A full lesson: a cartoon movie to start, then the NCERT experiment where you take your own readings, the forward curve and its knee, dynamic resistance from a tiny triangle, the full graph with reverse saturation and breakdown, a diode guarding a torch and taming ac, NEET shortcuts, problems to solve yourself, and 12 MCQs.
The turnstile turns only one way, and only after a firm push (the knee voltage). From the wrong side only a trickle gets through (reverse saturation current), until an enormous push breaks it (breakdown). Plot push against flow and you get the diode's V-I graph.
At the metro station Kabir leans lightly on the turnstile: nothing happens. He pushes a little harder: still stuck. Then, past a certain push, the arms swing freely and people stream through, faster and faster the harder the crowd pushes. Meera tries from the exit side: the arms will not move, and only a small child squeezes through underneath now and then. Only if a huge crowd shoves from the wrong side does the gate finally break.
A semiconductor diode behaves exactly like this. Forward bias: almost no current until the threshold (knee, cut-in) voltage, then a steep rise. Reverse bias: a tiny, nearly constant reverse saturation current, until breakdown. The graph of current against voltage that captures all this is the V-I characteristic, and NEET asks about every part of it.
"A resistor's graph is a straight line through the origin. A diode's graph is bent: flat, then a knee, then almost vertical. A bent graph means the resistance changes with voltage, and that is why we need dynamic resistance."
"Will a silicon diode in series with a 1.5 V cell and a bulb light the bulb? What about with a 0.3 V cell?"
With 1.5 V, yes (if the diode is forward biased): about 0.7 V drops across the diode and the rest drives the bulb. With 0.3 V the cell cannot even reach the knee of a silicon diode, so almost no current flows and the bulb stays dark.
Take an LED, a 3 V coin cell and two wires. Touch the long leg to + and the short leg to −: it glows. Turn it round: dark. Now try a single 1.5 V AA cell the right way round: a red LED stays dark because 1.5 V is below its knee (about 1.8 V). You have just found the threshold voltage.
NCERT circuit: a battery, a rheostat (potential divider) to vary the voltage, the diode, a voltmeter across the diode and a meter in series. Forward bias uses a milliammeter; reverse bias uses a microammeter and a larger voltage. Slide the rheostat (or drag its contact), then press Record to add a reading.
| # | Bias | V across diode | Current |
|---|---|---|---|
| No readings yet. Press Record. | |||
Not across the battery: part of the battery voltage drops on the rheostat and wires.
The reverse current is a thousand times smaller. A milliammeter would read zero.
Above the rated forward current the diode overheats and is destroyed, so a series resistor is always used.
"Look at your forward readings. From 0.3 to 0.5 volt the current hardly moves. From 0.65 to 0.75 volt it jumps several times. Equal steps in voltage, wildly different steps in current: that is the knee."
In forward bias the current stays very small until the applied voltage nearly cancels the barrier potential. After this threshold (cut-in, knee) voltage the current rises steeply, in fact exponentially. Drag the dot along the curve (or use the slider) and switch the material.
Below it the barrier still stops most carriers. Above it the current grows exponentially.
Beyond NCERT: every 60 mV or so of extra forward voltage multiplies the current about ten times.
In many NEET circuits the diode is "ideal": zero resistance forward, infinite reverse. If they say silicon, subtract 0.7 V.
"Cut-in voltage is a forward-bias idea. Breakdown voltage is a reverse-bias idea. Never mix the two in an answer."
Because the curve bends, V/I changes from point to point. The useful quantity is the dynamic (ac) resistance: the ratio of a small change in voltage to the resulting small change in current, r_d = ΔV/ΔI. Move the operating point and the step size and compare r_d with the static V/I.
In a NEET graph question they are usually read off the straight-looking part above the knee.
0.7 V → 0.8 V while 10 mA → 30 mA: r_d = 0.1 / 0.02 = 5 Ω.
r_d is the inverse of the slope of the I-V graph. Forward above the knee: a few ohms. Reverse (flat): megaohms.
"Static resistance V over I uses one point and the origin. Dynamic resistance uses a small change. NEET almost always asks for the dynamic one: change over change."
"As the forward current increases, does the dynamic resistance increase or decrease?"
It decreases. The curve gets steeper, so a given ΔV produces a much bigger ΔI. In fact r_d is roughly 0.05 V divided by the current: about 10 Ω at 5 mA, 1.5 Ω at 33 mA.
The complete characteristic is drawn with different scales in each quadrant: forward current in mA, reverse current in µA, forward voltage in tenths of a volt, reverse voltage in tens of volts. Slide across the whole range and switch on the heater.
A small reverse voltage already sweeps every available minority carrier across. More voltage cannot make more carriers.
Ordinary diodes are never used beyond the reverse saturation region; Zener diodes are built to work there (Lesson 8).
More heat, more minority carriers, more reverse current. The knee also moves slightly lower (about 2 mV per °C).
"Low resistance forward, very high resistance reverse. This one-way property is what lets a diode turn ac into dc. Remember the reason for the flat reverse part: it depends on the number of minority carriers, not on the voltage."
Many torches and remotes have a diode in series with the battery, so that a battery put in the wrong way round cannot damage the circuit. Tap the battery to put it in either way. Then feed the same circuit from ac and watch the diode let only one half of each cycle through.
The diode is reverse biased and only a few µA leak through: nothing is harmed.
From a 3 V battery the LED gets only about 2.3 V. Designers sometimes use Schottky diodes (≈ 0.3 V) instead.
The output is one-way but pulsating. Lesson 7 builds full-wave rectifiers and smooths them with a capacitor filter.
If a question says "silicon diode", subtract 0.7 V from the supply before using Ohm's law on the resistor.
5 V, Si diode, 430 Ω: I = 4.3/430 = 10 mA. Ideal diode: I = E/R.
Convert mA to A. 0.1 V / 20 mA = 5 Ω.
The whole supply voltage appears across the reverse-biased diode; none across the series resistor.
The two quadrants of the graph use different scales, so the graph is not "continuous" through the origin in slope.
All forward: subtract 0.7 V for each silicon diode.
"For circuit questions, first decide whether each diode is forward or reverse biased. Forward: replace it by a 0.7 volt drop, or a plain wire if ideal. Reverse: remove it. Then use Ohm's law. That one habit solves most NEET diode circuits."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
On the forward characteristic of a Si diode, the current is 10 mA at 0.7 V and 30 mA at 0.8 V. Find the dynamic resistance in this region.
r_d = ΔV/ΔI, with ΔI in amperes.
At V = 0.75 V a diode carries 15 mA. Find its static resistance. Is the dynamic resistance at this point larger or smaller?
Static R = V/I at one point. Compare slopes.
A Si diode (cut-in 0.7 V) is connected in forward bias in series with a 430 Ω resistor across a 5 V battery. Find the current and the voltage across the resistor.
Replace the forward-biased diode by a 0.7 V drop.
In Problem 3 the battery is turned round. Find the current and the voltages across the resistor and the diode.
Reverse biased: treat the diode as an open switch.
Two Si diodes in series, both forward biased, are connected with a 1 kΩ resistor across a 6 V supply. Find the current. What happens if one diode is turned round?
Subtract 0.7 V for each forward diode.
Take the cut-in voltage of silicon as 0.7 V unless the diode is called ideal. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "cut-in / knee / threshold voltage" | Si ≈ 0.7 V, Ge ≈ 0.2–0.3 V (forward) | Confusing it with breakdown voltage |
| "dynamic resistance from the graph" | r_d = ΔV / ΔI | Using V/I at one point, or leaving I in mA |
| "reverse current almost constant because" | limited by minority carriers | Saying "because the barrier is zero" |
| "meter for reverse current" | microammeter | Choosing a milliammeter |
| "Si diode in series with R across E" | I = (E − 0.7)/R | Forgetting the 0.7 V drop |
| "ideal diode" | forward: short circuit · reverse: open | Still subtracting 0.7 V |
| "effect of temperature on reverse current" | increases (more minority carriers) | Saying it decreases like in metals |
| "which graph is a diode's V-I" | knee in 1st quadrant, small flat line in 3rd | Picking a straight line through the origin |
Lesson 7 uses this one-way property: a half-wave rectifier with one diode, a full-wave rectifier with two (centre-tap) or four (bridge), and a capacitor filter to smooth the output into nearly steady dc.
Curves use an illustrative diode model I = I₀(e^(V/0.052 V) − 1) with I₀ = 7 nA (Si) and 16 µA (Ge); the full characteristic uses a reverse saturation current of about 2 µA and breakdown near 50 V for clarity. Real diodes differ; the shapes and rules are what matter.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.