Why do Batteries Have Voltage?

shintashi said:

I'm still not sure why 9v or 12v of 200mA fries a diode where 3.7v of 200mA doesn't. I thought all the electrons were moving at near light speed. How can they be moving faster? I think this is the part of pressure I don't get, because in fluid dynamics, if you increase the pressure through a pipe, you get more speed, but electrons are all transferring at the same speed, so I thought you were simply transferring MORE electrons, but since Amperage is how many electrons, I dont understand why Voltage matters as long as you already have your Amps calculated.

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It seems like a contradiction doesn't it? But the laws are absolute, so what actually happens?

Pretend you have a laser diode that has a forward voltage drop of 3.5V, and you use a 3.7V battery to power it. It works, and doesn't burn up. Then you hook up a 9V battery with the same current, why did it go up in smoke?

Well it's because you have two conflicting voltages. On one hand, your diode is supposed to only drop 3.5V; if the diode is passing current, you should be able to measure 3.5V across it. On the other hand, you have a 9V battery attached to the ends. This creates a voltage conflict: it should be 3.5V and should be 9V. This conflict means that something has got to give (i.e. break), and it can often mean that your diode (the weakest link) burns up when trying to eat up all that extra voltage.

So remember, your absolute assumptions (laws, parameters, etc.) MUST stay true in ALL cases. If there is an unavoidable contradiction (even if your professor writes the problem on the test wrong) there will be catastrophic damage.

In the 3.7V vs 3.5V case, that 0.2V just isn't enough to cause much damage. It might cause some heat in the diode, but probably not enough to damage it.

Bionic-Badger said:

It seems like a contradiction doesn't it? But the laws are absolute, so what actually happens?

Pretend you have a laser diode that has a forward voltage drop of 3.5V, and you use a 3.7V battery to power it. It works, and doesn't burn up. Then you hook up a 9V battery with the same current, why did it go up in smoke?

Well it's because you have two conflicting voltages. On one hand, your diode is supposed to only drop 3.5V; if the diode is passing current, you should be able to measure 3.5V across it. On the other hand, you have a 9V battery attached to the ends. This creates a voltage conflict: it should be 3.5V and should be 9V. This conflict means that something has got to give (i.e. break), and it can often mean that your diode (the weakest link) burns up when trying to eat up all that extra voltage.

So remember, your absolute assumptions (laws, parameters, etc.) MUST stay true in ALL cases. If there is an unavoidable contradiction (even if your professor writes the problem on the test wrong) there will be catastrophic damage.

In the 3.7V vs 3.5V case, that 0.2V just isn't enough to cause much damage. It might cause some heat in the diode, but probably not enough to damage it.

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so in some sense, voltage is the sideways pressure on the walls of the pipes, and pipe diameter is related to ohms, but when your circuit piece has thin walls - i.e., low "forward" voltage, it's the weakest link, and likely to blow out.

that makes me think wires have forward voltage. What is forward voltage, and do copper wires have it?

They do, but it is extremely small... but they do have a decent sized resistance, estimating in the 1ohm range for 26 gauge wire for a 1m length.

shintashi said:

so in some sense, voltage is the sideways pressure on the walls of the pipes, and pipe diameter is related to ohms, but when your circuit piece has thin walls - i.e., low "forward" voltage, it's the weakest link, and likely to blow out.

that makes me think wires have forward voltage. What is forward voltage, and do copper wires have it?

Click to expand...

Water in pipes does actually have direct analogies to electricity.

Think of the strength of the walls of your pipe as the ability of the wire to withstand heat. Like too much current in a wire, if there is too much water passing through a small pipe it will burst.

Forward voltage is a parameter of diodes, resistors and other passives, not of wires. Wires are considered just conductors, in other words, any given point of a continuous wire will have the same voltage potential. It's like thinking of a pipe of water: the pressure at any given location of the pipe is the same. Pipes and wires still do have "friction"/resistance in them, even if extremely small, and this is why they can still burst or burn out if too much flow is through them. Superconductors have no resistance at all--like a frictionless pipe.

For a resistor, think of it as a pipe that is in the same shape as the resistor symbol: --/\/\/\/\/\--. If that is the case, the water entering the resistor would need to bounce back and forth between the sides to get out, expending some of the energy potential in the water, but maintaining the same amount of flow because what goes in must also go out.

The water analogy for a capacitor is a piston in a tube with a spring attached. The water can push the plate only so far compressing the spring. When the pressure is released, the spring will push water back out. This kind of thing occurs with the charge contained in a capacitor. Discharging the capacitor is akin to the spring being released.

An inductor is like a flywheel in the water. The water pushes it along and it starts to spin, but it is hard to slow the flywheel down if it is already spinning, and it'll continue to force the flow of water in the pipe.