AST20112 Electronic Devices and Circuit
Tutorial 2 Solution
Q1.
What is a depletion region in a PN junction?
A depletion region in a PN junction is a region has two sides. One side which belongs to the Nside contains net immobile positive ions and the other side which belongs to the P-side contains
net immobile negative ions.
Q2.
Briefly explain why there is a fixed size depletion region in a PN junction when an N-type silicon
crystal and P-type silicon are joined together.
Initially, an N-type silicon crystal which is doped with electron-donors contains abundant mobile
electrons while a P-type silicon crystal which is doped with hole-donors contains abundant
mobile holes. When an N-type silicon crystal and P-type silicon crystal are joined together,
electrons would diffuse from N-side to P-side and holes would diffuse from P-side to N-side. The
neutralizations of mobile holes and mobile electron occur near to the junction results in charge
separation near to the junction. Hence a barrier voltage is developed across the junction. As long
as there is charge diffusion, the barrier voltage keeps increasing. When the barrier voltage is large,
only those energetic holes and free electrons can move across the junction. So the Barrier Voltage
opposes the diffusion of some low energy holes and low energy free electrons. The Barrier
Voltage continues to grow and, up to a certain point, the Barrier Voltage is large enough to stop
any diffusion. At that point, no holes and free electrons flow across the depletion region. The
Barrier Voltage stops growing. That reaches an equilibrium point. Therefore a fixed size depletion
region with a constant barrier voltage forms eventually.
Q3.
Explain why a forward biased PN junction conducts current from its P-side to N-side.
In forward bias, the P-side of a PN junction has higher potential than its N-side. Holes in the Pside are driven and move to the depletion region. Similarly electrons in the N-side are driven and
move to the depletion region. When holes and electrons move in, charge neutralization occurs
that reduces the number of immobile charges (-ve charge B ions and +ve charge P ions) in the
depletion region. Hence the barrier voltage is reduced. With the external source and lower
barrier voltage, more holes/electrons having enough energy can overcome the reduced barrier
voltage, move across the junction and restore back those immobile +ve P ions/-ve B ions in the
depletion. As a result the depletion region cannot be entirely eliminated. A fixed barrier voltage
always exists in an equilibrium state. Since the external source supplies power to holes and
electrons, so there are always holes and electrons with sufficient energy flowing through the
depletion region. Hence the forward biased PN junction conducts current from its P-side to Nside.
Q4.
Explain why a reverse biased PN junction does not conduct current from its N-side to P-side.
In reverse bias, the N-side of a PN junction has higher potential than its P-side. Holes in the Pside are driven by the external voltage source. They leave the P-side, leave behind -ve immobile
B ions and move towards the lowest potential point of the external source. Electrons in the Nside are driven by the source. They leave the N-side, leave behind +ve immobile P ions and move
towards the highest potential point of the source. As a result the number of net charges in the
depletion region increases. The depletion region contains more immobile charges and becomes
wider. Hence the barrier voltage Vo increases. Vo cannot grow indefinitely, it will reach an
equilibrium state and stop growing. At the equilibrium state, Vo becomes so large that electrons
in N-side and holes in P-side do not have enough energy to overcome the large Vo. As a result,
no carrier diffusion across the depletion region hence no current flow from the N-side to the Pside in a reverse-biased PN junction.