Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
A 6.2 V Zener regulator is fed from 20 V through 100 Ω and supplies a 2200 Ω load. Find the Zener current IZ.
An NPN transistor with β = 100 switches a 220 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 2 times the minimum needed for saturation. Find RB in kΩ.
Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 2.
The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage Vout.
The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 80 ms, in mJ.
Just before the switch closes at t = 0, the capacitor holds 12 V with the polarity shown. How long does it take for vC to reach 6.4 V? Find t.
Find the equivalent resistance Req seen by the source V1.
Find the current IR2 through R2.
An ideal boost converter turns 36 V into 120 V for a 15 Ω load. Find the average inductor current IL.
A buck converter turns 15 V into 7.5 V at 200 kHz with L = 220 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
A bridge has four 250 Ω arms and Vs = 9 V. The sensor R₂ increases by ΔR = 50 Ω. Find the exact output Vab in mV.
A bridge with 350 Ω arms is excited with 2.5 V and has four active gauges, two in tension and two in compression (full bridge). The gauge factor is 2.12 and the strain is 750 µε. Find the bridge output Vo in mV.
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
A synchronous mod-11 counter counts 0, 1, …, 10 and then returns to 0. It starts at 6. Which count Q (in decimal) does it show after 24 clock pulses?
Answer: IZ = 135.2 mA
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
Answer: RB = 2.424 kΩ
BJT as a switch: to switch a load fully on, the transistor is driven into saturation, where VCE = VCE(sat) and the collector current is set by the load: IC(sat) = (VCC − VCE(sat))/RC. Saturation needs at least IB = IC(sat)/β. If the base current is smaller, the transistor stays active and IC = β·IB.
Answer: Rf = 4.7 kΩ
Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.
V₋ = V₊ = Vin, and the current through R₁ also flows through Rf, so Vout = Vin·(1 + Rf/R₁).
Answer: Vout = 1.251 V
Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.
V₋ = 0, so each input resistor has its full input voltage across it.
Answer: W = 1496 mJ
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
Answer: t = 109.4 ms
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.
Answer: Req = 21.96 Ω
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
Answer: IR2 = 0.3916 A
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.
Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.
Answer: IL = 26.67 A
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Answer: Io,min = 0.04261 A
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
Answer: Vab = 409.1 mV
Wheatstone bridge: two voltage dividers share one supply Vs. The left divider R1–R2 gives Va = Vs·R2/(R1 + R2) and the right divider R3–R4 gives Vb = Vs·R4/(R3 + R4); the output is Vab = Va − Vb. The bridge is balanced (Vab = 0, no current in the meter) when R1/R2 = R3/R4, that is R1·R4 = R2·R3. Seen from the output, the bridge is a Thévenin source with Rth = R1∥R2 + R3∥R4.
The linear estimate Vs·ΔR/(4R) = 450 mV is 10 % too high: a quarter bridge is not quite linear.
Answer: Vo = 3.975 mV
Strain gauges: a gauge changes its resistance with strain: ΔR/R = GF·ε, where the gauge factor GF is about 2 for metal foil. In a bridge with all arms equal to R, one active gauge (quarter bridge) gives Vo ≈ Vs·GF·ε/4. Two active gauges with opposite strains in adjacent arms (half bridge) give Vs·GF·ε/2, and four (full bridge) give Vs·GF·ε. Stress follows from Hooke’s law, σ = E·ε. Load cells state their sensitivity in mV/V: the output at full load per volt of excitation.
Answer: AB = 11, CD = 00
Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.
A 1 with a neighbouring 1 can join a group of two or more.
Answer: Q = 8
Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.
Every 11 pulses bring the counter back to the same state.
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