How unit 5 is examined
Digital basics (number systems, gates, adders, flip-flops), then the diode and the BJT. Marks sit in the CE and CB characteristics, the R-S and J-K flip-flops, logic gates and the diode V-I curve.
Number systems & Their conversion used in digital electronics
Low weight
Definition. A number system of base $r$ uses $r$ digits, and each digit has a place value that is a power of $r$; digital electronics uses binary (2), octal (8), decimal (10) and hexadecimal (16).
Key points.
- Decimal to another base: divide the integer part repeatedly by the base and read the remainders upward; multiply the fraction repeatedly by the base and read the carried-out integers downward.
- Octal to binary: replace each octal digit by its 3-bit code; hex to binary: replace each hex digit by its 4-bit code (A=1010, B=1011, C=1100).
- Hex to octal: convert to binary first, then regroup in 3 bits from the binary point outward, padding zeros at the ends.
- To find an unknown base, expand the number as powers of $x$ and equate it to the decimal value.
Example (Dec 2024).
| Part | Working | Answer |
|---|---|---|
| i) $(257)_8$ | 2=010, 5=101, 7=111 | $(10101111)_2$ |
| ii) $(21.625)_{10}$ | 21 = 2×8+5 gives 25; 0.625×8 = 5.0 gives .5 | $(25.5)_8$ |
| iii) $(BC.2)_{16}$ | 1011 1100 . 0010, regroup: 010 111 100 . 001 000 | $(274.1)_8$ |
| iv) $(33)_{10}=(201)_x$ | $2x^2+0x+1=33$, $x^2=16$ | $x=4$ |
Answer frame. Show each conversion as its own line of working, regroup in 3s for octal, and state the answer with the base as a subscript.
Asked: [7 marks] (Dec 2024) Solve for $x$: i) $(257)_8=(x)_2$ ii) $(21.625)_{10}=(x)_8$ iii) $(BC.2)_{16}=(x)_8$ iv) $(33)_{10}=(201)_x$
De Morgan's theorem
Not asked since 2022
Definition. ==De Morgan's theorems state that the complement of a sum is the product of the complements, $\overline{A+B}=\bar A\cdot\bar B$, and the complement of a product is the sum of the complements, $\overline{A\cdot B}=\bar A+\bar B$.==
Key points.
- The first theorem says a NOR gate equals a bubbled-input AND gate; the second says a NAND gate equals a bubbled-input OR gate.
- To apply it, break the bar over the expression and change the operator (AND to OR, OR to AND).
- It extends to any number of variables and is used to convert circuits to NAND-only or NOR-only form.
- Truth-table proof: for $A=0,B=1$, $\overline{A+B}=0$ and $\bar A\bar B=1\cdot0=0$; the two columns match for all four rows.
Asked: [14 marks] (Nov 2022, Jun 2023, Dec 2023) Short note on De Morgan's theorem (options in "any two" notes; state both theorems with truth-table verification)
Logic Gates
High weight
Definition. A logic gate is a digital circuit with one or more inputs and a single output that follows a Boolean rule; AND, OR and NOT are basic gates, and NAND and NOR are universal gates.
Key points.
- AND gives 1 only when all inputs are 1 ($Y=A\cdot B$); OR gives 1 when any input is 1 ($Y=A+B$).
- NOT inverts its single input ($Y=\bar A$).
- NAND is AND followed by NOT ($Y=\overline{AB}$) and NOR is OR followed by NOT ($Y=\overline{A+B}$).
- NAND and NOR are universal because any other gate can be built from either one alone.
- XOR gives 1 when the inputs differ ($Y=A\oplus B$); XNOR gives 1 when they are equal.
- Symbols: AND has a D-shaped body, OR a curved-back body, and a small bubble on the output marks inversion (NOT, NAND, NOR).
Truth table.
| A | B | AND | OR | NAND | NOR | XOR | NOT A |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Answer frame. Open with the definition; draw the symbol of each of AND, OR, NOT, NAND and NOR beside its Boolean expression; give the combined truth table; close by saying NAND and NOR are universal.
Pitfall: NAND and NOR are the universal gates; do not call AND or OR universal.
Asked: [14 marks] (Jun 2025) Write short notes on any two: i) Logic gates ii) R-S flip flop iii) Compare electric and magnetic circuit iv) B-H curve
Half and full adder circuits
Low weight
Definition. An adder is a combinational circuit that adds binary numbers; a half adder adds two bits, and a full adder adds two bits plus a carry-in.
Key points.
- Half adder: $Sum=A\oplus B$ and $Carry=A\cdot B$, built from one XOR and one AND gate.
- Full adder has inputs $A$, $B$, $C_{in}$ and outputs $Sum$ and $C_{out}$.
- $Sum=A\oplus B\oplus C_{in}$ and $C_{out}=AB+C_{in}(A\oplus B)$.
- A full adder is two half adders plus an OR gate: the first adds $A,B$; the second adds its sum to $C_{in}$; the two carries are ORed.
Truth table (full adder).
| A | B | Cin | Sum | Cout |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
Diagram.
Answer frame. Open with the definition of an adder; draw the two-half-adder circuit; give the truth table and both equations; close with the two-half-adders-plus-OR statement.
Asked: [7 marks] (Dec 2023) What do you understand by Adder circuit? Draw and explain the working principle of full adder circuit.
R-S flip flop
High weight
Definition. An R-S (Set-Reset) flip-flop is a bistable memory element with two inputs, S and R, and complementary outputs $Q$ and $\bar Q$, that stores one bit.
Key points.
- It is built from two cross-coupled NOR (or NAND) gates, the output of each feeding an input of the other.
- S=0, R=0 is the no-change (memory) state: the output holds its previous value $Q_n$.
- S=1, R=0 sets the flip-flop, giving $Q=1$; S=0, R=1 resets it, giving $Q=0$.
- S=1, R=1 is invalid for NOR: both outputs go to 0, violating $Q=\bar Q'$, and the final state is unpredictable when both inputs return to 0 (race condition).
- A NAND latch has active-low inputs, so its invalid state is S=R=0.
- It is the basic memory cell of sequential circuits; a clocked version changes state only on the clock pulse.
Truth table (NOR latch).
| S | R | $Q_{n+1}$ | State |
|---|---|---|---|
| 0 | 0 | $Q_n$ | No change |
| 0 | 1 | 0 | Reset |
| 1 | 0 | 1 | Set |
| 1 | 1 | ? | Invalid |
Diagram.
Answer frame. Open with the definition; draw the NOR-gate circuit with the feedback; give the truth table; explain the four states, stressing the invalid one; close with the memory role.
Pitfall: Write S=R=1 as invalid or forbidden, not as a valid state.
Asked: [14 marks] (Nov 2022, Jun 2023, Jun 2025) Write short notes on R-S flip-flop (options with J-K flip-flop, De Morgan's theorem, star-delta transformation, torque-slip characteristics, logic gates, electric and magnetic circuit, B-H curve)
J-K flip flop
High weight
Definition. ==A J-K flip-flop is a clocked R-S flip-flop with feedback from the outputs to the inputs, which removes the invalid state by making J=K=1 toggle the output.==
Key points.
- It has inputs J (set) and K (reset), a clock input, and outputs $Q$ and $\bar Q$; $Q$ feeds back to the gate of K and $\bar Q$ to the gate of J.
- J=0, K=0 gives no change; J=0, K=1 resets ($Q=0$); J=1, K=0 sets ($Q=1$).
- J=1, K=1 toggles the output, $Q_{n+1}=\bar Q_n$, so the invalid condition of the R-S flip-flop does not occur.
- Characteristic equation: $Q_{n+1}=J\bar Q_n+\bar K Q_n$.
- Race-around: if the clock pulse stays high longer than the propagation delay while J=K=1, the output toggles repeatedly and the final state is uncertain.
- It is cured by a master-slave arrangement or edge triggering, so the output changes only once per clock.
Truth table.
| J | K | $Q_{n+1}$ | State |
|---|---|---|---|
| 0 | 0 | $Q_n$ | No change |
| 0 | 1 | 0 | Reset |
| 1 | 0 | 1 | Set |
| 1 | 1 | $\bar Q_n$ | Toggle |
Diagram.
Answer frame. Open with the definition; draw the symbol or the gated circuit with feedback; give the truth table; explain toggling, then race-around and its cure; close with the master-slave remedy.
Asked: [14 marks] (Dec 2023, Nov 2022, Jun 2023) Write a short note on J-K flip flop (options with dependent and independent sources, open and short circuit test, De Morgan's theorem)
Introduction to Semiconductors
Not asked since 2022
Definition. A semiconductor is a material whose conductivity lies between a conductor and an insulator, such as silicon and germanium, and rises with temperature and doping.
Key points.
- Intrinsic semiconductors are pure crystals in which thermal energy creates equal numbers of free electrons and holes.
- Extrinsic semiconductors are doped: n-type uses pentavalent donors (P, As) with electrons as majority carriers; p-type uses trivalent acceptors (B, Ga) with holes as majority carriers.
- Silicon has band gap about 1.1 eV and germanium about 0.7 eV.
- Unlike metals, a semiconductor has a negative temperature coefficient of resistance.
Diodes
Not asked since 2022
Definition. A diode is a two-terminal device formed by a p-n junction that conducts current easily in the forward direction and blocks it in the reverse direction.
Key points.
- Joining p and n type material creates a depletion region with a barrier potential of about 0.7 V (Si) and 0.3 V (Ge).
- Forward bias (anode positive) narrows the depletion region, so the diode conducts once the barrier is overcome.
- Reverse bias widens the depletion region and only a tiny leakage current flows.
- Its main use is rectification, converting AC to DC; the Zener diode is used for voltage regulation in breakdown.
V-I characteristics
Medium weight
Definition. The V-I characteristic of a p-n junction diode is the graph of diode current against applied voltage, with a forward region and a reverse region.
Key points.
- Forward bias: up to the cut-in (knee) voltage, 0.7 V for Si and 0.3 V for Ge, the current is negligible because the applied voltage has not overcome the barrier potential.
- Beyond the knee the current rises exponentially in mA, $I=I_0(e^{V/\eta V_T}-1)$, and the diode acts almost as a closed switch.
- Reverse bias: only a small reverse saturation current $I_0$ (in $\mu$A for Ge, nA for Si) flows, due to minority carriers, and it is nearly constant with voltage.
- At the reverse breakdown voltage $V_{BR}$ the current rises sharply by Zener or avalanche breakdown, and the diode can be damaged unless the current is limited.
- Axes: forward current $I_F$ in mA and voltage $V_F$ in the first quadrant; reverse voltage $V_R$ and current $I_R$ in $\mu$A in the third quadrant.
Diagram.
Answer frame. Open with the definition; draw the two-quadrant curve with the knee, $I_0$ and $V_{BR}$ labelled (mA scale forward, $\mu$A reverse); explain forward, reverse and breakdown in that order; close with the Si/Ge knee values.
Pitfall: Use different scales for the two quadrants, mA forward and $\mu$A reverse, and label them.
Asked: [7 marks] (Jun 2023, Jun 2025) Draw and explain the V-I characteristics of a PN junction / diode.
Bipolar junction transistors (BJT) and their working
Low weight
Definition. A bipolar junction transistor is a three-terminal, current-controlled device with two p-n junctions, in which a small base current controls a large collector current.
Key points.
- It is npn or pnp, with three regions: emitter (heavily doped), base (thin, lightly doped) and collector (moderately doped, largest area).
- For amplification the emitter-base junction is forward biased and the collector-base junction reverse biased.
- In an npn transistor the emitter injects electrons into the thin base; only about 1% recombine to form the base current, and the rest are swept across to the collector.
- Currents: $I_E=I_B+I_C$, with $I_C=\beta I_B$, $\alpha=I_C/I_E$ and $\beta=\alpha/(1-\alpha)$.
- Its characteristics are drawn in the CE configuration: input $I_B$ vs $V_{BE}$ (like a forward diode) and output $I_C$ vs $V_{CE}$ (see the CE section below).
Asked: [7 marks] (Jun 2022) Explain principle of operation and characteristics of BJT.
Introduction to CC, CB & CE transistor configurations
High weight
Definition. A transistor configuration is named after the terminal common to input and output: common base (CB), common emitter (CE) or common collector (CC).
Key points.
- CE input characteristic is $I_B$ against $V_{BE}$ at constant $V_{CE}$; it resembles a forward diode curve with knee near 0.7 V (Si).
- CE output characteristic is $I_C$ against $V_{CE}$ for fixed values of $I_B$; each curve rises steeply, then flattens.
- Active region: emitter junction forward and collector junction reverse biased; $I_C\approx\beta I_B$, the transistor works as a linear amplifier.
- Saturation region ($V_{CE}<V_{CE(sat)}\approx0.2$ V, the knee): both junctions forward biased, $I_C$ is maximum, and the transistor acts as a closed switch.
- Cutoff region ($I_B=0$): both junctions reverse biased, only the small leakage $I_{CEO}$ flows, and the transistor acts as an open switch.
- CB: input characteristic is $V_{EB}$ against $I_E$ at constant $V_{CB}$; it is a forward diode curve. Output is $I_C$ against $V_{CB}$ at constant $I_E$: nearly flat lines in the active region, with $I_C\approx I_E$.
- In CB, $\alpha=\Delta I_C/\Delta I_E$ (at constant $V_{CB}$), about 0.95-0.99, so there is no current gain, but the voltage gain is high.
- CB output regions: active ($V_{CB}$ reverse, $I_C\approx\alpha I_E$), saturation (small forward $V_{CB}$, $I_C$ falls rapidly), cutoff ($I_E=0$, $I_C=I_{CBO}$).
Diagram (CE test circuit).
Diagram (CB test circuit).
Comparison.
| Property | CB | CE | CC |
|---|---|---|---|
| Common terminal | Base | Emitter | Collector |
| Input, output | Emitter, collector | Base, collector | Base, emitter |
| Current gain | $\alpha<1$ | $\beta$ (high, 20-500) | $\gamma=\beta+1$ |
| Voltage gain | High | High | Less than 1 |
| Input resistance | Low | Medium | Very high |
| Phase shift | 0 degrees | 180 degrees | 0 degrees |
| Use | High-frequency amplifier | General amplifier | Buffer (emitter follower) |
Answer frame.
- CE output: open by defining CE and the active region; draw the circuit, then the $I_C$-$V_{CE}$ family for rising $I_B$ with the three regions marked; explain active, saturation, cutoff; if asked for both characteristics, add the input curve first; close by saying active is used for amplifiers and saturation/cutoff for switching.
- CB: open with the definition of CB; draw the CB circuit and both curves; give the biasing, then input, output regions and $\alpha$; close with $\alpha<1$ and $I_E=I_B+I_C$.
Asked: [7 marks] (Nov 2022, Dec 2024) Draw and explain the output characteristic of a NPN transistor operation in CE configuration / with input and output characteristics explain the operation of BJT in common emitter configuration.
Asked: [7 marks] (Jun 2023, Jun 2025) Draw the circuit and explain the characteristics of CB configuration / explain the working principle of common base transistor.
Different configurations and modes of operation of BJT
Low weight
Definition. A BJT works in cutoff, active or saturation mode, set by its bias, and its operating (Q) point is the dc $I_C$ and $V_{CE}$ at zero signal.
Key points.
- Cutoff: both junctions reverse biased, $I_C\approx0$ (switch OFF).
- Active: emitter junction forward and collector junction reverse biased (amplifier).
- Saturation: both junctions forward biased, $V_{CE}\approx0.2$ V (switch ON).
- Fixed bias: $I_B=\dfrac{V_{CC}-V_{BE}}{R_B}$, $I_C=\beta I_B$, $V_{CE}=V_{CC}-I_CR_C$.
Example (Dec 2023). Given $V_{CC}=24$ V, $R_B=220$ k$\Omega$, $R_C=4.7$ k$\Omega$; assume Si, $V_{BE}=0.7$ V, $\beta=100$.
| Step | Working |
|---|---|
| $I_B$ | $(24-0.7)/220\text{k}=105.9\ \mu$A |
| $I_C=\beta I_B$ | $100\times105.9\ \mu\text{A}=10.59$ mA |
| $V_{CE}$ | $24-10.59\text{m}\times4.7\text{k}=-25.8$ V, impossible |
| Check | $I_{C(sat)}=24/4.7\text{k}=5.1$ mA is less than 10.59 mA |
The transistor is driven into saturation, so $I_C$ is limited to 5.1 mA and $V_{CE}\approx0.2$ V. Q-point: $I_B=105.9\ \mu$A, $I_C\approx5.1$ mA, $V_{CE}\approx0.2$ V (saturation). If the examiner's expected active-region formula is used, write $Q(V_{CE},I_C)$ in terms of $\beta$ ($I_C=\beta\times105.9\ \mu$A) and state the saturation check.
Asked: [7 marks] (Dec 2023) In a fixed bias circuit using n-p-n transistor, find the operating point if $V_{CC}=24$ V, $R_B=220$ k, $R_C=4.7$ k.
Last-minute revision
- Binary to octal groups 3 bits; binary to hex groups 4 bits; $(257)_8=(10101111)_2$; $(21.625)_{10}=(25.5)_8$; $(BC.2)_{16}=(274.1)_8$; $(33)_{10}=(201)_4$.
- De Morgan: $\overline{A+B}=\bar A\bar B$ and $\overline{AB}=\bar A+\bar B$.
- NAND and NOR are universal gates; XOR output is 1 when inputs differ.
- Full adder: $Sum=A\oplus B\oplus C_{in}$, $C_{out}=AB+C_{in}(A\oplus B)$; two half adders plus OR.
- R-S flip-flop: S=R=1 is invalid; S=R=0 holds; S=1 sets; R=1 resets.
- J-K flip-flop: J=K=1 toggles; race-around cured by master-slave or edge triggering.
- Diode knee: 0.7 V Si, 0.3 V Ge; reverse current $I_0$ in $\mu$A; breakdown at $V_{BR}$.
- BJT: $I_E=I_B+I_C$, $\alpha=I_C/I_E$, $\beta=I_C/I_B$, $\beta=\alpha/(1-\alpha)$.
- CE regions: active (amplifier), saturation (ON), cutoff ($I_B=0$, OFF).
- Fixed bias: $I_B=(V_{CC}-V_{BE})/R_B$, $V_{CE}=V_{CC}-I_CR_C$; check saturation with $V_{CC}/R_C$.
Memory hooks
- NAND and NOR are "Not-And, Not-Or" and can build everything: universal.
- De Morgan: break the bar, change the sign.
- J-K: J=K=1 "Jumps" to the opposite state (toggle).
- BJT amplifier bias: Emitter Forward, Collector Reverse.
- CE = 180 degrees phase shift, CC = emitter follower, CB = smallest input resistance.
Coverage checklist
- Number systems & Their conversion used in digital electronics: Dec 2024 conversions.
- De morgan's theorem: covered in the short notes of Nov 2022, Jun 2023, Dec 2023.
- Logic Gates: Jun 2025 short note.
- half and full adder circuits: Dec 2023 full adder.
- R-S flip flop: Nov 2022, Jun 2023, Jun 2025 short notes.
- J-K flip flop: Dec 2023 short note (and Nov 2022 option).
- Introduction to Semiconductors: definition and key points.
- Diodes: definition and key points.
- V-I characteristics: Jun 2023, Jun 2025.
- Bipolar junction transistors (BJT) and their working: Jun 2022.
- introduction to CC, CB & CE transistor configurations: Nov 2022, Dec 2024, Jun 2023, Jun 2025.
- different configurations and modes of operation of BJT: Dec 2023 operating point.