Demo 3.1: Diodes, from Switch to Curve
Every component so far has been described by a number. A resistor has a resistance, and once you know it, Ohm’s law tells you everything: double the voltage and the current doubles, because the graph of current against voltage is a straight line through the origin.
A diode has no such number. Its graph is a curve, the current rises exponentially with voltage, and there is no single value of resistance you could write on it that would be true at more than one operating point. That sounds like a nuisance and it is the reason the device is useful: a component whose behaviour depends on which way round you apply the voltage is the only way to build a rectifier, and a component whose current rises steeply at a repeatable voltage is the only way to build a reference.
This demonstration takes the diode twice. First as a switch, which is wrong in detail and right often enough to design with, and then as a curve, which is what it actually is.
What is actually in there
Section titled “What is actually in there”Silicon has four electrons in its outer shell and shares all four in covalent bonds, which leaves nothing free to carry a current. Pure silicon is a poor conductor. Add a trace of an element with five outer electrons and the fifth has no bond to join, so it wanders off and leaves a fixed positive ion behind: that is n-type, and electrons are its majority carrier. Add an element with three instead and one bond is left short, and that vacancy moves as neighbouring electrons drop into it: that is p-type, and holes are its majority carrier.
Both are electrically neutral. Doping adds no net charge, it only makes some of the charge mobile, and it is worth being firm about that because the phrase “positive material” invites the idea that a p-type block would attract a balloon.
Join the two and electrons diffuse across the boundary into the holes waiting for them. Both carriers vanish in the process, and what is left either side of the junction is the fixed, charged dopant ions with nothing mobile among them. That layer is the depletion region, the field between its charged ions is the barrier, and the whole behaviour of a diode is the story of that region getting wider or narrower.
How to use it
Section titled “How to use it”Inside the junction has a slider that sweeps the applied voltage from reverse bias, through zero, to forward. Watch the depletion layer: reverse bias widens it and the mobile carriers are swept further back, forward bias squeezes it out of existence and current flows. The layer’s width is computed from the abrupt-junction result rather than drawn to look convincing.
The diode as a switch is the ideal model. Swing the supply through zero and the diode is either a piece of wire or a break in the circuit, nothing between. Underneath is the biasing exercise from the lecture, with the six cases from the slides and two extra where the diode is drawn the other way round.
The characteristic curve is the real device. You can put the ideal model, the constant-drop model and the Shockley equation on the same axes and see how the approximations relate to the truth, switch between silicon, germanium, Schottky and an LED to watch the knee move, and open the view out to all three regions of the characteristic including breakdown.
Load lines is the payoff: a supply, a resistor and a diode. Drag the supply and the resistance and watch the line move across the curve. It opens on the worked example from the lecture, 10 V through 500 Ω.
Diodes, from Switch to Curve
What is inside a p-n junction, the switch it pretends to be, the curve it really is, and how to solve a circuit containing one.
Walkthrough
Section titled “Walkthrough”Step 1: Sweep the junction and watch the layer
Section titled “Step 1: Sweep the junction and watch the layer”Open Inside the junction and press No bias.
There is already a depletion layer with nothing applied, built by diffusion alone, and there is already a barrier across it. Some majority carriers still cross by diffusion, an equal drift current of minority carriers crosses the other way, and the net current is exactly zero. It has to be. A diode in a drawer with a voltmeter across it reads zero, and anything else would be a source of free energy.
Now press Reverse bias. The layer more than doubles in width. Both kinds of carrier are being pulled away from the junction rather than towards it, so the region empties further and the barrier gets taller. The current is a few femtoamps of minority carriers wandering across the other way.
Then press Forward bias. The layer collapses. Carriers pour in from both sides, recombine, and once the applied voltage overcomes the built-in barrier there is a continuous conducting path.
Now drag the slider slowly from 0.4 V up to 0.8 V rather than jumping. The layer thins gradually. There is no click, no threshold, no moment where the device decides to switch on. That is the single most important observation on this tab, because it is the reason the characteristic on the third tab is a curve rather than a step, and the reason the “0.7 V” you will be quoted for the rest of your career is a convention rather than a physical constant.
Step 2: Use the switch model, and find where it lies
Section titled “Step 2: Use the switch model, and find where it lies”Open The diode as a switch and swing the supply from +10 V down to −10 V.
Above zero the diode is a closed switch with no resistance and the loop is complete. Below zero it is an open switch with infinite resistance and the loop is broken. That is the whole model, and it is enough to answer the only question that usually matters, which is whether current flows at all.
Now set the supply to 0.5 V and read the caution. The ideal model says the diode is conducting and passing half a milliamp through the kilohm. A real silicon diode at 0.5 V passes about 0.4 microamps, because it has barely started, so the model is out by a factor of more than a thousand. It gets the direction right and the magnitude badly wrong near the knee, which is exactly the region the third tab is about.
Step 3: Do the biasing exercise
Section titled “Step 3: Do the biasing exercise”Underneath is the set of eight. For each one, find the bar, work out the voltage from anode to cathode, and answer before you press anything.
Case (e) is the one that catches people: both terminals sit at large negative potentials, which looks alarming until you notice that the anode at −12 V is three volts above the cathode at −15 V. A diode has no connection to zero volts and no idea what it means. All it can respond to is the difference between its own two terminals.
Cases (g) and (h) have the diode drawn the other way round. They are there because a student who has learned “the bigger number on the left means forward” has learned something that is about to fail them. The rule is about the symbol, not about the layout.
Step 4: Meet the real curve
Section titled “Step 4: Meet the real curve”Open The characteristic curve with silicon selected, and drag the cursor along the forward region.
Below about 0.5 V there is effectively nothing. Between 0.6 and 0.8 V the current goes from microamps to tens of milliamps. Read the panel showing the Shockley equation with your numbers substituted, and note the shape of it: since kT/q is about 26 mV at room temperature, every extra 60 mV across the diode multiplies the current by roughly ten.
That single fact explains why the knee looks like a fixed voltage. Nothing in the equation is fixed, but the curve is so steep that over any sensible range of currents the voltage barely moves, so pretending it is 0.7 V costs you very little.
Now turn on all three models at once. The ideal switch is a vertical line at 0 V, the constant-drop model is a vertical line at 0.7 V, and the real curve bends between them. Each is useful, and choosing the simplest model that answers your question is most of what analogue design is.
Then switch parts. Germanium’s knee is at 0.3 V, Schottky’s at 0.35 V, and the red LED needs 2.0 V, which is the same red LED and the same forward voltage as in the demonstration on pull-up and pull-down resistors. That is not a coincidence being reused: an LED is a diode, and its forward voltage is its band gap expressed in volts, which is why a blue one needs more than a red one.
Step 5: Look at all three regions
Section titled “Step 5: Look at all three regions”Press All three regions.
There is the forward region you have been working in, the reverse region where the current is a flat few nanoamps however much voltage you apply, and at the far left the breakdown region where the current runs away.
Read the caution about the axes, because it matters. The forward side runs to about a volt and twenty milliamps; the reverse side runs to seventy-five volts and a couple of microamps. Drawn honestly on one linear scale, the entire forward curve would be a vertical line one pixel wide. The two halves of this picture are at different scales, every textbook does the same thing, and the shape you are looking at is partly a drawing convention.
Breakdown itself is the avalanche effect: minority carriers crossing the junction pick up enough kinetic energy to knock further carriers out of the lattice, those do the same to others, and the process runs away. In an ordinary silicon diode it happens somewhere around −50 V to −100 V and destroys the device, because nothing limits the current. A Zener diode is built to survive it at a precise, low voltage, which turns the steepest part of the curve into a voltage reference, and that has a demonstration of its own.
Step 6: Solve a circuit you cannot solve
Section titled “Step 6: Solve a circuit you cannot solve”Open Load lines, which starts on the lecture example: 10 V, 500 Ω, silicon.
Here is the difficulty. Kirchhoff gives you V_SS = R·I + V. The diode gives you its exponential. Both are true at once, so you would have to solve an exponential and a straight line simultaneously, and that has no neat closed form.
The load line sidesteps it entirely. Take the circuit equation and find its two intercepts:
- Point A: set the current to zero. No current means no drop across the resistor, so the whole supply appears across the diode. That is 10 V on the voltage axis.
- Point B: set the diode voltage to zero. Then the resistor has the whole supply across it, so the current is V_SS/R = 10/500 = 20 mA on the current axis.
Join them. That line is a statement about the supply and the resistor only, with a slope of −1/R, and it knows nothing about what component is in the other position. The curve is a statement about the diode only. The circuit must satisfy both, so it sits where they cross, and the demonstration reads that off at 0.779 V and 18.44 mA.
Check it yourself. 9.221 V across the resistor plus 0.779 V across the diode makes 10 V, and 9.221 V over 500 Ω is 18.44 mA, the same current as the diode’s. Both laws hold at that point, which is what made it the operating point.
Step 7: Move the line two different ways
Section titled “Step 7: Move the line two different ways”Press Double the resistance and watch the line pivot about Point A. The voltage-axis end is fixed, because with no current the supply appears across the diode whatever the resistor is, while the current-axis end moves as V_SS/R.
The current falls from 18.44 mA to 9.24 mA, roughly halving as you would expect. Now look at the diode voltage: it has moved from 0.779 V to about 0.761 V. The current changed by a factor of two and the voltage moved by eighteen millivolts. That is the exponential doing its work, and it is the real justification for treating a diode drop as a constant.
Then press Double the supply instead. The line slides sideways without changing its slope, because the slope is −1/R and R has not moved.
Finally, put a Red LED in and set the resistance to 200 Ω. The demonstration tells you the part is past its rating, because 40 mA through a 20 mA LED is twice what it will take. Raise the resistance until the warning clears. That is the pull-resistor demonstration arrived at from the other direction: the resistor is not protecting the LED by some special property, it is taking the voltage that would otherwise appear across it and fixing the current at a value the part can survive.
Check your understanding
Section titled “Check your understanding”A diode has −12 V on its anode and −15 V on its cathode. Is it forward biased?
Why is a silicon diode so often modelled as a constant 0.7 V drop when its actual equation is exponential?
Match each item to what it is
On a load line plot, what happens when you increase the series resistance while keeping the supply the same?
Which of these are true of the reverse-biased region? Select all that apply.
Wrap-up
Section titled “Wrap-up”Five things to take away.
- A diode is a depletion region with wires on it. Forward bias narrows it and current flows, reverse bias widens it and almost nothing does, and everything else is detail.
- Both doped materials are electrically neutral. Doping does not add charge, it makes charge mobile.
- The switch model answers whether current flows, and that is often the whole question. Which terminal is more positive is all that decides it, and node voltages that are both negative decide nothing on their own.
- The real characteristic is exponential, at roughly a decade of current for every 60 mV. That steepness is why a constant 0.7 V is such a good approximation, and it is a convention rather than a physical threshold.
- A load line is the circuit’s own straight line, drawn from two intercepts and depending only on the supply and the resistor. Where it crosses the device’s curve is the operating point, and this method works for any non-linear device, not just diodes.
The next demonstration puts diodes to work. A single diode passing only one half of an alternating waveform is a half-wave rectifier; four of them in a bridge redirect both halves the same way and give a full-wave one, and a capacitor across the load turns the result into something a circuit can actually run on. All of it rests on the one-way behaviour you have just been sweeping a slider across.
© 2026 Derek Molloy, Dublin City University. All rights reserved.