Understanding the second approximation of diodes

[JC2000]

TL;DR Summary

Need help putting a finger on what I have misunderstood as I am seem to feel the polarity of the battery we are replacing the diode with should be reversed.

My understanding of barrier potential is as follows : Due to the holes and electrons combining in the diode a depletion region is formed. Essentially this region has no mobile charge carriers (as the holes and electrons have combined) and thus the ions (whose holes/ electrons have combined) develop an electric field which prevents more charge carriers from 'neutralising' the entire diode. Thus the potential developed across the depletion region can be represented as in B (?).

My understanding of the second approximation of a pn diode is as follows : An ideal diode would act as a closed switch when forward biased and as an open switch when reverse biased (fig D.). By extension, an approximation could be made where the forward characteristics are observed only after barrier potential is overcome (fig E.).

Q1 . However it is at this point that my reasoning seems to go off the rails. If the diode is to be represented as C1 when the applied voltage exceeds barrier potential and as C2 when applied voltage is less than barrier potential, why is the polarity of the battery not the other way around (given the polarity of the electric field across the diode as in B.)?

 

[Joshy]

Hi! It's difficult for me to follow the model or drawings with the question.

Your question is making references to applied voltage, voltage barriers, and then the electric field, but it's difficult for me to follow which one is which in the drawings; furthermore: their polarity and direction (the electrical field). Is C1 and C2 flipped with respect to B if both are referring to the barrier?

By chance: Has your class or the textbook you're looking at used band diagrams yet? The book I'm looking at I'm sure is a classic I used it in my coursework. I saw several copies of it lying around during internships at various locations... it's called Semiconductor Device Fundamentals by Pierret. I'll scan the page I'm looking at because I think it's a really elegant way to qualitatively approach the question compared to drawing what looks like voltage source.

The left side of the drawing is still the p-side and the right is the n-side. E below is for energy band and not for E field (E_C condition, E_V valence, E_F fermi, and E_i intrinsic band). I am sure it is overkill for this question if the topic is new, but I would imagine if you have touched the topic at all, have access to the book or a similar book, then just a little more digging you might be able to approach your question by this method instead. It might also just be fascinating if you're wanting to dig into fundamentals.

I think another common thing that throws off a lot of people is the direction of the current. It's following the holes! So if the E field in your question is pointing one way we'll say from left to right, then the electrons will want to travel towards the positive side of the E field and will go from right to left (the opposite!). This could be another cause for confusion.

 

[JC2000]

Hi! It's difficult for me to follow the model or drawings with the question.

Your question is making references to applied voltage, voltage barriers, and then the electric field, but it's difficult for me to follow which one is which in the drawings; furthermore: their polarity and direction (the electrical field). Is C1 and C2 flipped with respect to B if both are referring to the barrier?

Thank you very much for taking the time to answer my question. My understanding of band diagrams is sketchy and based on your suggestion I will delve deeper into the matter. Let me also try and clarify my question...

(B) represents the electric field developed across the depletion region (it is not an applied voltage).
Is this representation correct?

(C1) Represents the 'Second Approximation of a PN Diode' under forward bias.
(C2) Represents the 'Second Approximation of a PN Diode' under reverse bias.

My understanding is that the diode is being replaced with a battery whose voltage is the same as that of the diodes voltage barrier. Is my understanding correct?

If so, shouldn't the polarity of this battery be reversed (reversed with respect to what my book suggests) since in the depletion region the negative ions are on the p-side and the positive ions on the n-side of the depletion region? My book seems to suggest otherwise. Hence my third and final question : What is the reasoning behind polarity of the battery that is replacing the diode?

Thank you for your patience!

 

[JC2000]

I think this bit addresses my questions. So my understanding at this point is that the battery is not a voltage source but merely represents the voltage that must be 'overcome' in order to achieve forward bias in a circuit. My new (and hopefully final) question is : Why is the battery symbol used in this case? Also, how should this voltage be treated during a circuit analysis?

 

[Joshy]

I see.

In circuit analysis it's just a fixed voltage across the diode. Here's a small example and we can simulate it through LTSpice.

So if I make R and R_L the same, then we'll have a voltage divider that divides by 2 until around 1V4 because the voltage across the diode will be 700 mV (that's when it roughly turns "on"). After the 700 mV it doesn't seem to want to budge anymore and it sits at around 700 mV. The current increases across the diode (plotted as I(D1) in red). You can see in the graph below that the model isn't exactly ideal diode, but you can see the approximation works.

Here's the schematic in LTSpice (the same as the hand drawing above)

 

[JC2000]

Thank you for your painstaking response! My circuit analysis fundamentals seem to be shaky as well so I have a few more trivial questions...

So if I make R and RL the same, then we'll have a voltage divider that divides by 2 until around 1V4 because the voltage across the diode will be 700 mV (that's when it roughly turns "on"). After the 700 mV it doesn't seem to want to budge anymore and it sits at around 700 mV. The current increases across the diode (plotted as I(D1) in red). You can see in the graph below that the model isn't exactly ideal diode, but you can see the approximation works.

1. I am not clear on as to why current through $R_{L}$ is fixed if voltage V is increased. Since an increase in V should cause an increase in $V_{0}$ which should result in the currents across both branches increasing (as the resistances don't change)...(?)

2. What I gather is as follows :
a. We are choosing $R_{L} = R$ for simplicity. (Reduces a variable in the equation/ gives a simpler relationship between $V_{in}$ and $V_{0}$?)
b. We increase $V (V_{in})$. Until $V_{in}$ is less than 1.4 A, $V_{0}$ will be less than 0.7 A (as indicated by the equation) and the circuit acts as a voltage divider (with negligible current flowing across the diode)
c. Once $V_{in}$ is greater than or equal to 1.4 A, barrier potential is overcome and the diode is forward biased and now the circuit acts as the one in the first diagram ("on").
d. For a large increase in current (ID1) , $V_{0}$ does change much from 0.7 A ($Vvdiode$) which corresponds to what the VI characteristics of a forward biased diode ought to be.

Is my understanding correct?

3. So if a diode is just connected to an external circuit (only a wire linking the p region and n region without any external voltage source), no current would flow. Why is this so? (Shouldn't the electric field across the depletion zone do something?

4. I am struggling to use LTSpice, where can I learn the basics. On a related note, would the circuit analysis be the same if we were to replace the diode D1 with the equivalent voltage?

Thank you for your patience!

 

[Joshy]

No problem!

You're right: I chose R = R_L to simplify the mathematics. A voltage divider is probably one of the most classic or among the first few circuits that is covered. When R = R_L, then the output voltage will be halved. Here's the derivation using KCL.

R_L is fixed because the voltage across it is also fixed (only when or if we're assuming the diode is "on", which we're assuming is approximated by a voltage source). You may recall Ohm's law is:

$$I = {{V} \over {R}}$$

Since V neither R is changing, then the current (I) couldn't be changing. Sorry for squeezing responses for 1 and 2a together.

2b is correct.
2c is correct.
2d is correct.

All very good :) great job.

3 is a great question (only a diode and voltage source). I cannot say I immediately know the answer to this. I'll take a stab at it with my thoughts or a guess, but if I were you just kind of take it with a grain of salt (I'll also need to look further into this myself- thanks!). I feel like there would be some violation with KVL- so the voltage source approximation clearly wouldn't work very well anymore. I was surprised to see there were no complaints in spice when trying this with a diode. What I did notice is that the current did "explode" (look at the units), which makes sense because the current characteristics for the diode is exponential. With a real diode I think it would break.

When you zoom into a smaller current range you can see the exponential behaviour much better.

I struggled with LTSpice a bit myself. Something that helped me was a shortcuts table. I had to browse around a bit for the one that I used and it asked me to agree to their conditions: https://www.analog.com/media/en/sim...pice_ShortcutFlyer.pdf?modelType=spice-models

4. Yes, for the voltage and only after the diode is "on" (after the 1V4). For before 1V4 then the equivalent one would be where there is no element where the diode is (an open circuit) and only the voltage divider would remain. Here's the simulation just to see.

The schematic

 

[sophiecentaur]

My understanding is that the diode is being replaced with a battery whose voltage is the same as that of the diodes voltage barrier. Is my understanding correct?

I think you could be trying to be too literal in your interpretation of an 'Equivalent Circuit' for the Diode. An equivalent circuit doesn't always imply that you could actually build that circuit and get the same results without using some very 'smart' components.

The diagram B seems to me, very confusing. The 'battery' that you are mentally putting inside the diode would have to be in series and not in parallel with the middle region. Also, it is only there when the diode is forward biased. The knee voltage could be considered as an emf subtracted from the applied voltage. So that battery would need to have a variable emf until the knee voltage is reached.

 

[Dullard]

I am not normally a huge fan of mechanical analogs, but for a rectifier diode (from a circuit perspective), there is a nearly perfect one:

A check valve. The bias voltage corresponds to the cracking pressure. The increase in voltage (with increased current) corresponds to friction losses. All of that is absolutely no help if you also find check valves mysterious.