4
\$\begingroup\$

I'm still trying to wrap my head around current, voltage and resistance, and slowly I have the feeling that I start to get some things. Hence, I would like to describe my understanding so far and ask whether it's correct so far.

Current, voltage and resistance

First of all, inside of a circuit there's a stream of electrons. While technically they go from the negative to the positive end, we usually consider it to be the other way round.

Now since it's a stream, we can count how many electrons pass an arbitrary point in the circuit in a specific amount of time. This is what is called current, and it's measured in Ampere.

So: 1A is so and so many electrons per time frame.

Something has to make the electrons move. The force that makes them move is the voltage, and it's measured in Volt. Finally, there is resistance, and it's measured in Ohm. While voltage "drives" electrons, resistance slows them down.

We have: 1V is the force needed to move 1A over a resistance of 1 Ohm.

If we increase the voltage, we get a higher current; and if we increase the resistance, we get a lower current. If we multiply current and voltage, we get the actual power, which is measured in Watt.

Is this correct so far?

Short circuits

Now, if we have a 9V battery, and we connect both of its poles directly using a wire, there is almost no resistance. Using Ohm's law, we get a very high current due to the extremely low resistance. This results in a huge amount of energy moved in a very short time, hence the wire and the battery heat up, and this is what we call a short circuit.

Adding an LED

Now let's make an LED brighten up. Since an LED is a diode, electrons can only flow through it in one direction. So let's connect the LED's anode and cathode to the battery's poles, and see what is going to happen.

The LED causes a voltage drop of, let's say 2.5V, and is able to handle a maximum of 20 mA. Now if we connect it directly to the battery, yes, well, what is going to happen?

I know that we should add a resistor, and I even know how to calculate it, but I can not explain why this is, or to put it another way round: What would happen if we did not add a resistor.

Let me explain this:

I know that if we have a 9V battery, and the LED drops 2.5V, there still need 6.5V to be dropped. Since the LED handles 20 mA, and current is everywhere the same in a circuit, we need to calculate:

6.5 / 0.02 = 325

Hence, we need a resistance with 325 Ohm. Since the resistor now drops 6.5V voltage, overall we drop

6.5V + 2.5V = 9V

which is exactly the potential of the battery. Also, it doesn't matter whether we put the resistor before or behind the LED into the circuit, as all that counts is the drop overall. So everything is fine.

Open questions

But what happens if we do not add the resistor? In this case I have some questions that I am not able to answer:

  • The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V? Where are they? What happens to them? What happens to the battery, and what to the LED?
  • If we do not add a resistor, I guess the current will be way too high for the LED, so I guess we will have a short circuit, but now with an additional blowing LED, right?
  • What is the actual current in the circuit?
\$\endgroup\$
  • 2
    \$\begingroup\$ If instead of "Somebody has to make the electrons move. The power that makes them move is the voltage", it read "Something has to make the electrons move. The force that makes them move is the voltage", it'd be more accurate. Power=Voltage x Current. Similarly "1V is the power needed to move 1A over a resistance of 1 Ohm" becomes "1V is the force needed to move 1A against a resistance of 1 Ohm" \$\endgroup\$ – gbulmer Dec 12 '15 at 21:31
  • \$\begingroup\$ Ah perfect, thanks. I'm not a native English speaker, so I wasn't too sure about the wording, I will change that. \$\endgroup\$ – Golo Roden Dec 12 '15 at 21:33
  • 1
    \$\begingroup\$ Oh, and thinking that the electrons flow around in the circuit will lead to a lot of pain and misunderstanding in the future. The electrons bounce around at a speed of an ant. It's easer to just think of "current" as an abstract entity, and then you don't have to start thinking of positive and negative electron directions and such things. \$\endgroup\$ – pipe Dec 12 '15 at 21:38
  • 1
    \$\begingroup\$ The important point about an LED connected to 9V without a resistor is not that it drops 2.5V and handles 20mA, under 'well controlled conditions'. The point is an LED has a low resistance, so it looks like a short circuit. So the 9V battery will force a large current through the LED, causing it to heat rapidly and be destroyed. You stop this by using a resistor in series to limit the current through the LED. \$\endgroup\$ – gbulmer Dec 12 '15 at 21:38
5
\$\begingroup\$

But what happens if we do not add the resistor? ... The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V?

If you have a battery and LED with no current-limiting resistor, then you basically have the same scenario as when you just connect a wire across the two battery leads, but now with only 6.5 V instead of 9 V.

Basically, in this scenario you can no longer neglect the internal resistance of the battery or the forward resistance of the LED (and possibly the resistance of the wire) if you want to determine the actual current. A very large current is produced.

What happens to the battery, and what to the LED?

Like in the short-circuited battery case, the battery heats up. If the LED is only rated for 20 mA, and you put 100 mA or 1 A through it like you might in this situation, then very quickly the light-emitting diode becomes a smoke-emitting diode (at least momentarily) and then it usually becomes an open circuit.

What is the actual current in the circuit?

As I mentioned above, you need to consider the internal resistance of the battery and the equivalent resistance of the LED, and maybe the wire resistance. From these you can estimate the actual current. You probably can't calculate it exactly because the numbers depend on temperature, which will be rapidly changing as the circuit heats up and then cools down after the LED pops.

\$\endgroup\$
2
\$\begingroup\$

You have a good understanding of how to calculate the value for the resistor. Now, what happens when you don't use it? Take a look at the datasheet for any LED. It will have a nice graph of the voltage drop vs. the current. In theory you can extrapolate this to your 9 volt, and estimate roughly the current that will go through the LED. Assuming that your LED could withstand this current, that is what would happen.

What will happen in reality is that the battery will see a near short circuit for a very short time, until the LED burns out (in a fraction of a second), and then there will most likely be an open circuit. During that short time, the current is difficult to calculate because the LED will not be characterized for such gross abuse.

\$\endgroup\$
2
\$\begingroup\$

Current, voltage and resistance

First of all, inside of a circuit there's a stream of electrons. While technically they go from the negative to the positive end, we usually consider it to be the other way round.

Now since it's a stream, we can count how many electrons pass an arbitrary point in the circuit in a specific amount of time. This is what is called current, and it's measured in Ampere.

This is correct, but not about the quantity of electrons passing through a given point, if not, the amount of electrons passing through the transverse surface. Consider this, will help you calculate wire size required for a given current.

So: 1A is so and so many electrons per time frame.

Something has to make the electrons move. The force that makes them move is the voltage, and it's measured in Volt. Finally, there is

Technically, the voltage is a measure of the amount of energy available per unit of electric charge (electrons), not the amount of force. The force is related to the electric field, and this is closely related to the voltage, but are not the same.

resistance, and it's measured in Ohm. While voltage "drives" electrons, resistance slows them down.

We have: 1V is the force needed to move 1A over a resistance of 1 Ohm.

1 volt, it means we have available one Joule of energy to move one Coulomb of charge in 1 second, through a resistance of 1 Ohm. Think energy.

Open questions

But what happens if we do not add the resistor? In this case I have some questions that I am not able to answer:

The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V? Where are they? What happens to them? What happens to the battery, and what to the LED?

Think in terms of energy. The battery has 9 Joules per Coulomb of electric charge, while the LED, needs only 2.5 Joules per Coulomb. The rest of the unused energy is transformed, I would say in heat and sound.

If we do not add a resistor, I guess the current will be way too high for the LED, so I guess we will have a short circuit, but now with an additional blowing LED, right?

There will be short, excess energy destroys the LED, leaving open circuit. There will be a transitional behavior, where the LED will shine with maximum intensity before being destroyed.

What is the actual current in the circuit?

The current in the circuit does not have a stable value. It will vary from zero to the maximum value that supports the LED before destroyed within milliseconds, and then will again be zero.

\$\endgroup\$
0
\$\begingroup\$

In addition to pipe. In reality you can add a small resistance, the so called parasitic resistance, between the LED and battery, because of the wires, battery terminals etc, which have in the non-ideal case a small resistance. The voltage will then be divided across this resistance and the LED. The current can then again be calculated using that resistance. The exact value might be difficult to determine, since you will find some problems when determining the resistance.

What pipe already said. You should always look up the characteristics of the LED in the datasheet. The datasheet gives some useful insight on how to model your LED.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.