Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
A 5.1 V Zener regulator is fed from 24 V through 680 Ω and supplies a 2200 Ω load. Find the Zener current IZ.
An NPN transistor with β = 50 switches a 330 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 = 3.3 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.
Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 11.
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 = 1.2 V is applied at t = 0. Find the output voltage Vout at t = 20 ms.
The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 80% of its final value? Find t.
Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. How long does it take for vC to reach 9.6 V? Find t.
Find the current IR2 through R2.
Find the current IR2 through R2.
A buck converter turns 48 V into 19.2 V for a 20 Ω load. It switches at 250 kHz with L = 330 µH. Find the peak inductor current Imax.
An ideal buck converter turns 24 V into 9.6 V for a 20 Ω load at 50 kHz. Find the smallest inductance Lcrit for continuous conduction.
A bridge has four 470 Ω arms and Vs = 6 V. The sensor R₂ increases by ΔR = 94 Ω. Find the exact output Vab in mV.
A quarter bridge (one active gauge, GF = 2) is excited with 2.5 V. The largest strain to be measured is 400 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 product-of-sums (POS) form of F(A, B, C, D) = Σm(7, 9, 10, 12, 14). How many sum terms (factors) N does it have?
Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 5, 8, 10, 11, 12) + d(9, 14, 15), where d marks the don't-care conditions. How many product terms N does it have?
Answer: IZ = 25.48 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: IC = 39.39 mA
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.
β·IB is smaller than IC(sat), so the transistor stays active.
The transistor is active, so the load is not fully switched on.
Answer: Rf = 470 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 = −5.106 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.
kΩ times µF gives milliseconds.
The ideal value lies inside ±Vsat, so the integrator has not clipped yet.
Answer: t = 0.7082 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.
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
Answer: t = 37.09 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: IR2 = 1.29 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: IR2 = 0.09703 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: Imax = 1.03 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: Lcrit = 120 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.
Set the average inductor current equal to half the ripple current and solve for L.
Answer: Vab = 272.7 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) = 300 mV is 10 % too high: a quarter bridge is not quite linear.
Answer: G = 10000
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.
Another grouping of the same size exists; it has the same number of terms and literals.
Answer: N = 3
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.
The largest groups that can be formed from 1s and Xs.
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