p–n Junction Diode

PHYp–n Junction Diodes

A p-n junction diode conducts easily in forward bias once its threshold (knee) voltage is exceeded and blocks current in reverse bias. ISC numericals focus on two ideas: the dynamic (a.c.) resistance from the slope of the V-I characteristic, and the d.c.

circuit current when a silicon diode (with its fixed $\approx 0.7\,\text{V}$ drop) is in series with a resistor across a battery.

Dynamic (a.c.) resistance

r_{ac} = \dfrac{\Delta V}{\Delta I}

\Delta V

= change in forward voltage,

\Delta I

= corresponding change in forward current; slope of the V-I curve at the operating point.

Series-circuit current

I = \dfrac{V_{supply} - V_d}{R}

V_{supply}

= battery voltage,

V_d

= forward voltage drop (

\approx 0.7\,\text{V}

for Si,

\approx 0.3\,\text{V}

for Ge),

R

= series resistance.

Voltage across the resistor

V_R = V_{supply} - V_d

By Kirchhoff's voltage law in the series loop; this voltage drives the current through

R

Forward bias narrows the depletion region and lowers the barrier, so current rises steeply above the knee voltage.
Reverse bias widens the depletion region; only a tiny saturation current flows until breakdown.
Knee/threshold voltage $\approx 0.7\,\text{V}$ for silicon and $\approx 0.3\,\text{V}$ for germanium.
Dynamic resistance $r_{ac}=\Delta V/\Delta I$ is small in forward conduction and very large in reverse bias.
Static (d.c.) resistance $V/I$ differs from dynamic resistance — the question wording tells you which to use.
In a series circuit subtract the diode drop first, then divide the remaining voltage by $R$ .
Convert mA to A before computing resistance in ohms ( $1\,\text{mA} = 10^{-3}\,\text{A}$ ).

Where the marks go

Forgetting to subtract the $0.7\,\text{V}$ diode drop and using the full supply voltage across $R$ .
Confusing dynamic resistance $\Delta V/\Delta I$ with static resistance $V/I$ .
Leaving currents in mA so the resistance comes out $1000\times$ too small.
Treating the depletion width backwards — it decreases in forward bias, not increases.

How the board asks it

Numericalseries-circuit current after subtracting the diode drop
A silicon diode is connected in series with a $200\,\Omega$ resistor across a $5\,\text{V}$ battery. Taking the diode drop as $0.7\,\text{V}$ , calculate the current in the circuit and the voltage across the resistor.
Numericaldynamic resistance from the slope of the v-i characteristic
For a junction diode, the forward current changes from $5\,\text{mA}$ to $15\,\text{mA}$ when the forward voltage changes from $0.70\,\text{V}$ to $0.72\,\text{V}$ . Calculate the dynamic (a.c.) resistance of the diode.
Diagram / graphforward and reverse v-i characteristic curve
Draw the V-I characteristic of a p-n junction diode and mark on it the knee (threshold) voltage and the region used to determine the dynamic resistance.
Give reasonsdepletion-region width and barrier potential on biasing
Explain why a p-n junction diode conducts heavily in forward bias but allows only a very small current in reverse bias, in terms of the depletion region and the potential barrier.
Distinguishstatic $V/I$ versus dynamic $\Delta V/\Delta I$ resistance
Distinguish between the static (d.c.) resistance and the dynamic (a.c.) resistance of a p-n junction diode, stating how each is obtained from its characteristic curve.
Define / statethreshold/knee voltage values for silicon and germanium
Define the knee (threshold) voltage of a junction diode and state its approximate value for a silicon diode and for a germanium diode.

Practise this topic

PreviousIntrinsic & Extrinsic Semiconductors NextRectifiers

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.