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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 with four 500 Ω arms is supplied with 15 V. When the sensor R₂ changes, the output reads Vab = 357.1 mV. Find the exact change ΔR of R₂.
Vs = 15 VR = 500 ΩVab = 357.1 mV
ΔR = Ω
2.Wheatstone bridge & strain gaugesHard
A bridge with 120 Ω arms is excited with 10 V and has four active gauges, two in tension and two in compression (full bridge). The gauge factor is 2 and the strain is 250 µε. Find the bridge output Vo in mV.
Vs = 10 VGF = 2ε = 250 µε
Vo = mV
3.Data acquisition & samplingMedium
A 3 kHz sine wave is sampled at 4 kHz without an anti-aliasing filter. At which frequency fa does it appear in the samples? Give it in Hz.
f = 3 kHzfs = 4 kHz
fa = Hz
4.Data acquisition & samplingHard
A temperature sensor gives 10 mV/°C (0 V at 0 °C). Its signal is amplified 2 times and fed to an 8-bit ADC with a range of 0 to 5 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.
S = 10 mV/°CG = 2N = 8VFSR = 5 V
ΔTLSB = °C
5.Op-amp circuitsMedium
The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.
Vin = 1.5 VR1 = 22 kΩRf = 100 kΩ
Vout = V
6.Op-amp circuitsHard
The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage Vout.
V1 = 1 VV2 = −1.7 VR1 = 2 kΩR2 = 1 kΩRf = 10 kΩ
Vout = V
7.Series & parallel resistorsMedium
Find the current IR1 delivered by the source, which flows through R1.
Find the Thévenin resistance Rth seen from terminals a–b.
V1 = 6 VR1 = 8 ΩR2 = 2 ΩR3 = 15 Ω
Rth = Ω
10.Thévenin & NortonHard
Find the Thévenin voltage Vth seen from terminals a–b.
V1 = 30 VIs = 1.5 AR1 = 5 ΩR2 = 20 ΩR3 = 2 Ω
Vth = V
3Solutions
1.Wheatstone bridge & strain gauges
Answer: ΔR = 50 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.
Node voltage
Va = 7.5 + 0.3571 = 7.857 V
Solve the divider
R₂ = 500·7.857/(15 − 7.857) = 550 Ω
ΔR = 550 − 500 = 50 Ω
Answer
ΔR = 50 Ω
2.Wheatstone bridge & strain gauges
Answer: Vo = 5 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 = 2·250·10⁻⁶ = 0.0005
Bridge output
Vo = Vs·GF·ε = 10·0.0005 = 5 mV
Answer
Vo = 5 mV
3.Data acquisition & sampling
Answer: fa = 1000 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.
Nyquist frequency
fs/2 = 2 kHz < f = 3 kHz
Nearest multiple of the sampling rate
k = round(3000/4000) = 1
Alias frequency
fa = |3000 − 1·4000| = 1000 Hz
Answer
fa = 1000 Hz
The dashed curve in the figure is the alias: it passes through exactly the same samples.
4.Data acquisition & sampling
Answer: ΔTLSB = 0.9766 °C
ADC resolution: an N-bit converter divides its full-scale range VFSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = VFSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊Vin/LSB⌋; a rounding ADC is never more than ½ LSB away from the true value, its quantisation error. Every extra bit halves the LSB and adds about 6 dB to the ideal signal-to-quantisation-noise ratio SQNR = 6.02·N + 1.76 dB for a full-scale sine. To reach a resolution r over a span, choose the smallest N with 2ᴺ ≥ span/r.
Step size
1 LSB = 5000/256 = 19.53 mV
Sensitivity at the ADC
10·2 = 20 mV/°C
Temperature resolution
ΔTLSB = 19.53/20 = 0.9766 °C
Answer
ΔTLSB = 0.9766 °C
The ADC reaches full scale at 250 °C.
5.Op-amp circuits
Answer: Vout = 8.318 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₊ = 1.5 V
I = V₋/R1 = 1.5/22 = 0.06818 mA
Output voltage
The current through R₁ also flows through Rf.
Vout = V₋ + I·Rf = 1.5 + 0.06818·100 = 8.318 V
Answer
Vout = 8.318 V
6.Op-amp circuits
Answer: Vout = 12 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 = 1/2 = 0.5 mA
I2 = V2/R2 = −1.7/1 = −1.7 mA
If = I1 + I2 = −1.2 mA
Output voltage
Vout = −Rf·If = −10·(−1.2) = 12 V
Answer
Vout = 12 V
7.Series & parallel resistors
Answer: IR1 = 1.759 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 = 6 + 8 = 14 Ω
R2 ∥ 14 = 5·14/(5+14) = 3.684 Ω
Req = R1 + 3.684 = 2 + 3.684 = 5.684 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 = 10/5.684 = 1.759 A
Answer
IR1 = 1.759 A
8.Series & parallel resistors
Answer: IR2 = 0.7096 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 = 5 + 6 = 11 Ω
R4 ∥ 11 = 3.882 Ω
R3 + 3.882 = 11.88 Ω
R2 ∥ 11.88 = 1.712 Ω
Req = R1 + 1.712 = 20 + 1.712 = 21.71 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 = 18/21.71 = 0.829 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 = 18 − 0.829·20 = 1.419 V
IR2 = VA/R2 = 1.419/2 = 0.7096 A
Answer
IR2 = 0.7096 A
9.Thévenin & Norton
Answer: Rth = 16.6 Ω
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 = 8·2/(8+2) = 1.6 Ω
Rth = (R1 ∥ R2) + R3 = 1.6 + 15 = 16.6 Ω
10.Thévenin & Norton
Answer: Vth = 33 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 = 5·20/(5+20) = 4 Ω
Rth = (R1 ∥ R2) + R3 = 4 + 2 = 6 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) = 24 + 1.5·4 = 30 V