Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
An NPN transistor with β = 120 has voltage-divider bias from VCC = 20 V: R₁ = 56 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 1 kΩ and the emitter resistor is RE = 470 Ω. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector–emitter voltage VCE.
A 3.3 V Zener regulator must supply up to 50 mA to a load. The input can drop to 15 V, and the Zener needs at least 1 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.
The difference amplifier uses an ideal op-amp with ±15 V supplies, with R3 = R1 and R4 = R2. Find the output voltage Vout.
The op-amp is ideal and runs from ±15 V supplies. The resistor ratios are not matched. Find the output voltage Vout.
The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 50 ms.
Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. How long does it take for vC to reach 20 V? Find t.
Find the current IR1 delivered by the source, which flows through R1.
Find the current IR1 delivered by the source, which flows through R1.
An ideal buck converter turns 36 V into 21.6 V at a switching frequency of 200 kHz. Find the inductance L that keeps the peak-to-peak inductor ripple current at 0.5 A.
An ideal boost converter turns 48 V into 160 V for a 2 Ω load at 25 kHz. The output capacitor is 22 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
A bridge has four 350 Ω arms and Vs = 12 V. The sensor R₂ increases by ΔR = 0.7 Ω. Using the small-change approximation, find the output Vab in mV.
A bridge with 120 Ω arms is excited with 2.5 V and has one active gauge (quarter bridge). The gauge factor is 2 and the strain is 750 µε. Find the bridge output Vo in mV.
Simplify F = BC′ + A′BC′ + A′B′C′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?
A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 3. Which count Q (in decimal) does it show after 29 clock pulses?
Answer: VCE = 12.5 V
BJT bias point: in the active region VBE = 0.7 V, IC = β·IB and IE = (β + 1)·IB. Find IB from the base loop with KVL, then IC, then VCE from the collector–emitter loop. A voltage divider at the base is replaced by its Thévenin equivalent VTH = VCC·R₂/(R₁ + R₂), RTH = R₁∥R₂; an emitter resistor appears (β + 1) times larger in the base loop.
KVL around the base loop; the emitter current is (β + 1)·IB.
Answer: Rmax = 229.4 Ω
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: Vout = −4.255 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.
R₃ and R₄ carry the same current, so they form a voltage divider; feedback makes V₋ = V₊.
Answer: Vout = 2.349 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.
With V₂ = 0, V₁ sees an inverting amplifier. With V₁ = 0, V₊ is amplified by the non-inverting gain 1 + R₂/R₁.
Answer: vC = 1.113 V
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.
Short the source to find Rth; open the capacitor to find the voltage it will settle at.
Answer: t = 471.3 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: IR1 = 0.766 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.
Answer: IR1 = 0.3849 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.
Answer: L = 86.4 µH
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: ΔVo = 101818 mV
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: Vab = 6 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 exact value is 5.994 mV; the approximation is good while ΔR/R is small.
Answer: Vo = 0.9375 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: L = 4
Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.
Answer: Q = 2
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 10 pulses bring the counter back to the same state.
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