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1Formulas and methods
Wheatstone bridge & strain gauges
Divider voltages
Va = VsR2/(R1 + R2), Vb = VsR4/(R3 + R4)
Output
Vab = Va − Vb
Balance
R1/R2 = R3/R4
Slide-wire bridge
R/Rx = l/(100 − l)
Thévenin resistance
Rth = R1∥R2 + R3∥R4
Gauge factor
ΔR/R = GF·ε
Bridge output
quarter Vs·GF·ε/4, half Vs·GF·ε/2, full Vs·GF·ε
Small change, one arm
Vab ≈ Vs·ΔR/(4R)
Stress
σ = E·ε
Method
Treat the bridge as two voltage dividers across the same supply.
For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
For strain gauges, convert strain to ΔR/R = GF·ε and use Vs·GF·ε/4, /2 or ×1 for a quarter, half or full bridge.
Data acquisition & sampling
Nyquist criterion
fs > 2·fmax, fN = fs/2
Alias frequency
fa = |f − k·fs|, k = round(f/fs)
Step size
1 LSB = VFSR/2ᴺ
Quantisation error (rounding)
±½ LSB
Truncating ADC code
D = ⌊Vin/LSB⌋
Bits for a resolution r
2ᴺ ≥ span/r
Ideal SQNR, full-scale sine
SQNR = 6.02·N + 1.76 dB
Data rate
channels × fs × N bit/s
Multiplexed ADC
fs = 1/(n·(tc + tset))
UART
Tb = 1/baud; 8N1 = 10 bits per byte; bytes/s = baud/bits per frame
Baud rate divider
baud = fclk/(16·D)
Method
Write down what is given: sampling rate, signal frequencies, number of bits, input range, baud rate.
For sampling questions, compare each frequency with fs/2; a higher frequency appears at fa = |f − k·fs| with the nearest multiple of fs.
For resolution questions, find 1 LSB = VFSR/2ᴺ, or the smallest N with 2ᴺ ≥ span/resolution.
For data rates and UART links, count everything that is sent: channels, samples per second, bits or whole bytes per sample, and the start, parity and stop bits of every frame.
Op-amp circuits
Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0
Series & parallel resistors
Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R
Method
Start at the end of the network furthest from the terminals.
Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
Redraw the circuit after each step.
Repeat until one resistor is left between the terminals.
Thévenin & Norton
Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)
Method
Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is Vth.
Switch off all independent sources and find the resistance seen from a–b: this is Rth.
Put the load back on the equivalent circuit: Vth in series with Rth.
For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.
2Practice problems
1.Wheatstone bridge & strain gaugesMedium
A bridge has four 330 Ω arms and Vs = 12 V. The sensor R₂ increases by ΔR = 0.825 Ω. Using the small-change approximation, find the output Vab in mV.
Vs = 12 VR = 330 ΩΔR = 0.825 Ω
Vab = mV
2.Wheatstone bridge & strain gaugesHard
A bridge with 120 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 2.1 and the strain is 750 µε. Find the bridge output Vo in mV.
Vs = 3 VGF = 2.1ε = 750 µε
Vo = mV
3.Data acquisition & samplingMedium
A recorder samples 2 channels at 50 kHz each with a 10-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 5 min recording, in MB (1 MB = 10⁶ bytes).
n = 2fs = 50 kHzN = 10t = 5 min
M = MB
4.Data acquisition & samplingHard
A 1.44 kHz square wave is sampled at 8 kHz with no anti-aliasing filter. A square wave contains the odd harmonics 3f₁, 5f₁, 7f₁, … At which frequency fa does its harmonic 5f₁ appear in the sampled data? Give it in Hz.
f₁ = 1.44 kHzfs = 8 kHzharmonic: 5f₁
fa = Hz
5.Op-amp circuitsMedium
The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.
Vin = −0.5 VR1 = 2 kΩRf = 4.7 kΩ
Vout = V
6.Op-amp circuitsHard
The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±12 V. The capacitor is uncharged when a constant input Vin = 0.7 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.
Vin = 0.7 VR = 100 kΩC = 0.1 µFVsat = ±12 V
t = ms
7.Series & parallel resistorsMedium
Find the current IR2 through R2.
V1 = 24 VR1 = 4 ΩR2 = 15 ΩR3 = 8 ΩR4 = 6 Ω
IR2 = A
8.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
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.
Relative change
ΔR/R = 0.825/330 = 0.0025
Linearised output
Vab ≈ 12·0.0025/4 = 7.5 mV
Answer
Vab = 7.5 mV
The exact value is 7.491 mV; the approximation is good while ΔR/R is small.
2.Wheatstone bridge & strain gauges
Answer: Vo = 2.363 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.
Relative resistance change
ΔR/R = 2.1·750·10⁻⁶ = 0.001575
Bridge output
Vo = Vs·GF·ε/2 = 3·0.001575/2 = 2.363 mV
Answer
Vo = 2.363 mV
3.Data acquisition & sampling
Answer: M = 60 MB
Data rates: every channel delivers fs samples per second of N bits, so the raw rate is channels × fs × N bit/s. In memory a sample takes whole bytes (a 12-bit sample is stored in 2 bytes), so the memory for a recording is channels × fs × bytes per sample × time. A multiplexed ADC converts the channels one after another: each conversion (with settling) takes t, so one round of n channels takes n·t and each channel is sampled at 1/(n·t).
Bytes per sample
⌈10/8⌉ = 2 B
Byte rate
2·50000·2 = 200000 B/s
Memory
M = 200000·300 = 6 × 10⁷ B = 60 MB
Answer
M = 60 MB
4.Data acquisition & sampling
Answer: fa = 800 Hz
Sampling: an ADC measures the signal fs times per second. By the Nyquist–Shannon theorem a frequency f is captured correctly only while f < fs/2, the Nyquist (folding) frequency, so the sampling rate must exceed twice the highest signal frequency. A higher frequency is not lost: it shows up as an alias at fa = |f − k·fs|, with the whole number k that brings the result between 0 and fs/2. The samples cannot tell the two apart, so an analogue anti-aliasing filter must remove everything above fs/2 before the ADC.
Harmonic frequency
5f₁ = 5·1.44 kHz = 7.2 kHz
Nearest multiple of the sampling rate
m = round(0.9) = 1
Alias frequency
fa = |7200 − 1·8000| = 800 Hz
Answer
fa = 800 Hz
The alias lands inside the signal band, where no digital filter can remove it. Only an analogue filter before the ADC helps.
5.Op-amp circuits
Answer: Vout = −1.675 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.
Inverting input follows the input
V₋ = V₊ = −0.5 V
I = V₋/R1 = −0.5/2 = −0.25 mA
Output voltage
The current through R₁ also flows through Rf.
Vout = V₋ + I·Rf = −0.5 + (−0.25)·4.7 = −1.675 V
Answer
Vout = −1.675 V
6.Op-amp circuits
Answer: t = 171.4 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.
Input current and ramp rate
kΩ times µF gives milliseconds.
RC = 100 kΩ · 0.1 µF = 10 ms
ramp rate = −Vin/(RC) = −0.7/10 = −0.07 V/ms
Time to saturation
|Vin|·t/(RC) = Vsat → t = Vsat·RC/|Vin| = 12·10/0.7 = 171.4 ms
Answer
t = 171.4 ms
7.Series & parallel resistors
Answer: IR2 = 1.031 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.
Combine resistors from the far end
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.
R3 + R4 = 8 + 6 = 14 Ω
R2 ∥ 14 = 15·14/(15+14) = 7.241 Ω
Req = R1 + 7.241 = 4 + 7.241 = 11.24 Ω
Current from the source
The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.
IR1 = V1/Req = 24/11.24 = 2.135 A
Current through R₂
Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.
VA = V1 − IR1·R1 = 24 − 2.135·4 = 15.46 V
IR2 = VA/R2 = 15.46/15 = 1.031 A
Answer
IR2 = 1.031 A
8.Series & parallel resistors
Answer: Req = 13.78 Ω
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.
Combine resistors from the far end
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.
R5 + R6 = 20 + 12 = 32 Ω
R4 ∥ 32 = 14.04 Ω
R3 + 14.04 = 16.04 Ω
R2 ∥ 16.04 = 1.778 Ω
Req = R1 + 1.778 = 12 + 1.778 = 13.78 Ω
Answer
Req = 13.78 Ω
9.Thévenin & Norton
Answer: IN = 1.034 A
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Find Rth with every source switched off
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
R1 ∥ R2 = 2·8/(2+8) = 1.6 Ω
Rth = (R1 ∥ R2) + R3 = 1.6 + 10 = 11.6 Ω
Find Vth, the open-circuit voltage
With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.
Vth = V1·R2/(R1+R2) = 15·8/(2+8) = 12 V
Norton equivalent
A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.
IN = Vth/Rth = 12/11.6 = 1.034 A
10.Thévenin & Norton
Answer: IL = 0.5234 A
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Find Rth with every source switched off
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
R1 ∥ R2 = 15·12/(15+12) = 6.667 Ω
Rth = (R1 ∥ R2) + R3 = 6.667 + 30 = 36.67 Ω
Find Vth, the open-circuit voltage
With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.
VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 4 + 0.5·6.667 = 7.333 V
Vth = VB = VA + Is·R3 = 7.333 + 0.5·30 = 22.33 V
Connect the load to the Thévenin equivalent
The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.
IL = Vth/(Rth + RL) = 22.33/(36.67 + 6) = 0.5234 A