PMDC Unit 49 · 2 outcomes

Electronics

Unit 13 on the PMDC list is two outcomes only: rectification with diodes, and the PN junction with forward and reverse bias. No transistors, no gates.

Open the Unit 49 quiz Worked numericals

13.1

Rectification

PMDC: Define rectification and describe the use of diodes for half- and full-wave rectification.

Rectification is turning AC into one-way current (DC). A diode is a one-way valve: easy in the arrow direction, almost open the other way.

Half-wave: one diode. One half of the sine reaches the load; the other half is blocked. Pulse frequency stays f. Average (DC) voltage is $V_{m} / π$ .

Full-wave: both halves are steered the same way through the load. Pulse frequency is 2f. Average is 2 $V_{m} / π$ . A bridge uses four diodes and no centre tap. A centre-tap uses two diodes and a tapped transformer.

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full-wave: every half-cycle, same sign

half-wave: $f_{o u t}$ = f $V_{d c} = V_{m} / π$
full-wave: $f_{o u t}$ = 2 f $V_{d c}$ = 2 $V_{m} / π$
bridge: 4 diodes, no tap PIV ≈ $V_{m}$
centre-tap: 2 diodes + tap PIV ≈ 2 $V_{m}$

Hold these

A transformer changes voltage. It does not rectify. Diodes do.
Full-wave is twice the DC of half-wave for the same peak $V_{m}$ .
PIV is the reverse peak the off diode must survive.

13.2

PN junction

PMDC: Describe the PN junction and discuss its forward and reverse biasing.

Join p-type (holes) to n-type (electrons) in one crystal. Majority carriers diffuse across, leave ions behind, and a depletion layer forms. Those ions make a barrier field from n toward p.

Forward bias: + on p, − on n. The battery fights the barrier. The layer thins. After the knee (Si ≈ $0.7 V$ , Ge ≈ $0.3 V$ ) a large current flows.

Reverse bias: + on n, − on p. The battery adds to the barrier. The layer widens. Only a tiny minority current flows — until breakdown.

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P N = p - t y p e

| n-type

barrier field: n → p
forward: p to +, barrier thins, I large after the knee
reverse: n to +, barrier thickens, I ≈ $I_{s}$ (tiny)
Si knee ≈ $0.7 V$ Ge knee ≈ $0.3 V$

Hold these

Anode is p, cathode is n. Conventional I enters the anode when ON.
Bulk p and n are electrically neutral. The name is the majority carrier, not a net charge.

Unit 49 formula sheet

Write these until they are cheaper than thinking, then open the quiz.

half-wave: $f_{o u t}$ = f $V_{d c} = V_{m} / π$
full-wave: $f_{o u t}$ = 2 f $V_{d c}$ = 2 $V_{m} / π$
bridge: 4 diodes, no tap PIV ≈ $V_{m}$
centre-tap: 2 diodes + tap PIV ≈ 2 $V_{m}$
barrier field: n → p
forward: p to +, barrier thins, I large after the knee
reverse: n to +, barrier thickens, I ≈ $I_{s}$ (tiny)
Si knee ≈ $0.7 V$ Ge knee ≈ $0.3 V$

Open the Unit 49 quiz