CLASS 12 · CHAPTER 14 · MODERN PHYSICS
Semiconductor Electronics
Drag this diode's voltage from strongly reverse to mildly forward. Across the entire reverse range, almost nothing happens to the current. Cross into forward bias, and that same device suddenly explodes into an exponential current source.
Watch it happen
No applied voltage, no net current. This is the starting point: the depletion region's built-in field is unopposed and uncompensated.
Start at the leftmost (most reverse) end of the slider and drag slowly to the right. For a long stretch, the current readout barely moves at all, it’s pinned near −I₀, the tiny reverse saturation current. Then, past about 0.4–0.5 V, watch the curve bend sharply upward: the same device that looked like an open switch moments ago is now carrying current that climbs by orders of magnitude for every small step of the slider.
Where the formula comes from
A p-n junction is formed by doping one side of a semiconductor crystal to be p-type (trivalent impurities contribute mobile positive “holes”) and the other n-type (pentavalent impurities contribute mobile free electrons). The moment the junction forms, electrons near the boundary diffuse into the p-side and holes diffuse into the n-side, leaving behind fixed, immobile charged ions. This strip of exposed ions is the depletion region: it has no free carriers of its own, and its internal field opposes any further diffusion, bringing the process to a self-limiting equilibrium.
Applying an external voltage disturbs that equilibrium. Under forward bias (p-side to +, n-side to −), the applied field opposes the junction’s own built-in field, so the depletion region narrows, and once it’s narrow enough, carriers flood across the junction in large numbers. Under reverse bias (p-side to −, n-side to +), the applied field reinforces the built-in field, so the depletion region widens further, and essentially nothing but a trickle of thermally-generated minority carriers can cross.
A full quantum/statistical treatment of carrier crossing probabilities isn’t expected at this level; the result is simply stated and used. The current through the junction as a function of applied voltage is given by the diode equation (the Shockley equation):
where is the reverse saturation current (tiny, set by the minority-carrier supply), and is the thermal voltage at room temperature. For negative (reverse bias) and much larger in magnitude than , , so : a small, constant, reverse current no matter how reverse you go. For positive (forward bias), grows explosively, so , an exponential rise.
Where the shortcut stops working
A diode is dramatically non-Ohmic and asymmetric. Ohm’s law, , assumes current scales with voltage through a fixed, constant resistance. A diode does nothing of the sort, and it behaves completely differently depending on which way you connect it.
Under reverse bias, the current stays essentially flat and minuscule across a huge range of voltage. At V=−1 V, I≈−1.000×10⁻¹² A. At V=−5 V, five times more reverse voltage, I≈−1.000×10⁻¹² A, the same value, to the precision this equation can show. A resistor subjected to five times the voltage would carry five times the current; this device carries the same current either way, because the reverse current isn’t limited by voltage at all, it’s limited by the tiny supply of thermally generated minority carriers available to cross the junction.
Then, crossing into forward bias, the exact same device’s current explodes. At V=0.5 V, I≈2.248×10⁻⁴ A (0.2248 mA). At V=0.6 V, not even double the voltage, I≈1.052×10⁻² A (10.52 mA), roughly 47 times more current. At V=0.7 V, I≈0.4927 A, another ~47-fold jump. Going from 0.5 V to 0.7 V (not even doubling the voltage) multiplies the current by roughly 2000 times. The device behaves like an open switch in one direction and a voltage-controlled exponential current source in the other; (real diodes also eventually break down and conduct heavily at some large enough negative reverse voltage, the Zener or avalanche breakdown region, but that is a distinct breakdown mechanism, not a return to Ohmic behaviour).
Apply it under exam conditions
Q1. For a silicon diode with I₀ = 1×10⁻¹² A and V_T = 0.026 V, find the forward current at V = 0.6 V.
Q2. Compare the reverse current at V = −1 V and V = −5 V for the same diode, and comment on what this shows.
In both cases, , so (e.g. at V=−1 V, , utterly negligible):
The two currents are identical to the precision this equation can show, despite a 5× change in reverse voltage. This is the signature of non-Ohmic behaviour: in reverse bias, the diode’s current is set by carrier availability, not by how hard the voltage pushes.
Quick answers
Doesn't more reverse voltage mean more reverse current, the way it would for a resistor?+
No, and that's precisely the non-Ohmic behaviour this chapter is built around. For a resistor, V=IR means current scales with voltage. For a diode under reverse bias, the current is essentially capped at I₀, the reverse saturation current, set by how many minority carriers are thermally available to cross the junction, not by how hard you're pushing them electrically. Going from −1 V to −5 V (five times more reverse voltage) barely changes the current at all: both sit at roughly −1×10⁻¹² A in this simulation. Eventually, at some large enough reverse voltage, real diodes do break down (Zener or avalanche breakdown) and conduct heavily, but that's a completely different mechanism, not Ohm's law reasserting itself.
What is a diode actually used for?+
Its most basic use is rectification: converting AC (which reverses direction every half-cycle) into DC (which flows one way only). Because the diode conducts easily in forward bias but blocks current in reverse bias, it only lets through the half-cycles that forward-bias it, chopping an alternating input into a one-directional, pulsing output. Diodes also show up as voltage regulators (Zener diodes), light sources (LEDs), light sensors (photodiodes), and solar cells.
Why does forward current rise so fast with voltage, instead of smoothly like in a resistor?+
Because the diode equation is exponential, not linear: I = I₀(e^(V/V_T) − 1), with V_T ≈ 0.026 V at room temperature. Since V_T is so small, even a modest increase in V (a few tenths of a volt) corresponds to a huge value of V/V_T in the exponent, and e raised to a larger number grows explosively. Going from V=0.5 V to V=0.6 V multiplies the current by roughly 47 times; going from 0.6 V to 0.7 V multiplies it by roughly another 47 times. This is why diodes appear to have a fairly sharp 'turn-on' voltage in practice, even though the underlying curve is smooth.
If I is capped in reverse bias, is the diode acting like an open circuit?+
For most practical purposes, yes. A reverse current of order 10⁻¹² A (a picoamp) is so small compared to the milliamp-to-amp currents a forward-biased diode carries that engineers routinely treat a reverse-biased diode as an open switch. It's not literally zero current, but it's close enough that the asymmetry (near-open one way, a steep exponential conductor the other way) is what makes diodes useful as one-way valves for current.
Related concepts
Physics doesn’t stay inside chapter boundaries. Neither should you.