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I have always had a basic understanding of electronics. I am now starting to learn a bit more, using an Arduino as a test platform, and I have a question about resistors that I can not seem to solve through research.

Why do we use them? I understand that they limit current. (In the case of an LED, too much current would heat it up and burn it.) But how is this measured/calculated/chosen? I am not specifically asking about an LED use case, or how to use an LED. I am trying to understand "why" resistors are needed on a physics level.

  1. What happens to the rest of the current not used (because of the resistor)?
  2. Does the LED then use ALL of the current available in the circuit? If not, where does the rest go? (Recycled back into power source?)
  3. Why does an LED "drop voltage" by a certain amount? And what happens to the rest of the components in series, does the voltage drop for every component, up until there is nothing left? This would make sense, but an LED does not have internal resistance (so it is explained), so why does it drop voltage?
  4. I recently watched a video, where the guy explaining resistors, drew a sketch showing 12 V → resistor → LED --- 0 V (Do you choose your resistor to the extent of "using up all the current/voltage" before it gets to the end of the circuit? YouTube video
  5. Why does a battery go into a dead short if you connect the terminals directly, but if you add a light bulb (resistor), it does not?
  6. I have done hours and hours of research, and I understand what a resistor does, but I do not understand why it is needed (to not dead short a battery? ... Does this mean it "eats" all of the power before it returns to the anode?)
  7. Why do different light bulbs work on the same battery (different resistance, but no dead short?)

I know these questions are broad, and I am not specifically looking for answers to each of them individually. I am mentioning these multiple questions above to demonstrate that I do not have a firm grasp on the concept of why a circuit needs resistance. This would be the question to answer.

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closed as too broad by old_timer, Dmitry Grigoryev, pipe, PeterJ, Mitu Raj Feb 25 '18 at 15:23

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Nick Alexeev Feb 23 '18 at 21:26

12 Answers 12

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Your understanding of how power flows through a circuit needs adjusting.

1. How much power flows through a circuit, and is taken from the battery or power source, depends on how much current flows through that circuit.

2. How much current flows though the circuit is dictated by how conductive the circuit is. If a circuit has a high resistance, it is less conductive, and less current/power flows.

So, putting those two together and looking at your questions...

1.What happens to the rest of the current not used (because of the resistor)?

There is no "rest of the current" the current is defined by the resistance of the circuit.

2.Does the LED then use ALL of the current available in the circuit? If not, where does the rest go? (Recycled back into power source?)

Again, the LED and its resistor define the current they will take. There is no "rest".

3.Why does an LED "drop voltage" by a certain amount? And what happens to the rest of the components in series, does the voltage drop for every component, up until there is nothing left?

The LED has a more or less fixed forward voltage at a given current. The rest of the voltage is dropped across the resistor. That defines the current through the LED.

4.I recently watched a video, where the guy explaining resistors, drew a scetch showing 12v --> Resistor --> LED --- 0V (Do you choose your resistor to the extent of "using up all the current/voltage" before it gets to the end of the circuit? Youtube Video

In any series circuit, the applied voltage gets divided among the elements of that series circuit. The current is defined by what the circuit elements demand and is constant throughout the series circuit.

Keep in mind that voltage is simply a measurement of the potential for electrons to flow between two points. It is always measured between two points, and a value of 0 volts tells us that there would be no current between those same two points.

5.Why does a battery go into a dead short if you connect the terminals directly, but if you add a light bulb (resistor), it does not?

A dead short has virtually zero resistance and take a lot of current from the supply. A bulb has a resistance and takes much less current.

6.I have done hours and hours of research, and I understand what a resistor does, but I do not understand why it is needed (to not dead short a battery? .. does this mean it "eats" all of the power before it returns to the Anode?)

Resistors are needed to set currents and adjust voltage levels through a series circuit. They are used for other functions too, like part of frequency filters, oscillators etc. etc.

7.Why do different light bulbs work on the same battery (different resistance, but no dead short?)

Different light bulbs have different resistances.


In order to understand all this you need to familiarize yourself with Ohm's Law and Kirchoff's Voltage Law.


EDIT: Adding comment question since it's useful all on its own and may get migrated.

Am I correct in stating the following: "If I put an LED directly on a 600maH power source, it will "use" everything that is available (600maH). Do I then calibrate for the resistor to resist enough current in order to feed the LED only what it needs?

A 600mAh power source is fairly meaning less here. mAh is a measure of how much charge and effectively total power a battery will supply in any given time. If your circuit takes 1mA the battery will last 600 hours. If your circuit takes 1A the battery will only last 36 minutes. Note the units... mA * Hours.

A bigger battery, of the same technology and voltage, has more mAh.

How much power it can deliver at any given time is dependent on the terminal resistance of the battery and how fast the chemistry inside the battery can react. A 3.7V 600mAh Li-Ion battery will deliver much more raw power than a 1.5V 600mAh Alkaline. Power and Energy is not the same thing. Ultimately, though, the load, the circuit, dictates how much it sucks out of the battery and how fast, assuming it's not drawing too fast, at which point the battery voltage will drop off.

You have to think of a battery like the gas tank on your car. How quickly the gas goes down depends on how hard and fast you are driving. 600mAh only defines how big the "gas tank" is. The gas has to go from the tank to the engine through a pipe and the injectors. If you demand too much gas, it won't make it through those fast enough and the engine gets starved of gas.

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    \$\begingroup\$ Thank you. That actually puts things into better perspective. Am I correct in stating the following: "If I put an LED directly on a 600maH power source, it will "use" everything that is available (600maH). Do I then calibrate for the resistor to resist enough current in order to feed the LED only what it needs? \$\endgroup\$ – Louis van Tonder Feb 22 '18 at 15:38
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    \$\begingroup\$ @LouisvanTonder um, not quite. mAh is how much energy is stored in the battery. A bigger battery has more mAh. How much power it can deliver at any given time is dependent on the terminal resistance of the battery and how fast the chemistry inside the battery can react. A 1.5V 600mAh Lion battery will deliver much more raw power than a 1.5V 600mAh Alkaline. Power and Energy is not the same thing. Ultimately, though, the load, the circuit, dictates how much it sucks out of the battery and how fast, assuming it's not drawing too fast, at which point the battery voltage will drop off. \$\endgroup\$ – Trevor_G Feb 22 '18 at 15:51
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    \$\begingroup\$ @LouisvanTonder continued: You have to think of a battery like the gas tank on your car. How quickly the gas goes down depends on how hard and fast you are driving. 600mAh only defines how big the "gas tank" is. The gas has to go from the tank to the engine through a pipe and the injectors. If you demand too much gas, it wont make it through those fast enough and the engine gets starved of gas. \$\endgroup\$ – Trevor_G Feb 22 '18 at 15:55
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    \$\begingroup\$ @Trevor: The traditional analogy is that voltage === pressure; current === flow rate; resistance === pipe bore. If I had a five-inch pipe from the header tank in my loft to the bath, then it would be unnecessarily full in a matter of seconds while the water heater couldn't come close to keeping up. There is also the spring and damper model for inductors and capacitors, which is mainly for AC applications. And then we get into imaginary maths! \$\endgroup\$ – Borodin Feb 23 '18 at 20:07
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    \$\begingroup\$ @Borodin there are a million analogies. SOme of the plumbing ones are lost on folks that don't understand plumbing either LOL \$\endgroup\$ – Trevor_G Feb 23 '18 at 20:11
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Here's a physics-based intro to the EE concepts you're trying to understand.

Your questions are answered at the bottom.


Everything derives from the flow of "charge"

Electronics, as its root word electron denotes, is very much a study of the flow of electrons in a particular system.

Electrons are the fundamental "carriers" of charge in a typical circuit; i.e., they are how charge gets "moved around" in most circuits.

We adopt a signing convention saying that electrons have a "negative" charge. Moreover, an electron represents the smallest unit of charge at the atomic (classical physics) scale. This is called the "elementary" charge and sits at \$-1.602{\times}{10}^{-19}\$ Coulombs.

Conversely, protons have a "positive" signed charge of \$+1.602{\times}{10}^{-19}\$ Coulombs.

However, protons cannot move around so easily because they are typically bound to neutrons within the atomic nuclei by the nuclear strong force. It takes far more energy to remove protons from atomic nuclei (the basis for nuclear fission technology, by the way) than to remove electrons.

On the other hand, we can dislodge electrons from their atoms pretty easily. In fact, solar cells are based entirely on the photoelectric effect (one of Einstein's seminal discoveries) because "photons" (particles of light) dislodge "electrons" from their atoms.


Electric fields

All charges exert an electric field "indefinitely" into space. This is the theoretical model.

A field is simply a function that produces a vector quantity at every point (a quantity containing both magnitude and direction... to quote Despicable Me).

An electron creates an electric field where the vector at each point in the field points towards the electron (direction) with a magnitude corresponding to Coulomb's law:

$$\lvert \vec E\rvert~~=~~\underbrace{\frac{1}{4\pi\epsilon_{0}}}_{ \begin{array}{c} \text{constant} \\ \text{factor} \end{array} }~~\underbrace{\frac{\lvert q\rvert}{r^{2}}}_{ \begin{array}{c} \text{focus on} \\ \text{this part} \end{array} }$$

The directions can be visualized as:

Electric Field Directions

These directions and magnitudes are determined based on the force (direction and magnitude) that would be exerted upon a positive test charge. In other words, the field lines represent the direction and magnitude a test positive charge would experience.

A negative charge would experience a force of the same magnitude in the opposing direction.

By this convention, when an electron is near an electron or a proton near a proton, they will repulse.


Superposition: collections of charges

If you sum up all the electric fields exerted individually by all charges in a region on a particular point, you get the total electric field at that point exerted by all the charges.

This follows the same principle of superposition used to solve kinematics problems with multiple forces acting on a singular object.


Positive charge is the absence of electrons; negative charge is the surplus of electrons

This specifically applies to electronics where we are dealing with charge flow through solid materials.

To re-iterate: electronics is the study of the flow of electrons as charge carriers; protons are not the primary charge carriers.

Again: for circuits, electrons move, protons do not.

However, a "virtual" positive charge can be created by the absence of electrons in a region of a circuit because that region has more net protons than electrons.

Recall Dalton's valence electron model where protons and neutrons occupy a small nucleus surrounded by orbiting electrons.

The electrons that are the farthest away from the nucleus in the outermost "valence" shell have the weakest attraction to the nucleus based on Coulomb's law which indicates that electric field strength is inversely proportional to the square of the distance.

By accumulating charge e.g. on a plate or some other material (say, by rubbing them vigorously together like in the good ol' days), we can generate an electric field. If we place electrons in this field, the electrons will macroscopically move in a direction opposite the electric field lines.

Note: as quantum mechanics and Brownian motion will describe, the actual trajectory of an individual electron is quite random. However, all electrons will exhibit a macroscopic "average" movement based on the force indicated by the electric field.

Thus, we can accurately calculate how a macroscopic sample of electrons will respond to an electric field.


Electric potential

Recall the equation based on Coulomb's law indicating the magnitude of force \$\lvert \vec E\rvert\$ exerted on a positive test charge:

$$\lvert \vec E\rvert = \frac{1}{4\pi\epsilon_{0}} \frac{\lvert q\rvert}{r^{2}}$$

From this equation, we see as \$r \to 0\$, \$\lvert \vec E\rvert \to \infty\$. That is, the magnitude of force exerted on a positive test charge becomes larger the closer we get to the origin of the electric field.

Said in the opposite, as \$r \to \infty\$, \$\lvert \vec E\rvert \to 0\$: as you get infinitely far away from the origin of an electric field, the field strength tends to zero.

Now, consider the analogy of a planet. As the total cumulative mass of the planet increases, so does its gravity. The superposition of the gravitational pulls of all the matter contained in the planet's mass produces gravitational attraction.

Aside: the mass of your body exerts a force on the planet, but the mass of the planet so far exceeds your body's mass \$\left(M_{\text{planet}} \gg m_{\text{you}}\right)\$ that your gravitational attraction is eclipsed by the planet's pull.

Recall from kinematics that gravitational potential is the amount of potential an object has owing to its distance from the planet's gravitational center. The planet's gravitational center can be treated as a point gravity source.

Similarly, we define electric potential as how much energy is required to move a positive test charge \$q\$ from infinitely far away to a specific point.

In the case of gravitational potential, we assume that the gravity field is zero infinitely far away from the planet.

If we have a mass \$m\$ that starts infinitely far away, the planet's gravitational field \$\vec g_{\text{planet}}\$ does work to pull the mass closer. Therefore, the gravitational field "loses potential" as a mass approaches the planet. Meanwhile, the mass accelerates and gains kinetic energy.

Similarly, if we have a positive test charge that starts from infinitely far away from a source charge \$q_{\text{source}}\$ which generates an electric field \$\vec E_{\text{source}}\$, the electric potential at a point is how much energy would be required to move the test charge to some distance \$r\$ from the source charge.

This results in:

  • Negative charges gain electric potential when moving in the direction of the electric field \$\vec E\$ and away from a positive source charge.
  • Negative charges lose electric potential when moving opposite the direction of electric field \$\vec E\$ and towards a positive source charge.
  • Conversely, positive charges lose electric potential when moving in the direction of the electric field \$\vec E\$ and away from a positive source charge.
  • Positive charges gain electric potential when moving in the direction opposite the electric field \$\vec E\$ and towards a positive source charge.

Electric potential in conductors

Consider the model of conductors or transition metals like copper or gold having a "sea of electrons". This "sea" is composed of valence electrons which are more loosely coupled and sort of "shared" amongst multiple atoms.

If we apply an electric field to these "loose" electrons, they are inclined, on a macroscopic average, to move in a specific direction over time.

Remember, electrons travel in the direction opposite the electric field.

Similarly, placing a length of wire conductor near a positive charge will cause a charge gradient across the length of wire.

The charge at any point on the wire can be calculated using its distance from the source charge and known attributes of the material used in the wire.

Positive charge owing to the absence of electrons will appear farther away from the positive source charge, while negative charge owing to the collection and surplus of electrons will form closer to the source charge.

Because of the electric field, a "potential difference" will appear between two points on the conductor. This is how an electric field generates voltage in a circuit.

Voltage is defined as electric potential difference between two points in an electric field.

Eventually, the charge distribution along the length of wire will reach "equilibrium" with the electric field. This doesn't mean charge stops moving (remember Brownian motion); only that the "net" or "average" movement of charge approaches zero.


Non-ideal batteries

Let's make up a galvanic or voltaic cell power source.

This cell is powered by the electrochemical redox reaction of Zinc and Copper rods in an aqueous solution of ammonium nitrate salt \$\left(\text{NH}_{4}\right)\left(\text{NO}_{3}\right)\$.

Ammonium nitrate is an ionically bonded salt that dissolves in water into its constituent ions \$\text{NH}_{4}^{+}\$ and \$\text{NO}_{3}^{-}\$.

Useful terminology:

  • cation: a positively charged ion
  • anion: a negatively charged ion
  • cathode: the cations accumulate at the cathode
  • anode: the anions accumulate at the anode

Useful mnemonic: "anion" is "an ion" is "A Negative ion"

If we examine the reaction for the Zinc-Copper galvanic cell above:

$$\text{Zn}\left(\text{NO}_{3}\right)_{2}~~+~~\text{Cu}^{2+} \quad\longrightarrow\quad \text{Zn}^{2+}~~+~~\text{Cu}\left(\text{NO}_{3}\right)_{2}$$

The movement of cations \$\text{Zn}^{2+}\$ and \$\text{Cu}^{2+}\$ is the flow of positive charge in the form of ions. This movement goes towards the cathode.

Galvanic Cell Charge Flow

Note: Earlier we said that positive charge is the "absence" of electrons. Cations (positive ions) are positive because stripping away electrons results in a net positive atomic charge owing to the protons in the nucleus. These cations are mobile in the galvanic cell's solution, but as you can see, the ions do not travel through the conductive bridge connecting the two sides of the cell. That is, only electrons move through the conductor.

Based on the fact that positive cations move and accumulate towards the cathode, we label it negative (positive charges are attracted to negative).

Conversely, because electrons move towards and accumulate at the anode, we label it positive (negative charges are attracted to positive).

Remember how you learned that current flows from \$\textbf{+}\$ to \$\textbf{-}\$? This is because conventional current follows the flow of positive charge and cations, not negative charge.

This is because current is defined as the flow of virtual positive charge through a cross-sectional area. Electrons always flow opposite to current by convention.

What makes this galvanic cell non-ideal is that eventually the chemical process generating the electric field through the conductor and causing electrons and charge to flow will come to equilibrium.

This is because ion buildup at the anode and cathode will prevent the reaction from proceeding any further.

On the other hand, an "ideal" power source will never lose electric field strength.


Ideal voltage sources are like magic escalators

Let's return to the analogy of gravitational potential.

Assume you're on a hill and you have some arbitrary path down the hill constructed with cardboard walls. Let's say you roll a tennis ball down this path with cardboard walls. The tennis ball will follow the path.

In circuits, the conductor forms the path.

Now let's say you have an escalator at the bottom of the hill. Like a Rube Goldberg machine, the escalator scoops up tennis balls you roll down the path, then drops them off at the start of the path at the top of the hill.

The escalator is your ideal power source.

Now, let's say you almost completely saturate the entire path (escalator included) with tennis balls. Just a long line of tennis balls.

Because we didn't completely saturate the path, there are still gaps and spaces for the tennis balls to move.

A tennis ball that is carried up the escalator bumps into another ball, which bumps into another ball which... goes on and on.

The tennis balls going down the path on the hill gain energy owing to the potential difference in gravity. They bounce into each other until finally, another ball is loaded onto the escalator.

Let's call the tennis balls our electrons. If we follow the flow of electrons down the hill, through our fake cardboard "circuit", then up the magical escalator "power source", we notice something:

The "gaps" between tennis balls are moving in the exact opposite direction of the tennis balls (back up the hill and down the escalator) and they are moving much faster. The balls are naturally moving from high potential to low potential, but at a relatively slow speed. Then they are moved back to a high potential using the escalator.

The bottom of the escalator is effectively the negative terminal of a battery, or the cathode in the galvanic cell we were discussing earlier.

The top of the escalator is effectively the positive terminal of a battery, or the anode in a galvanic cell. The positive terminal has a higher electric potential.


Current

Okay, so the direction that positive charge flows in is the direction of electrical current.

What is current?

By definition, it is: the amount of charge that passes through a cross-sectional area per second (units: Coulombs per second). It is directly proportional to the area of the cross section of the wire/conducting material and the current density. Current density is the amount of charge flowing through a unit of area (units: Coulombs per meter-squared).

Here's another way to think of it:

If you have a tennis ball launcher spitting positively charged balls through a doorway, the number of balls it gets through the door per second determines its "current".

How fast those balls are moving (or how much kinetic energy they have when they hit a wall) is the "voltage".


Conservation of charge and voltage

This is a fundamental principle.

Think of it like this: there are a fixed number of electrons and protons. In an electrical circuit, matter is neither created nor destroyed... so the charge always stays the same. In the tennis-ball escalator example, the balls were just going in a loop. The number of balls remained fixed.

In other words, charge does not "dissipate". You never lose charge.

What happens is that charge loses potential. Ideal voltage sources give charge its electric potential back.

Voltage sources do NOT create charge. They generate electric potential.


Current flowing in and out of nodes, resistance

Let's take that conservation of charge principle. A similar analogy can be applied to the flow of water.

If we have a river system down a mountain that branches, each branch is analogous to an electric "node".

          / BRANCH A
         /
        /
MAIN ---
        \
         \
          \ BRANCH B

-> downhill

The amount of water that flows into a branch must be equal to the amount of water flowing out of the branch by the conservation principle: water (charge) is neither created nor destroyed.

However, the amount of water that flows down a particular branch is dependent on how much "resistance" that branch puts up.

For example, if Branch A is extremely narrow, Branch B is extremely wide, and both branches are the same depth, then Branch B naturally has the larger cross-sectional area.

This means Branch B puts up less resistance and a larger volume of water can flow through it in a single unit of time.

This describes Kirchoff's Current Law.


You're still here? Awesome!

1. What happens to the rest of the current not used?

Because of the conservation principle, all charge into a node must flow out. There is no "unused" current because current isn't used. There is no change in current in a single series circuit.

However, different amounts of current can flow down different branches in an electrical node in a parallel circuit depending on the resistances of the different branches.

2. Does the LED use all the current?

Technically, the LED and resistor(s) don't "use" current, because there is no drop in current (the amount of charge passing through the LED or resistor(s) in a unit of time). This is because of the conservation of charge applied to a series circuit: there is no loss in charge throughout the circuit, hence no drop in current.

The amount of current (charge) is determined by the behavior of the LED and resistor(s) as described by their i-v curves

3. Why does the LED "drop voltage" by a certain amount?

Here's a basic LED circuit.

An LED has an activation voltage, usually around ~1.8 to 3.3 V. If you do not meet the activation voltage, practically no current will flow. Refer to the LED i-v curves linked below.

If you attempt to push current in the direction opposite the LEDs polarity, you will be operating the LED in a "reverse-bias" mode in which almost no current passes through. The normal operating mode of an LED is forward-bias mode. Beyond a certain point in reverse-bias mode, the LED "breaks down". Check out the i-v graph of a diode.

LEDs are actually PN junctions (p-doped and n-doped silicon squashed together). Based on Fermi levels of the doped silicon (which is contingent on the electron band-gaps of the doped material) the electrons require a very specific amount of activation energy to jump to another energy level. They then radiate their energy as a photon with a very specific wavelength/frequency as they jump back down to a lower level.

This accounts for the high efficiency (well over 90% of energy dissipated by an LED is converted to light, not heat) of LEDs compared to filament and CFL bulbs.

This is also why LED lighting seems so "artificial": natural light contains a relatively homogeneous mix of a broad spectrum of frequencies; LEDs emit combinations of very specific frequencies of light.

The energy levels also explain why the voltage drop across an LED (or other diodes) is effectively "fixed" even as more current goes through it. Examine the i-v curve for an LED or other diode: beyond the activation voltage, the current increases a LOT for a small increase in voltage. In essence, the LED will attempt to let as much current flow through it as it possibly can, until it physically deteriorates.

This is also why you use an inline current-limiting resistor to limit the current flow through a diode / LED to a specific rated milliamp based on the LED spec.

3(b). And what happens to the rest of the components in series, does the voltage drop for every component, up until there is nothing left?

Yep, Kirchoff's voltage law is that the sum of all voltage drops in a loop around a circuit is zero. In a simple series circuit, there is only one loop.

4. Do you choose your resistor to the extent of "using up all the current/voltage" before it gets to the end of the circuit?

No. You choose your resistor based on the LED current rating (say 30 mA = 0.03 A) and Ohm's law as described in the LED circuit article.

Your voltage will get used up. Your current remains the same throughout a single series circuit.

5. Why does a battery go into a dead short if you connect the terminals directly, but if you add a light bulb (resistor), it does not?

I'm not sure what you mean by "dead short".

Connecting the terminals of a battery together results in a large current discharged at the voltage of the battery. That voltage is dissipated through the battery's internal resistance and the conductor wire in the form of heat -- because even conductors have some resistance.

This is why shorted batteries get super hot. That heat can adversely affect a chemical cell's composition until it blows up.

6. Why are resistors necessary?

Here's the rhetoric: imagine there's this amazing concert. All your favorite bands are going to be there. It's going to be a smashing good time.

Let's say the event organizers have no concept of reality. So they make the entry fee to this amazing concert almost completely free. They put it in an extremely accessible area. In fact, they're so disorganized, they don't even care if they oversell and there aren't enough seats for everyone who buys tickets.

Oh, and this is in NYC.

Pretty quickly, this amazing concert turns into a total disaster. People are sitting on each other, spilling beer everywhere; fights are breaking out, the restrooms are jammed, the groupies are freaking everyone out, and you can barely hear the music above all the commotion.

Think of your LED as that amazing concert. And think of how messed up your LED is going to be if you don't have more resistance there to prevent EVERYONE and their moms from showing up to the concert.

In this dumb example, "resistance" translates into "cost of entry". By simple economic principles, raising the cost of the concert decreases the number of people who will attend.

Similarly, raising the resistance in a circuit prevents charge (and subsequently current) from going through. This means your LED (concert) doesn't get completely wrecked by all the people (charge).

Yeah, electrical engineering is a real party.

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    \$\begingroup\$ Offtopic: How many hours did it take you to write this answer? \$\endgroup\$ – Harry Svensson Feb 23 '18 at 2:30
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    \$\begingroup\$ As demonstrated by this answer, the OP question is too broad. \$\endgroup\$ – StainlessSteelRat Feb 23 '18 at 2:53
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    \$\begingroup\$ @HarrySvensson ... I was so proud of myself until I pondered your question \$\endgroup\$ – afeique Feb 23 '18 at 3:09
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    \$\begingroup\$ Wow, I almost feel compelled to change the correct answer to this one purely on the layman's type of answers to my questions. That is perfect for a beginner such as myself to take in. Thank you for this awesome answer. \$\endgroup\$ – Louis van Tonder Feb 23 '18 at 7:56
  • \$\begingroup\$ Please do not accept an answer too fast. That discourages others from answering it, whilst there might be even better answers out there! Just wait one or two days and then pick an answer. \$\endgroup\$ – csg Feb 24 '18 at 6:29
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What's the quickest way to gain understanding of basic electricity? Just focus on "hot button" issues like the following. Fix your mental concepts, and everything snaps into place and makes sense.

Conductors are materials which are composed of "movable electricity." They don't conduct electricity, instead they contain electricity, and their electricity can move along. Beware of the widespread incorrect definition of conductors:

WRONG: conductors are transparent to current, like empty water pipes? Nope.

CORRECT: all conductors contain movable charge, like water-filled pipes.

Wires are like pre-filled hoses, where the electrons of the metal are like water already inside the hose. In metals, the atoms' own electrons are constantly jumping around and 'orbiting' all throughout the entire metal bulk. All metals contain a 'sea' of movable fluid-like electricity. So, if we hook some metal wires in a circle, we've created a kind of concealed drive-belt or flywheel. Once the loop is formed, the circular "electricity belt" is free to move inside the metal. (If we grab and wiggle our wire circle, we'll actually produce a tiny electric current by inertia, just as if the wire was a hose full of water. Search: Tolman effect.)

The path for current is a complete circle, including the power supply. Power supplies don't supply any electrons. (In other words, the circle has no beginning. It's a loop, like a movable flywheel.) The movable electrons are contributed by the wires themselves. Power supplies are just electricity-pumps. The path for current is through the power supply and back out. A power supply is just another part of the closed loop.

Electric currents are fairly slow flows. But, like wheels and drive-belts, when we push upon one part of the wheel, the entire wheel moves as a unit. We can use a rubber drive-belt to instantly transfer mechanical energy. We can use a closed loop of electricity to instantly transfer electrical energy to any part of the loop. Yet the loop itself doesn't move at the speed of light! The loop itself moves slow. And for AC systems, the loop moves back and forth while the energy moves continuously forward. Big hint: the faster the electrons, the higher the amps. Zero amperes? That's when the wires' own electrons come to a halt. Another hint: electrical energy is waves, and electrons are the "medium" along which the waves travel. The medium wiggles back and forth, while the wave propagates fast forward. Or, the medium jerks backwards, moving slowly, while the wave moves forward extremely rapidly. (In other words, no single "electricity" exists, since there were always two separate things moving inside circuits: the slow circular currents of electrons, and the fast one-way propagation of electromagnetic energy. They move with two entirely different speeds in circuits, and while currents flow in loops, the energy flows one-way from a source to a consumer.)

Batteries don't store electricity. They don't store electric charge. They don't even store electrical energy. Instead, batteries only store chemical "fuel" in the form of uncorroded metals such as Lithium, Zinc, Lead, etc. But then, how can batteries work? Easy: a battery is a chemically-powered charge-pump. As their metal plates corrode, chemical energy is released and they pump electricity through themselves. The path for current is through the battery and back out again. (Pumps aren't used for storing the stuff being pumped!) And, battery 'capacity' is just the quantity of chemical fuel inside. A certain amount of fuel is able to pump a certain total quantity of electrons before the fuel is used up. (It's a bit like rating your gas tank in miles of travel, rather than in gallons. Gas tanks don't store miles, and batteries don't store electricity!) Rechargable batteries? That's when we forcibly run them backwards, so their internal "exhaust products" get converted back into fuel: corrosion compounds get turned back into metal again.

Resistors don't consume electricity. When a light bulb is turned on, its own electrons begin moving, as new electrons enter one end of the filament, yet at the same time other electrons are leaving the far end. The filament is part of a complete ring of electrons which move like a drive-belt. The heating effect is a kind of friction, as when you push your thumb against the rim of a rotating tire. (Your thumb isn't consuming rubber, instead it's just heated up frictionally, and light bulbs don't consume electrons, they just "rub upon" the moving electrons, and heat up frictionally.) So, resistors are just friction devices. The path for electrons is through, and no electrons are consumed or lost. Note that the faster the electrons, the higher the amperes, and the greater the heating. "Low" current is just slow electricity.

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  • \$\begingroup\$ Thank you for this excellent write-up. You are not first, but well worth the extra reading to get my basic understanding where it should (have been..) ;-) Thanks. \$\endgroup\$ – Louis van Tonder Feb 23 '18 at 8:14
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I'm also a beginner but try to answer your questions:

  1. There is no 'rest' of the current. Current is used as much as needed. If you connect a wire from + (VCC) to - (GND) you get a short circuit. See it as there is no brake at how fast the electrons can run.

  2. If there is no resistor, the LED will use the electrons in the fastest 'speed' possible. Since this is too much, the LED will burn (sooner or later).

  3. I don't know the reason why it drops, probably the internal mechanism of the LED causes some voltage to be used. This means the remainder does have less voltage left. And yes, it will continue until nothing is left. This can cause further LEDs to be either not lighting at all, or blinking/behaving irregularly or being dimmed.

  4. Actually you should calculate it due to how bright you want your led to be. Thus a higher resistor makes the LED bright less.

  5. A light bulb has internal resistance, so a resistor is not needed.

  6. It does not eat battery, it just makes the flow of electrons go slower (at least that is an easy analogy).

  7. Each light bulb has an internal resistance, so it does not result in a short circuit. If you use too much voltage, it will break.

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    \$\begingroup\$ Thank you for your answers Michel. You have indeed given me some different perspectives. \$\endgroup\$ – Louis van Tonder Feb 22 '18 at 15:46
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    \$\begingroup\$ You're welcome, hope it helps; maybe some more experienced people can give more precise answers (greetings from the Netherlands). \$\endgroup\$ – Michel Keijzers Feb 22 '18 at 15:50
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    \$\begingroup\$ @already been done by Trevor_G I see :-) \$\endgroup\$ – Michel Keijzers Feb 22 '18 at 16:04
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Read about the electricity water model. It compares current with water flowing around and can help to understand what terms like current and voltage mean and how they act together.

Edit
I mentioned this model because it has helped me a lot to understand several things.
laptop2d is right, an explanation is better than "go look for that". But it's pretty lengthy to explain the whole thing here, when other sites already did that properly. I am not an expert and describing things in English may also not be the best idea... but let's try.

Correct me if I'm wrong!

Compare electricity with a water tank above - the source - and a water tank below - the sink. In the upper tank there is water that wants to run through a pipe into the lower tank. This is your battery. Charging the battery means to fill water from the bottom tank into the upper tank. Having an empty upper tank is an empty battery.
Imagine there is a pipe from top to bottom - the wire.
The water wants to flow down the pipe - the battery wants to produce electric current in the wire.
A valve in the pipe is compared to a switch.
Opening a valve to the half only can be understood as a resistor. It limits the flow of water.
A water wheel is a consumer and a resistor too. It limits the water flow too. If the valve is additionally used to create resistance, the wheel's rotation speed can be controlled.
The water pressure between the two tanks is the voltage. A higher placed tank has a higher pressure relative to the bottom tank.
The amount of water flowing in 1 second through the pipes is the current. Be aware of the time here!
Water pressure, the resistor and the amount of water flowing depend on each other. This is Ohm's law. A wide pipe with nothing else between lets water flow uncontrollably heavy - a short circuit. Tanks and pipe can be damaged.

With this model you can perhaps understand things better. For example that water not flowing through the wheel doesn't go anywhere else. It waits in the tank to be used later.

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  • \$\begingroup\$ It might be better to go on and describe the water model in your post so readers don't have to go look it up. \$\endgroup\$ – Voltage Spike Feb 22 '18 at 23:25
  • \$\begingroup\$ See the edit above with some basic things. \$\endgroup\$ – puck Feb 23 '18 at 9:29
  • \$\begingroup\$ +1 I was going to write this answer if you hadn't already posted it. A diode corresponds to a one-way valve. A one-way valve cannot be just empty pipe, so it will introduce some resistance to the water even if it is flowing in the right direction. It will also break if you push water through it too hard, in either direction, just like an actual diode will break if the voltage across it is too high. That being said, as with any analogy, this one has its weaknesses. One being the very complicated implementations you would eventually need for relatively simple electronic components. \$\endgroup\$ – Arthur Feb 23 '18 at 9:56
  • \$\begingroup\$ Not only is the water analogy a good one, but the formulas used in hydraulics and pneumatics are extremely similar to the ones used in electronics. \$\endgroup\$ – Kris Peeling Feb 23 '18 at 14:49
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The answers so far focus on the specific examples in the question, which are all fairly limited in scope. I believe the real misunderstanding stems from a greater familiarity with digital logic than traditional analogue circuits (leading to these limited examples).

Simplistically, a digital circuit (such as an MPU) can be built with just 'hard' on/off switching elements. Integrated circuits are built like this to improve power consumption.

Resistors are important whenever a circuit becomes analogue (or real as some people might express it). If the size of your signal is important, there are most likely resistors involved.

  • A classic op-amp circuit (unless the gain is -1) relies on the ratio of resistors.
  • A/D and D/A converters probably use resistors.
  • Default state control (pull-up / pull-down) use resistors.
  • Simple timing circuits use an R-C network. You might see this in a reset delay circuit.
  • Battery charge, voltage and current regulation use resistors, as identified in the question - in various types of feedback and stabilisation functions.

The analogue aspects of lots of modern circuitry is obscured, or contained in pre-packaged modules. The emergence of digital design has reduced the opportunities to understand the simple analogue concepts.

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TL;DR for the specific case of a LED (as asked about):

Any load connected to a DC constant voltage supply (eg a battery) that is not effectively a resistor of some description - is either unable to draw energy from the battery, or a short circuit.

Some electrical loads inherently do behave as resistors (and they ARE resistors, just not looking like the electronics component), eg lightbulbs, space heaters, ovens. These will, if correctly designed, self regulate their power consumption if fed from a constant voltage source (battery, mains, most power supplies).

Some (like motors, transformers), while not resistors, will behave equivalent to one when connected to a constant voltage AC source.

Other loads (like LEDs, bare flourescent tubes) do not in themselves behave as resistors, and are not able to regulate their own power consumption when fed from constant voltage sources. The ideal power supply for these loads is a constant current source, and the extra components needed around them are there to make your constant voltage supply behave sufficiently like a constant current supply.

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Hopefully the answers already posted give some clarification but unless I missed it there was one question that didn't quite get covered: "Why does a battery go into a dead short if you connect the terminals directly, but if you add a light bulb (resistor), it does not?"

Actually, when it's cold (i.e., not lit), an incandescent lamp very nearly is a dead short; its resistance is very low - but it'll generally have a lot more than the wires connected to it do. So we can approximate the situation as a very low-value resistor in an otherwise resistanceless circuit. Because of that, when the battery is first connected its entire potential difference (voltage) drops across the lamp's small resistance, making for a high current (Ohm's law at work). When we have a mostly stable voltage at high current across a component, it's going to consume a lot of power (P = IV) and so it will heat up (as an aside, the battery experiences the same potential difference and the exact same current so it heats up too - but it's a big heavy object while the lamp is a tiny coiled-up sliver of tungsten wire, so the latter heats up much, much more).

The thing about the lamp, though, is that its resistance is temperature-dependent. Ordinarily that's not a phenomenon that shows itself much because the temperature ranges we usually deal with are small, but a lamp filament will get upwards of 3000K and in the case of tungsten the resistance increases with temperature. So once the filament's temperature stabilizes after the battery is connected - as does its glow and its resistance - it's acting like a fairly hefty resistor. In fact, you can measure this yourself: using a DMM's resistance setting, measure the resistance across the lamp's terminals (the DMM uses a very low voltage for this and won't even come close to lighting the lamp) and then use the DMM to measure both the voltage across and then the current through the lamp when it's connected to a battery. Then use Ohm's law with those two numbers (V/I=R) and you'll get a much higher resistance number than you did when the lamp wasn't lit. In fact, the unlit lamp's resistance is so low that the quality of the contact between your DMM's probes and the lamp's terminals will matter and you might struggle to arrive at a stable reading.

As someone else said, shorting out a small battery doesn't immediately melt the wire you use to do it with because the battery has a fairly small internal effective resistance. You can measure what that is by taking V and I readings with first a small resistor (say, 25 ohms for a 9V battery) and then the V reading with no load on the battery. You'll note that the voltage you measure with the resistor present is slightly less than the near-open-circuit voltage the DMM reads by itself; that voltage difference divided by the current you read with the resistor connected is the effective internal resistance of the battery.

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Well first of all, you some times need to protect elements from high currents. For example, if you plug a diode on a 9 volt battery, the current will destroy it if it is connected the right way (A on +, C on -). To avoid that, we plug in a 600 ohms resistor to take some of the voltage on its ends, so the smaller voltage (+- 3.3 volts for an LED) will appear on the ends of the LED.

Second of all, we cant always choose the power supply. You can say "well there are IC converters and transformators" Yes but that is simply not practical since they cost more and are more difficult to operate (not to mention the difference between ideal and real transformators and their weight). Also we have dynamic resistors (resistors that change their resistance - sorry if this is not the term, i am Russian and only 1st year of highschool in electronics) which are much more practical since you cant change the number of wire rolls on a transformator.

Judging by the nature of this question, i am guessing you are just getting into electronics, so you dont need to worry a lot about what does what. Just learn the walls - Kirchoffs most importantly and you will understand how current works and how voltage works. The rest will follow. Other things you should focus on is understanding the elements. Walls come first, elements second... When you learn your theory you will be able to work with LSIC's and get your hands dirty. Or you can start working with an Arduino or something. I have the OSOYO and it is amazing. (this post is not branded by arduino)

ALSO REMEMbER THIS:

Current is equal to voltage over resistance.

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It may be helpful to get a handle on units and ratings:

  • mAh - milliampere-hour. A measure of electric charge. By itself, it doesn't say much. As a rating on a battery, it becomes meaningful in combination with the battery's nominal voltage as a measure of energy the battery can store. A milliampere-hour is the amount amount of charge represented by a one milliampere current flowing for one hour.
  • A - Amp (or Ampere). A measure of electric current - rate of flow of charge.
  • V - voltage. This is a measure of potential. Again, by itself, it isn't a complete specification for a battery, but it is an important one. An ideal battery will maintain a specified voltage and supply as much or as little current to a circuit as needed to maintain that voltage at its terminals. A real battery will have internal resistance, so it will have an "open circuit" (no load) voltage; the voltage will drop as load increases (it has to supply more current into a circuit). As most real batteries become depleted, the voltage declines too; the relationship between charge state and open circuit voltage depends on the design and chemistry of the battery. "Short circuit" current is the amount of current a battery will deliver when limited only by its internal resistance.
  • W - Watt. This is a measure of power (rate of energy delivered over some period of time). Watts can measure mechanical or electrical power; either way, it is a rate at which work is done. In electrical terms, power is a product of voltage and current (volts x amps).
  • kWh - kilowatt-hour. This is a measure of energy. A kilowatt-hour represents one thousand watts of power delivered for one hour, or 1 watt of power delivered for one thousand hours, 10 watts for 100 hours, etc. (Watts x hours).
  • Ohm - resistance. An ideal resistor will exhibit a proportional relationship between the current passing through it and the voltage applied to its terminals; double the voltage and you double the current (or vice-versa). This relationship can been seen as acting in either of two ways: if you apply a specific voltage across a resistor, it will pass a defined amount of current; if you force a specific amount of current through a resistor, it will create a defined voltage drop. Either way, the value of the resistance establishes a fixed relationship between voltage across its terminals and current through it. When you analyze a circuit, you can use this to solve for any one of the three values (current, voltage, resistance) if you know the other two. Ohms = Volts/Amps, or, Amps = Volts/Ohms, or, Volts = Amps x Ohms. Real resistors have an additional rating: wattage - this is the amount of power the resistor can dissipate without destroying itself. If you apply one Volt across a 1 Ohm resister, 1 Amp of current will flow through it, and it will dissipate 1 Watt of power as heat; if you double the voltage, you double the current, but now this 1 Ohm resistor will dissipate 2V x 2A = 4W of power as heat. If not rated for this, or the physical design does not allow for the removal of this heat, it will overheat, burn out, and potentially start a fire.

When you analyze circuits, you will have "knowns" and "unknowns". For example, you may know the voltage of a battery and the resistance of the load it is supplying. Given that, you can calculate the current which the circuit will draw. In a complex circuit, you may have numerous resistance values, and devices such as LEDs or transistors which will have certain properties:

  • diodes have characteristic forward voltages - they will maintain approximately the same voltage over a wide range of current. A real diode will have a characteristic non-linear curve relating forward current to forward voltage; over its normal operating range, the curve has such a shallow slope that for most purposes, it is deemed flat (constant voltage). To understand why this happens, you need to read up on semiconductor diodes
  • junction transistors have a characteristic base-emitter voltage - like a diode forward voltage, the base-emitter voltage is also nearly constant over a wide current range; it too has a non-linear curve relating voltage and current, and looks very similar to that of a diode. Again, to understand these properties, you need to read up on transistors.

You can use these properties to work through a circuit to calculate currents through paths where you know voltages, voltages at nodes where you know currents through certain paths, and equivalent resistances where you have resistors connected together. This is important because currents and voltages determine power consumption (or dissipation) which tells you whether a circuit will work at all, what ratings of components need to be selected, and how much power will need to be supplied.

Now... why do we need a resistor in series with our LED?

Let's say we have a 5V power supply and an LED for which the specs are 3.2V and 20mA, This means the LED will operate at a forward voltage of 3.2V, and should be driven with about 20mA of current; less and it won't give off as much light as spec'd, more and it will be brighter, run warmer, and might have a shorter life.

If we connect the LED with no resistor, the power supply will try to drive as much current as it can to maintain the 5V. The LED will pass a huge amount of current before the voltage across its terminals comes up to 5V. In all likelihood, the power supply will reach its current limit, and allow the voltage to drop, but at this point, too much current will flow through the LED and it will emit a bright flash and go up in a puff of smoke.

So... we want to limit the LED current to around 20mA while the voltage at the power supply remains 5V and the voltage across the LED is 3.2V. We need a resistor in series that will pass about 20mA (0.02A) of current at 1.8V (1.8 + 3.2 = 5). So, we calculate 1.8V/.02A = 90 Ohm. We could select a standard 82 Ohm resistor for this. 1.8V/82 Ohms = 21.9mA. A little above spec, but a margin of 10% shouldn't be an issue. Bear in mind that real devices cannot be assumed to have precisely defined properties; the resistor may be a little more or a little less than spec'd and the LED may operate at a voltage a little higher or a little lower than spec'd. We design for a nominal case knowing the actual performance of our circuit may be a bit different.

So... what have we done here? We've used a resistor to adjust what is going on in our circuit so that we can use the power supply we have available and operate the LED within its specifications.

What else can we do with a resistor?

Common uses of resistors are to adjust voltages or limit current flows. For example: you have a 5V power supply and need a 3V reference. Select two resistors from our parts bin: a 330 Ohm and a 220 Ohm, and connect them in series: the 220 between the 5V wire and our reference output, and the 330 between the reference output and 0V. There will be a constant current through these resistors of 5V/550 Ohm = ~10mA, but we will see a voltage of 3V at our reference terminal. This sort of thing is frequently used to design circuits like amplifiers where we need to establish a specific voltage, fraction of some other voltage, and so forth.

We can use resistors to define time constants. If you connect a resistor and capacitor in series, current will initially flow into the capacitor; this initial current will be determined by the circuit voltage and resistance value. But, the capacitor will charge; as it charges, it will create a voltage across its terminals; this will reduce the voltage across the resistor's terminals, reducing the current through it. This will reduce the rate at which the capacitor charges, reducing the rate at which its voltage rises, and so on and so on. Eventually, the capacitor will attain the circuit voltage, voltage across and current through the resistor will be zero. The resistance and capacitance values will determine the time it takes for the capacitor to charge to a certain fraction of the circuit voltage; the quantity known as time constant is the time taken for the capacitor voltage to charge to about 63% of circuit voltage. This is used to design circuits like oscillators and filters.

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Resistors exist and are used to limit virtual 'infinities'. In the sense that without a resistor a component would burn up or a fuse would blow, or a circuit would simply not operate as expected.

Less extreme examples would be to 'bias' a circuit to a specific voltage, in combination with other resistors or zener diodes. They also limit 'inrush' current to power supplies, thus extending the life of the power switch.

Due to the voltage drop across a resistors with current flowing through it, they make excellent and accurate current sensors.

Even more exotic reasons would be to stop parasitic oscillation or reflected waves in RF transmission lines. MOSFET's usually have a resistor at their gate to prevent ringing and overshoot at the drain, due to sharp rising/falling edges.

In combination with capacitors they create a 'time-constant' for use as a filter or delay. This can be for frequency tuning, or if more robust act as a ripple filter in power supplies.

To say they limit 'infinities' sounds kind of trite, but we would have no technology without them. Even the Model 'T' Ford had large resistors banks to select the proper charging current for the battery. It was not the precision charging we have today, but a 'just-get-by' solution was good enough back then.

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It sounds like you don't fully grasp how current flows and its relationship to voltage. If you understand this relationship then you can easily answer all of your questions.

Electrons want to move from a place of high voltage to a place of low voltage as quickly as possible, such as from one end of the battery to another. If the two ends of the battery are connected directly together by wire, the electrons will all jump incredibly quickly to the low voltage end, because there is nothing slowing them down.

The resistor slows down how fast the electrons can move through the circuit. Without the resistor, the battery will instantly burn out.

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    \$\begingroup\$ (1) "... the electrons will all jump incredibly quickly ...". No, the speed of the electrons is measured in mm/hour. See Electric drift. You are confusing it with the speed of the electric wave. (2) "Without the resistor, the battery will instantly burn out." No, the battery's internal resistance will limit the current. Try it yourself: put a short circuit on a AA for a few seconds. According to your answer the battery will "burn out". It will neither burn nor go instantly flat. \$\endgroup\$ – Transistor Feb 22 '18 at 21:26
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    \$\begingroup\$ (3) "Electrons want to move from a place of high voltage to a place of low voltage ..." The opposite is true. Electrons will move to the higher potential. \$\endgroup\$ – Transistor Feb 22 '18 at 21:26
  • \$\begingroup\$ It's just a laymans explanation, intended more to help form an intuition than be technically thorough. \$\endgroup\$ – DoctorMoose Feb 22 '18 at 22:02
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    \$\begingroup\$ It doesn't have to be technically thorough but it should be technically correct otherwise the OP will be as confused as you are. \$\endgroup\$ – Transistor Feb 22 '18 at 22:05

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