Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
3 green LEDs (forward voltage 2.2 V each) are connected in series with a resistor to a 12 V supply. Find the resistance R that sets the current to 15 mA.
An NPN transistor with β = 50 switches a 220 Ω load resistor connected between VCC = 24 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. Find the largest base resistor RB, in kΩ, that still drives the transistor just into saturation.
Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 2.
The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±14 V. The capacitor is uncharged when a constant input Vin = 1.4 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.
The inductor carries no current before the switch closes at t = 0. Find the current i at t = 5 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 19.2 V? Find t.
Find the current IR1 delivered by the source, which flows through R1.
Find the current IR2 through R2.
An ideal boost converter turns 9 V into 18 V at 50 kHz with L = 33 µH. Find the peak-to-peak inductor ripple current ΔIL.
A buck converter turns 48 V into 19.2 V at 250 kHz with L = 68 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
A galvanometer with a resistance of 50 Ω is connected between a and b of the bridge shown (Vs = 9 V). Find the galvanometer current Ig from a to b, in µA.
A quarter bridge (one active gauge, GF = 2.12) is excited with 3 V. The largest strain to be measured is 750 µε, and it must give 5 V at the input of an analogue-to-digital converter. Find the amplifier gain G.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 6, 8, 9, 10, 12, 14, 15). How many product terms N does it have?
A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 5 clock edges.
Answer: R = 360 Ω
LED and diode circuits: a forward-biased LED drops roughly its forward voltage VF (about 2 V for red, 3 V for blue or white), whatever the current. A series resistor sets the current: R = (Vs − n·VF)/I for n LEDs in series. With a more exact diode model V = V₀ + I·rd, the operating point is where the diode line meets the load line I = (Vs − V)/R, so I = (Vs − V₀)/(R + rd).
The resistor dissipates 81 mW.
Answer: RB = 1.202 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.
Any smaller base resistor also saturates the transistor; in practice you pick one a few times smaller.
Answer: Rf = 10 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: t = 220 ms
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.
kΩ times µF gives milliseconds.
Answer: i = 0.1318 A
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 = 20.16 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 = 0.9281 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: IR2 = 0.1353 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: ΔIL = 2.727 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.
While S is on, the inductor sits directly across Vin.
Answer: Io,min = 0.3388 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: Ig = −34706 µA
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.
Answer: G = 4193
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: N = 5
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.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Answer: Q = 4
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.
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