Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
An NPN transistor with β = 150 is biased by a base resistor RB = 820 kΩ from the supply VCC = 9 V. The collector resistor is RC = 1.2 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector–emitter voltage VCE.
An NPN transistor with β = 80 switches a 470 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 = 2.2 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.
The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.
The summing amplifier uses an ideal op-amp with ±15 V supplies. 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 = 40 ms.
A timer starts charging an uncharged capacitor through a resistor from a 15 V supply at t = 0. It must trigger when the capacitor voltage reaches 12 V, exactly 50 ms after the start. With C = 22 µF, find the resistance R needed.
Find the current IR1 delivered by the source, which flows through R1.
Find the current IR2 through R2.
A buck converter turns 15 V into 4.5 V for a 25 Ω load. It switches at 100 kHz with L = 150 µH. Find the peak inductor current Imax.
An ideal boost converter turns 9 V into 12 V for a 20 Ω load at 40 kHz. Find the smallest inductance Lcrit for continuous conduction.
A 1000 Ω strain gauge is one arm (R₂) of a bridge with four equal 1000 Ω arms and an excitation of 3 V. Find the power Pg dissipated in the gauge, in mW. Self-heating limits the excitation.
A load cell (a full bridge of strain gauges) has a rated output of 1 mV/V at its rated load of 500 kg. It is excited with 12 V. Find its output Vo under a load of 250 kg.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 8, 9, 10, 11, 12, 13). How many product terms N does it have?
Find the minimal SOP form of F(A, B, C, D) = Σm(2, 4, 7, 9, 14, 15) + d(0, 1, 11), where d marks the don't-care conditions. How many product terms N does it have?
Answer: VCE = 7.178 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.
Answer: IC = 50.64 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 larger than IC(sat), so the transistor saturates.
The transistor is saturated.
Answer: Vout = −2.238 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₋ = V₊, and the current through R₁ also flows through Rf.
Answer: Vout = −0.9583 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: vC = 3.118 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: R = 1.412 kΩ
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.
kΩ times µF gives milliseconds.
Answer: IR1 = 1.412 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 = 1.18 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 = 0.285 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 = 35.16 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: Pg = 2.25 mW
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: Vo = 6 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: 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.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
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.
The largest groups that can be formed from 1s and Xs.
Another grouping of the same size exists; it has the same number of terms and literals.
ExamProblems · examproblems.com/courses/electronics-1