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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 = 5 V. The sensor R₂ increases by ΔR = 2.64 Ω. Using the small-change approximation, find the output Vab in mV.
Vs = 5 VR = 330 ΩΔR = 2.64 Ω
Vab = mV
2.Wheatstone bridge & strain gaugesHard
A strain gauge (GF = 2.05) on a member with E = 210 GPa forms a quarter bridge excited with 5 V. The output reads 0.513 mV. Find the axial stress σ in MPa.
Vs = 5 VGF = 2.05Vo = 0.513 mVE = 210 GPa
σ = MPa
3.Data acquisition & samplingMedium
A 92.5 kHz sine wave is sampled at 50 kHz without an anti-aliasing filter. At which frequency fa does it appear in the samples? Give it in kHz.
f = 92.5 kHzfs = 50 kHz
fa = kHz
4.Data acquisition & samplingHard
A microcontroller with a 10 MHz clock sets its UART baud rate as fclk/(16·D), with D a whole number rounded to the nearest value. The target is 9600 baud. Find the percentage error of the real baud rate (positive when it is too fast).
fclk = 10 MHztarget = 9600 Bd
e = %
5.Op-amp circuitsMedium
The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±14 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.
R1 = 2 kΩRf = 47 kΩVsat = ±14 V
Vin,max = V
6.Op-amp circuitsHard
The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage Vout.
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 = 2.64/330 = 0.008
Linearised output
Vab ≈ 5·0.008/4 = 10 mV
Answer
Vab = 10 mV
The exact value is 9.96 mV; the approximation is good while ΔR/R is small.
2.Wheatstone bridge & strain gauges
Answer: σ = 42.04 MPa
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.
Strain
ε = 4·0.000513/(5·2.05) = 0.0002002
Stress
σ = 210·10³ MPa·0.0002002 = 42.04 MPa
Answer
σ = 42.04 MPa
3.Data acquisition & sampling
Answer: fa = 7.5 kHz
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.
Nyquist frequency
fs/2 = 25 kHz < f = 92.5 kHz
Nearest multiple of the sampling rate
k = round(92.5/50) = 2
Alias frequency
fa = |92.5 − 2·50| = 7.5 kHz
Answer
fa = 7.5 kHz
The dashed curve in the figure is the alias: it passes through exactly the same samples.
4.Data acquisition & sampling
Answer: e = 0.1603 %
UART: an asynchronous serial link sends each byte as a frame: one start bit (low), the data bits, an optional parity bit and one or two stop bits (high). The baud rate is the number of bits per second, so one bit lasts Tb = 1/baud and a frame of n bits lasts n/baud. The common 8N1 format has 10 bits per byte. The baud generator divides the clock: baud = fclk/(16·D) with a whole number D, so the real baud rate differs slightly from the target; an error below about 2 % is usually acceptable.
Divider
D = 10·10⁶/(16·9600) = 65.1 → 65
Real baud rate
10·10⁶/(16·65) = 9615 Bd
Error
(9615 − 9600)/9600·100 = 0.1603 %
Answer
e = 0.1603 %
An error below about 2 % is usually acceptable.
5.Op-amp circuits
Answer: Vin,max = 0.5714 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.
Gain
Av = 1 + Rf/R1 = 1 + 47/2 = 24.5
Largest input before clipping
Vin,max = Vsat/|Av| = 14/24.5 = 0.5714 V
Answer
Vin,max = 0.5714 V
6.Op-amp circuits
Answer: Vout = −0.2851 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.
Input currents
V₋ = 0, so each input resistor has its full input voltage across it.
I1 = V1/R1 = −1.3/47 = −0.02766 mA
I2 = V2/R2 = 0.8/4.7 = 0.1702 mA
If = I1 + I2 = 0.1426 mA
Output voltage
Vout = −Rf·If = −2·0.1426 = −0.2851 V
Answer
Vout = −0.2851 V
7.Series & parallel resistors
Answer: IR1 = 0.8864 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 = 12 + 20 = 32 Ω
R2 ∥ 32 = 20·32/(20+32) = 12.31 Ω
Req = R1 + 12.31 = 8 + 12.31 = 20.31 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 = 18/20.31 = 0.8864 A
Answer
IR1 = 0.8864 A
8.Series & parallel resistors
Answer: IR2 = 1.268 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.
R5 + R6 = 25 + 12 = 37 Ω
R4 ∥ 37 = 9.061 Ω
R3 + 9.061 = 39.06 Ω
R2 ∥ 39.06 = 5.201 Ω
Req = R1 + 5.201 = 3 + 5.201 = 8.201 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 = 12/8.201 = 1.463 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 = 12 − 1.463·3 = 7.61 V
IR2 = VA/R2 = 7.61/6 = 1.268 A
Answer
IR2 = 1.268 A
9.Thévenin & Norton
Answer: Vth = 1.385 V
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 = 10·3/(10+3) = 2.308 Ω
Rth = (R1 ∥ R2) + R3 = 2.308 + 5 = 7.308 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) = 6·3/(10+3) = 1.385 V
10.Thévenin & Norton
Answer: PL = 124.5 W
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 = 15·25/(15+25) = 9.375 Ω
Rth = (R1 ∥ R2) + R3 = 9.375 + 8 = 17.38 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) = 15 + 6·9.375 = 71.25 V
Vth = VB = VA + Is·R3 = 71.25 + 6·8 = 119.3 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.