Senior Physics | Semiconductors and Logic Gates Prep Guide (for tutors) | Diodes, rectification and truth tables in three steps

2026-11-05

Tutor lesson-prep guides · Secondary Physics

Students can copy a circuit diagram but freeze when asked why a diode only conducts one way. This lesson runs from doping and the depletion layer to truth tables, so rectification and logic gates become the same switch idea.

Objectives — Students can: (1) name the majority carriers in intrinsic, n-type and p-type silicon and explain one-way conduction; (2) use a 0.7 V forward drop to size a current-limiting resistor and to find the average of a half-wave rectified voltage; (3) complete AND, OR, NOT and NAND truth tables and explain how a transistor switch implements them.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Current direction	Which terminal does conventional current leave	The positive terminal
Ohm law	Current through 150 ohm at 5 V	33 mA
Binary	What do 0 and 1 mean in a circuit	Low and high voltage

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening check	Ask why a phone charger turns AC into DC	Hands up
8-22 min	Doping and one-way flow	n-type has electrons, p-type has holes; forward conducts, reverse blocks	Sketch the depletion layer
22-34 min	Current-limiting resistor	5 V supply, LED drop 2 V, R = 150 ohm, current 20 mA	Compute it themselves
34-46 min	Rectification	12 V AC has peak 17.0 V; half-wave average 5.4 V	Peak to average
46-56 min	Logic gates	Two inputs give 4 combinations; NAND inverts AND	Fill the truth table
56-60 min	Exit check	Collect and mark three items	Work alone

03 Teaching script (read it out)

Scene 1: Doping changes carriers, not direction

Intrinsic silicon has equal electrons and holes and conducts poorly. Adding a pentavalent atom such as phosphorus gives n-type material with electrons as majority carriers; adding a trivalent atom such as boron gives p-type material with holes.
At the junction a depletion layer and built-in field form. Forward bias (p to positive) narrows the layer and carriers cross; reverse bias widens it and almost no current flows.
Method: ask which terminal the positive rail reaches. That one sentence separates forward from reverse every time.

Students say: why does the lamp stay dark when the battery is reversed? They must mention the wider depletion layer.

Scene 2: Both resistor and rectifier start by subtracting a drop

An LED across 5 V destroys itself because it holds only about 2 V. The series resistor must take 3 V, and for 20 mA that means R = 3 divided by 0.020, which is 150 ohm, with 0.04 W in the LED.
Rectification: a 12 V AC supply has a peak of 17.0 V, since peak equals rms times root two. Half-wave rectification keeps one half, so the average is peak over pi, which is 5.4 V, not 12 V.
Passing 5 V through one more diode leaves 4.3 V, and across 1 kohm that is 4.3 mA. Every diode charges a 0.7 V entry fee.

Students do: change the supply to 9 V, recompute R and the current, and say why the resistor changes.

Scene 3: A logic gate is a switch in series or parallel

AND: two switches in series, both must close. OR: in parallel, either closes the path. NOT: a transistor inverts, so a high input gives a low output.
Two inputs give 4 rows. AND outputs 1 in exactly one row, OR outputs 0 in exactly one row, and NAND is the exact opposite of AND.
Transistor as a switch: base current turns it on and pulls the collector low. The row students miss is NOT and NAND, so walk the table line by line.

Students do: draw AND and OR with two switches, then write all four rows.

04 Question types and worked steps

Type	Example	Steps	Answer
Carriers	Majority carriers in n-type and p-type	Look at the dopant valency	Electrons and holes
Resistor	5 V, LED 2 V, 20 mA	Subtract the drop, then divide	150 ohm
Rectification	Peak and half-wave average of 12 V AC	Times root two, then divide by pi	17.0 V and 5.4 V
Logic	Write the NAND truth table	Write AND, then invert each row	Only 00 gives 1

Sanity check (teach this): Sanity check: 5 V minus 2 V leaves 3 V, and dividing by 150 ohm gives 20 mA, a normal LED current; 0.04 W will not destroy it; the peak 17.0 V of a 12 V rms supply must exceed 12 V; the half-wave average 5.4 V is below the 12 V rms value, which is the right direction; two inputs can only give 4 rows, never five.

05 Independent practice (3 questions)

Question	Answer	What to watch
Carriers, three items	Identify n-type and p-type carriers	Are electrons and holes kept apart
Resistor, three items	Different supplies and LED currents	Was the drop subtracted first
Rectification, two items	Peak and half-wave average	Were root two and pi applied
Truth tables, two items	AND and NAND, four rows	Any missing row or inverted output

06 Common mistakes and what to say

Mistake	Student writes	Say this
Dividing the supply directly	R = 5 divided by 0.020 = 250 ohm	The LED holds 2 V, so the resistor sees only 3 V; subtract first.
Using rms as peak	Treating 12 V AC as the peak	Peak is 12 times root two, about 17 V; the average then divides by pi. Order matters.
Calling reverse bias high resistance	Saying reverse bias just means more resistance	Reverse bias widens the depletion layer so carriers cannot cross; it is not a resistance value.
NAND treated as AND	Output 1 when both inputs are 1	NAND is AND followed by inversion, so the last row must flip.

07 Homework (paper, section and question numbers)

Secondary physics chapter index — All chapters at a glance
Prep guide master index — Compare similar chapters
Redo this chapter — Two resistor items, one rectification item and one NAND truth table

08 Exit check (last 5 minutes)

Question	Answer	Decision
Majority carriers in p-type	Holes	Answering electrons sends you back to act one
5 V, LED 2 V, 20 mA, what is R	150 ohm	Answering 250 ohm sends you back to act two
NAND output for inputs 00	1	Answering 0 sends you back to act three

09 Tutor FAQ

Must the 0.7 V drop be memorised?

Yes; a silicon diode drops about 0.7 V forward. Most questions give it, but you must know where to subtract it.

How many gates are needed?

DSE needs AND, OR and NOT; NAND and NOR are built from them, so remember the pattern of doing the base operation then inverting.

Are semiconductor calculations examined?

Mostly concepts and circuit arithmetic such as current-limiting resistors and rectified averages; band diagrams in depth are not.

Find a tutor — free Tutor Registration →

A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.