Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
2 green LEDs (forward voltage 2.2 V each) are connected in series with a resistor to a 9 V supply. Find the resistance R that sets the current to 10 mA.
An NPN transistor with β = 150 switches a 100 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 220 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 inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.
The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the current i2 through R2 at t = 100 ms.
Find the current IR2 through R2.
Find the current IR1 delivered by the source, which flows through R1.
An ideal buck converter turns 9 V into 2.7 V at 20 kHz with L = 47 µH. Find the peak-to-peak inductor ripple current ΔIL.
An ideal boost converter turns 36 V into 90 V for a 25 Ω load at 20 kHz. Find the smallest inductance Lcrit for continuous conduction.
Find the Thévenin resistance Rth of the bridge shown, seen from the output terminals a and b.
A bridge with 1000 Ω arms is excited with 3 V and has two active gauges in adjacent arms, one in tension and one in compression (half bridge). The gauge factor is 2.12 and the strain is 1200 µε. Find the bridge output Vo in mV.
A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = C, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.
A 4-bit shift register Q₃Q₂Q₁Q₀ holds 1010. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀, Q₀ is lost) and the serial input enters Q₃. The serial input is 1, 0, 0, 1, 0, 1 at the 6 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.
Answer: R = 460 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 46 mW.
Answer: IC = 2.932 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: Vout = 8.55 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.
The current through R₁ also flows through Rf.
Answer: Vout = −2.604 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: i = 0.2586 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: i2 = 0.3693 mA
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.
In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.
Answer: IR2 = 0.5882 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: IR1 = 0.2777 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: ΔIL = 2.011 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 between Vin and Vo.
Answer: Lcrit = 60 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: Rth = 54.09 Ω
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 = 3.816 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
Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.
Answer: Q = 10
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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