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A variant of this question has been asked a few times, but not received a response. Usually, we get an analogy with water pressure, or gravity, or a version of Ohm's law; not the actual cause-effect relationship.

Recently I have been reading about Zener diodes and how when reverse biased they drop a consistent amount of voltage. So for example, a simple circuit with only a battery and a diode would drop some fixed amount of voltage through the diode (to reverse bias it) and the rest of the electric potential would be dropped somewhere else.

In contrast, a resistor will drop as much voltage as required through it (as long as it is within its thermal limits).

I still do not understand why a resistor does not drop a fixed amount of voltage. What is the fundamental principle that causes this behaviour? Why doesn't this principle also affect diodes?

Thank you for your help.

Edit - addendum for clarity:

My main confusion is why a resistor does not drop a fixed amount of voltage but a variable one depending on battery voltage. If you use the analogy of gravity and friction, would a small amount of friction drop any amount of potential energy? Would it not drop a fixed amount depending on how significant the friction is? Since voltage is a per charge quantity, it would follow that an electron could drop as much or as little voltage as required at a resistor.. how does it do that regardless of resistor value?

Ohm's law offers an explanation of what happens, but not why it happens if that makes sense.

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    \$\begingroup\$ Ohm's law is pretty fundamental for describing how resistances work. Diodes can be modeled many ways, including simple models that do have a resistance and some treshold voltage. \$\endgroup\$
    – Justme
    Dec 3 '20 at 12:30
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    \$\begingroup\$ I understand the use of Ohm's law. What I do not understand is why the resistor behaves in this way? I am familiar with the PN junction, basics of how the electric field creates regions of higher and lower potential and why electrons move from high to low potential. Presumably, the electron will seek to drop all this potential energy if a conducting path is open. But in the diode example in drops only a fixed amount, not so in the case of a resistor. How come? \$\endgroup\$ Dec 3 '20 at 13:18
  • \$\begingroup\$ "How does resistance actually work" is clearly electrical physics, but it's also physics physics so it's worth having a look elsewhere, for example this might be interesting: physics.stackexchange.com/questions/406451/… Nominally this is the kind of thing for which basic references such as textbooks, etc should be the starting point, unfortunately wikipedia doesn't seem to have much that's readily found. Some additional web searching may be worthwhile. \$\endgroup\$ Dec 3 '20 at 18:35
  • \$\begingroup\$ Hi Chris, funnily enough I had looked at that already! Thank you for sharing in any case. He does mention there loss of kinetic energy and equates it to loss in potential energy which I do not understand, however my main conclusion is why a resistor does not drop a fixed amount of voltage but a variable one depending on battery voltage. If you use the analogy of gravity and friction, would a small amount of friction drop any amount of potential energy? Would it not drop a fixed amount depending on how significant the friction is? \$\endgroup\$ Dec 3 '20 at 19:30
  • \$\begingroup\$ This is one of those things where "how does it work really" is very complicated; for stackoverflow purposes might consider this a duplicate of the clarified question at electronics.stackexchange.com/questions/535548/… \$\endgroup\$
    – pjc50
    Dec 8 '20 at 14:35
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The voltage drop across a resistor is given by Ohm's law: $$V = IR$$

Where:

  • \$V\$ is voltage in volts
  • \$I\$ is current in amperes
  • \$R\$ is resistance in ohms

schematic

simulate this circuit – Schematic created using CircuitLab

Take the above circuit as an example. The total resistance \$R\$ is the sum of \$R1\$ and \$R2\$. That gives a current through both resistors of 5 amperes. That five amperes flows through both resistors. The voltage across \$R2\$ is therefore \$V = 10 \times 5 = 50 volts\$.

schematic

simulate this circuit

In comparison, the current through this second circuit is 1 ampere. By Ohm's law, there's now only 10 volts across \$R2\$. The voltage across \$R2\$ always varies depending on the total resistance and the total voltage.

Now replace \$R2\$ with a Zener diode:

schematic

simulate this circuit

Zener diodes do not follow Ohm's law. It is a diode, and has rules all its own.

When current flows through a reverse biased Zener diode, the voltage across the diode is set by the construction and the materials of the diode.

The voltage at \$V_{out}\$ for the Zener diode will stay pretty much the same, regardless of the value of \$R1\$ (over a fairly large range.) If I change \$R1\$ to 1000 or 10000 ohms, \$V_{out}\$ will stay in the vicinity of 5.1V.

The voltage drop across the diode is fixed by the construction and materials. As long as the reverse voltage is above the Zener voltage there will be a constanct voltage across the diode. The voltage drop is (largely) independent of the current.

The voltage drop across a resistor is fixed by the resistance and the current through the resistor. Anything that changes the current changes the voltage drop. That's pretty much what Ohm's law tells you: Voltage is dependent on current and resistance.


I see in comments that you "understand how Ohm's law works" and that you want to know "why the resistor behaves this way," and that you think that an "electron will seek to drop all this potential energy."

Ohm's law is descriptive - it puts mathematics to an observed behaviour.

If you want the physics behind how a resistor does what it does, then you're going to have to wait for someone else to answer. I don't have a clear model in my mind of how different types of resistors work. I know Ohm's law, and that I've used it quite a bit to get usable results. I don't know the actual physics of what happens to individual electrons, though.

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    \$\begingroup\$ I understand the use of Ohm's law. What I do not understand is why the resistor behaves in this way? I am familiar with the PN junction, basics of how the electric field creates regions of higher and lower potential and why electrons move from high to low potential. But none of that answers the question. \$\endgroup\$ Dec 3 '20 at 13:09
  • \$\begingroup\$ I'm not finished yet. Have to do something else right now. \$\endgroup\$
    – JRE
    Dec 3 '20 at 13:22
  • \$\begingroup\$ Thank you, no rush of course. \$\endgroup\$ Dec 3 '20 at 14:08
  • \$\begingroup\$ Please note that in practice, R1 and R2 should be in the several thousands of ohms. Not a few dozen. Else the energy loss will be excessive. \$\endgroup\$
    – Fredled
    Dec 3 '20 at 17:49
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    \$\begingroup\$ @ZhelyazkoGrudov So you are asking about the actual, physical reason why resistors follow ohm's law. Since everyone has misunderstood your question (they thought you wanted to know which law resistors follow) I suggest asking a new one. (It's bad to change what a question means, after it has been answered) \$\endgroup\$
    – user253751
    Dec 3 '20 at 18:07
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Because a resistor resists the current and the voltage (V_Out in the schematics posted by JRE) is the result of this resistance effect called Voltage Drop in combination of the other resistor R2. R1 is giving a limited amount of current to V_Out while R2 is removing a limited amount of current from V_Out. The word limited is important because the limit is defined by the resistor value.

The higher the impedance of the resistor, the less current will cross it and the more voltage will drop.

The higher the voltage at the resistor input, the higher the current will flow through and the less the voltage will drop. Hence, the higher the voltage at the input of the resistor, the higher voltage at its output.

Note that the output voltage is always the result of the current limiting resistor (R1) and the load resistance between the limiting resistor and ground (here R2 but not forcibly a resistor.) If the output of the current limiting resistor (R1) is left floating, the output voltage will be the same as the input voltage. And if it's connected directly to ground it will be zero.

A zener diode will always limit the voltage to a fixed value. However a zener always needs some current limiting device, such as a resistor, else the current flowing through the zener will be infinite and will burn it out. Current have to flow through the zener in order to maintain the desired voltage. This amount of current is defined by the current limiting resistor (R1).

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  • \$\begingroup\$ Note that a zener actually does not have a fixed voltage drop. It varies with current, but it varies only quite slightly over a useful range of currents. And like most things, it changes a little with temperature, too. \$\endgroup\$ Dec 3 '20 at 18:14
  • \$\begingroup\$ Thanks for answering. Yes I am aware of this, but do not know why which is my question. \$\endgroup\$ Dec 3 '20 at 19:43
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Draw the I_V curve for resistor and diode in series. Tie diode to 0 volts. Tie resistor to +10 volts, and tilted toward zero

(I would draw this --- as I have in the past here on stackX --- but stackX no longer lets my browser use the drawing/schem tool)

The interception of the diode and resistor line is the operating point for those 2-elements in series cicuit.

Now draw a steeper resistor curve, again starting at +10 vols. This is the model for a lower value resistor (allows more current)

Notice the intercept point moves up on the diode curve --- higher current.

Add this visualization method to your cognitive gifts.

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