and we use a 60W adapter delivering 12VDC, what will happen? The circuit will operate just fine. You are delivering the correct number of electrons to the circuit and the circuit will only take the electrons at a specific rate so any additional electrons will simply be sent back out of the circuit on the next train leaving the circuit. In fact, you may find that a 42W 12VDC circuit may slightly heat up a 12V @ 3.5A supply just a little bit (since most are not of the best quality) but a 12V @ 5A supply should not heat up at all and the circuit will operate just fine. If a supply adapter is properly made to operate a full-duty cycle (all the time), it should not heat up and be able to provide all the power needed.



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