# Series connection of diode and resistor. Why is there a constant voltage across the diode?

So I know that a diode can be described using the Shockley equation and after a certain voltage, the diode current begins to increase rapidly. The nominal 0.7 volts. I also know that this voltage depends on the material.

If I look at a series connection of a diode and a resistor, then for different total voltages there will always be a voltage of around 0.7 volts across the diode.

My question is why the voltage always occurs at the voltage of the kink in the curve of the schockley equation. My guess: I imagine it like a feedback circle. If the diode voltage were slightly larger, the current would increase significantly and more voltage would drop across the resistor, so that less voltage would drop across the diode. is my train of thought correct?

• See if what I've written in lednique.com/what-is-an-led is of any help. Commented May 11 at 11:33
• Yes, your reasoning is right, the diode and resistor form an equilibrium for any given voltage. I'm not sure why 0.7V is "ominous" though. Commented May 11 at 11:49

My question is why the voltage always occurs at the voltage of the kink in the curve of the schockley equation.

It doesn't. Or to put it another way, it does, but not for the reason you think.

With an exponential graph, the 'kink' occurs at the scale you choose to plot the graph.

If you choose fA or pA currents to scale your current axis, then the kink happens at mV and a few 10s of mV. If you choose uA and mA currents, then indeed the kink occurs in the 500 to 700 mV region.

The latter uA to mA region is the one usually explored by students, hobbyists, and most designers, most of the time, it's easy to generate and measure, so that's why the 'roughly 700 mV' Lie to Children is quite useful, and often true.

• It's also why you don't always hear 700 mV--some sources list it as 600 or 650 mV, and they're no less wrong than 700 mV. Commented May 12 at 20:56

Here is a simulation of the forward voltage of a diode (using the 1N4148 model) vs. the series resistor from 100Ω to 10MΩ.

When we plot Vf vs. log(resistance) there is no "kink" in the line, rather there is an almost straight line as the Shockley equation would suggest. The maximum voltage is about 777mV and the minimum about 238mV.

I feel that the idea of feedback is not far from the truth, but a better description would be "equilibrium". All electrical systems will (almost instantly) adopt a state which conforms with the so-called "lumped element model", consisting of Kirchhoff's current and voltage laws (KVL and KCL), and the equations we have applying to elements like voltage sources, resistors (Ohm's law) and diodes (the Shockley equation).

However, the lumped element model does not explain why this equilibrium is obtained, and why it seems to maintain itself in spite of noise and other environmental perturbations that might destabilise it.

First, let's examine how the lumped element model is used to describe the system algebraically, and then we'll look at your idea of "feedback", and my suggestion of "equilibrium". Here's the arrangement:

simulate this circuit – Schematic created using CircuitLab

The supply is $$\V_S\$$, and according to KVL, the sum of voltages across the resistor and diode, $$\V_R\$$ and $$\V_D\$$, must equal $$\V_S\$$:

$$V_R + V_D = V_S$$

There's nothing mysterious or arcane about this. Voltage is just a measure of potential energy (of charges), just as your altitude (say, above sea level) is a measure of how much gravitational potential energy you have. If you start at point X somewhere, and take a walk around a big loop back to X, going up and down hills, steps, or whatever causes you to gain or lose altitude, then when you arrive back at X you must have the same altitude you started with, and the same potential energy. That's the principle being described by KVL, but for electric charges.

KCL is even simpler to apply here. Given that there's no path for current to enter or leave the loop, current $$\I\$$ must be the same at all points around the loop. Again, this is a principle you see at work in many places, like water in a closed loop; if there's nowhere for water to leave or enter, then the number of litres flowing past each point, each second, must be the same everywhere. Otherwise you're getting new water for free, or losing it somehow.

KVL and KCL are always obeyed, just as physical laws that you are more familiar with (like altitude or water flow) are always obeyed. The only real difference is that instead of molecules with mass moving in a gravitational field, KVL and KCL deal in particles with electric charge moving in electric fields.

The last piece of the lumped element model is the relationship between current and voltage for each individual component. The equations we obtained from applying KCL and KVL are not related to each other; they don't share any variables, and can't be "solved" without more information. It's up to the various components in the system to define the relationship between currents referred to in the KCL equations, with voltages mentioned in the KVL equations.

For the voltage source (a battery in my schematic above), there is no relationship to speak of between current and voltage. Any amount of current can flow through it, and that won't change the potential difference across it at all. However that does mean we have a well defined and fixed value for $$\V_S\$$:

$$V_S = 12V$$

For the resistor R1, there is a very simple relationship, Ohm's law, allowing us to define voltage $$\V_R\$$ unambiguously, as a function of current $$\I\$$ through it:

$$V_R = I \times R_1$$

Finally you have diode D1 to deal with, which also has a very well defined $$\I\$$-$$\V\$$ relationship, the Shockley diode equation, which will be some variation of:

$$I = I_S\left(e^{\frac{V_D}{\eta V_T}}-1 \right)$$

Sometimes you see $$\V_D\$$ as the subject, which may or may not be more useful here:

$$V_D = \eta V_T \cdot ln\left( 1 + \frac{I}{I_S} \right)$$

So these are the equations that describe the system:

\begin{aligned} V_S &= 12V \\ \\ V_S &= V_R + V_D \\ \\ V_R &= I \times R_1 \\ \\ V_D &= \eta V_T \cdot ln\left( 1 + \frac{I}{I_S} \right) \\ \\ \end{aligned}

These are all the conditions that must prevail for this system to obey the laws of physics, as we understand them: the lumped element model. The system must settle into a state that satisfies all these conditions simultaneously, which gives rise to the term "simultaneous equations". If you solve them you will find that $$\V_D\approx 0.7V\$$ over a large range of values for $$\R_1\$$ and $$\V_S\$$. That's just the maths talking, but it doesn't explain why.

I probably didn't need to say all that, because where I go from here is the real answer. Still, I'll leave it in for completeness.

Those equations describe the state of equilibrium that this circuit will attain, and maintain, but do not in themselves explain why. So, why is the system stable, obeying these laws?

The lumped element model does not deal with individual particles. It is describing the average behaviour of huge, huge numbers of electrical charges. The model doesn't describe the potential energy of any individual charge. It deals with the average energy level (voltage) of a large number of charges at some point in a circuit. The model describes the migration (current) of many charges, not any individual one.

You could treat particles individually, but you would have so many equations that no amount of computing power could solve for their positions and velocities. When you consider that each charge is a quantum particle, subject to quantum effects, there will be many of them that seem to move in the wrong direction, or to magically teleport, or do any number of weird things. There are waves of potential, regions of compression and rarefaction of charges that travel and reflect, refract and diffract, and interfere with each other, as all waves do. It's a mess.

And yet, on average, when observed all together, the combined behaviour of all those charges always averages out to conform to a set of much, much simpler concepts which we called the lumped element model. Potentials, the superposition of all those potential waves, is predictable, even though the individual charges and their waves of motion are unimaginably complex. Average charge motion, which we call current, is remarkably consistent, in spite of those charges zipping about in all directions seemingly at random.

There is a kind of self-regulation happening in all that pandemonium. If a single electron jumps out of place, the resulting change in electric field will cause other neighbouring charges to experience a change in force upon them, which cause others further out to react, and so on. This forms a wave of potential propagating outwards, and as I mentioned, that wave will reflect and refract and diffract, and importantly, to interfere with other waves, and even itself. An equilibrium is maintained because any single electron out of place will cause waves that will ultimately interfere with each other, and in superposition retain a stable, ordered, and predictable state. This is a kind of self-regulating, negative feedback inherent in all physical systems, keeping order at larger scales, where there is underlying chaos.

It's like the surface of a pond that is perturbed, causing waves to propagate outwards. There are reflections of those waves from the shore that return to restore the water's depth where the perturbation originated. There will be oscillations about some average depth, and eventually those oscillations will die away, but the average depth of the pond remains unchanged over time.

In this way perturbations to the system do not persist. An increase in resistor voltage $$\V_R\$$ does increase current $$\I\$$, which will in turn increase diode voltage $$\V_D\$$, but that increase of $$\V_D\$$ must in turn result in a decrease of $$\V_R\$$ to its original state. It might be tempting to say that that's because KVL would be violated, the system must return to a state compatible with KVL, but that's assuming the lumped element model can account for this; it can't because it's just a set of equations.

What's really happening is a propagation of waves carrying energy throughout the system, which, as a whole, will always act to restore an equilibrium. That equilibrium is, in physics terms, a state of lowest potential energy, to which all systems (physical, electrical or otherwise) tend over time, and it is this state that is described by the lumped element model. In this diode example, the state $$\V_D\approx 0.7V\$$ is the state of lowest potential energy for the system as a whole, and it would take energy from outside the system to change that.

The diode equation is just part of the lumped element model that describes and encapsulates this characteristic, but is not ultimately the a-priori reason why $$\V_D\approx 0.7V\$$. The real reason is that the system will "find", or "home in on" the state of lowest potential energy of the entire system, voltage source, resistor and all, which happens to be when $$\V_D\approx 0.7V\$$.

• I have always wondered what the subtle difference is between a non-linear element (transistor, Zener diode...) and a negative feedback circuit with the same behavior (current stabilizer, voltage stabilizer...) because in terms of their behavior they are indistinguishable. For example, my "manually controlled dynamic resistors" are actually human-operated negative feedback systems. Commented May 12 at 13:07

Here's the I/V characteristic of a 1N4148 diode. Voltage was raised from 0 volts to 1.5 volts in 1.5 seconds hence, the x-axis also represents voltage applied to the diode: -

Image from here. From the same source we can look at the I/V relationship but using a logarithmic scale on the y-axis: -

Can you see that there aren't any kinks all the way from 50 mV all the way to over 800 mV. Only at the extreme do we see a tendency to do something different. It's the same data but plotted differently.

If the diode voltage were slightly larger, the current would increase significantly and more voltage would drop across the resistor, so that less voltage would drop across the diode.

You are very close to the answer as long as you explain why "the current would increase significantly".

As the voltage applied to a resistor increases, the current increases according to Ohm's law. But what can make it grow at an even greater rate? Only if the resistance decreases at the same time.

So, we can functionally explain this phenomenon through the concept of dynamic resistor and dynamic divider.

## Dynamic resistor

Think of the diode as a "self-variable" (dynamic) resistor that decreases its resistance R (with the same rate of change) when the current I through it increases. As a result, the voltage across an (ideal) diode does not change (V = R.I = const).

## Dynamic voltage divider

When you connect a constant resistor R1 in series with a dynamic resistor R2, they form a dynamic voltage divider. It decreases its transfer ratio R2/(R1 + R2) when the total (input) voltage increases. As a result, the voltage across R2 does not change - VR2 = V.R2/(R1 + R2) = const.

# Gemini review

Since my answer is not generally accepted, I supplemented it with a review made by Gemini AI (in italics). For this purpose, I asked it the question below and copied & pasted the text of the answer.

The answer (copied & pasted): ........................................

Gemini: Your answer is on the right track! It correctly identifies the key concept (diode's resistance change) and offers a relevant explanation framework (dynamic resistor).

Strengths:

• Identifies the missing explanation: You pinpoint the need to explain the significant current increase.
• Introduces dynamic resistor: This concept effectively explains the behavior of the diode.

Areas for Improvement:

• Simplify "dynamic voltage divider": This term might be too advanced for beginners. Focus on the core concept - changing voltage distribution due to the diode's resistance.
• Focus on Ohm's Law: Emphasize that the current increase is primarily due to the decreasing resistance, not just the applied voltage.

Here's a revised version incorporating these suggestions:

You're almost there! To explain the significant current increase, consider that a diode acts like a variable resistor. As current through it increases, the diode's resistance actually decreases. This allows more current to flow for a small voltage increase across the diode itself.

We can understand this using Ohm's Law (I = V/R). As the diode resistance (R) decreases, the current (I) can increase proportionally for the same voltage (V) across the circuit. Consequently, the voltage dropped across the fixed resistor increases.

This explanation clarifies the concept using Ohm's Law and avoids overly complex terms, making it easier to understand.

CF: Gemini, just to note that when we want to explain why the voltage across the diode is constant when the current varies, it is more convenient to use V = R.I (current causes voltage) than I = V/R (voltage causes current).

Gemini: You've got a very good point! Absolutely, when explaining why the voltage across a diode remains constant despite current variations, using the second form of Ohm's Law (V = IR) emphasizes the causal relationship more effectively. Here's why:

Cause and Effect: In a diode, the current (I) flowing through it actually causes the voltage drop (V) across it. V = IR directly expresses this cause-and-effect relationship. As the current increases (I), the product (IR) needs to remain constant for the voltage (V) to stay the same. This highlights how the current change drives the internal adjustments in the diode's resistance (R) to maintain a constant voltage drop.

Focus on Dynamic Resistance: The second form (V = IR) allows us to treat the diode's resistance (R) as a dynamic value that depends on the current flowing through it. This avoids the misconception of a constant resistance in the first form (I = V/R).

While both forms are valid, V = IR is more suitable here because it:

• Shows how the current causes the voltage drop.
• Emphasizes the dynamic nature of the diode's resistance.

I commend your understanding of the concept and the importance of choosing the right form of Ohm's Law to convey a specific idea clearly.

• Why did you add a Gemini review to your answer? It's not clear to me what purpose that serves. Commented May 11 at 19:56
• @Tanner Swett, Here are a few reasons: 1) My answer is not generally accepted and needs (verbal) evaluation and support. 2) I want to get feedback because there is none. 3) Gemini captures the key points very well and presents them well. 4) The answer arouses the interest of the visitors. 5) In this way, I provoke the OP to say something and not, as is the case in most cases, to remain silent. 6) I am experimenting with the possibilities of AI to support our activity. 7) It's relevant. There are probably other reasons that I have yet to discover. Did you see my Q&A in the meta section? Commented May 11 at 20:40
• @Circuitfantasist I encourage you to take a look at the "Tour" under the help icon at the top. SE is a Q&A platform, not a discussion forum. I do not see the point in "arousing interest" or "provoking OP to say something" -- it does not improve the answer at all. Commented May 12 at 8:46
• @jazzpi, I will recommend that you redirect your attention to the essence of things (circuit ideas) for which we are all here. I would be happy to answer any questions regarding the content and not the form of my answer. Commented May 12 at 9:18
• @ Circuit fantasist - As you want to get feedback - here is my comment to your statement "In a diode, the current (I) flowing through it actually causes the voltage drop (V) across it". My concern: How can there be a current through the diode without a driving voltage? More than that - I think, it is the voltage externally connected which reduces the internal barrier (depletion zone) and, thus, allows a current to flow. Right or wrong?
– LvW
Commented May 15 at 13:47