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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 galvanometer with a resistance of 20 Ω is connected between a and b of the bridge shown (Vs = 12 V). Find the galvanometer current Ig from a to b, in µA.
A quarter bridge (one active gauge, GF = 2.05) is excited with 10 V. The largest strain to be measured is 1200 µε, and it must give 10 V at the input of an analogue-to-digital converter. Find the amplifier gain G.
Vs = 10 VGF = 2.05εmax = 1200 µεVFS = 10 V
G =
3.Data acquisition & samplingMedium
A UART sends bytes at 115200 baud in the format 8N1: 8 data bits, no parity and 1 stop bit(s), plus the start bit. Find the largest number of data bytes per second it can transfer.
baud = 115200 Bd8N1
RB = B/s
4.Data acquisition & samplingHard
A microcontroller with a 20 MHz clock sets its UART baud rate as fclk/(16·D), with D a whole number rounded to the nearest value. The target is 19200 baud. Find the percentage error of the real baud rate (positive when it is too fast).
fclk = 20 MHztarget = 19200 Bd
e = %
5.Op-amp circuitsMedium
Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 51.
R1 = 22 kΩAv = 51
Rf = kΩ
6.Op-amp circuitsHard
The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±14 V. The capacitor is uncharged when a constant input Vin = −2.4 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.
Vin = −2.4 VR = 22 kΩC = 4.7 µFVsat = ±14 V
t = ms
7.Series & parallel resistorsMedium
Find the current IR1 delivered by the source, which flows through R1.
V1 = 6 VR1 = 8 ΩR2 = 8 ΩR3 = 15 ΩR4 = 4 Ω
IR1 = A
8.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 24 VR1 = 6 ΩR2 = 3 ΩR3 = 12 Ω
IN = A
10.Thévenin & NortonHard
Find the Thévenin voltage Vth seen from terminals a–b.
V1 = 24 VIs = 1.5 AR1 = 10 ΩR2 = 5 ΩR3 = 25 Ω
Vth = V
3Solutions
1.Wheatstone bridge & strain gauges
Answer: Ig = 46287 µA
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.
Thévenin voltage
Vth = 12·(800/920 − 80/680) = 9.023 V
Thévenin resistance
Rth = R₁∥R₂ + R₃∥R₄ = 174.9 Ω
Galvanometer current
Ig = 9.023/(174.9 + 20) = 46287 µA
Answer
Ig = 46287 µA
2.Wheatstone bridge & strain gauges
Answer: G = 1626
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.
Bridge output at full scale
Vo = 6.15 mV
Gain
G = 10/0.00615 = 1626
Answer
G = 1626
3.Data acquisition & sampling
Answer: RB = 11520 B/s
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.
Bits per frame
1 + 8 + 0 + 1 = 10
Byte rate
115200/10 = 11520 B/s
Answer
RB = 11520 B/s
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 = 20·10⁶/(16·19200) = 65.1 → 65
Real baud rate
20·10⁶/(16·65) = 19231 Bd
Error
(19231 − 19200)/19200·100 = 0.1603 %
Answer
e = 0.1603 %
An error below about 2 % is usually acceptable.
5.Op-amp circuits
Answer: Rf = 1100 kΩ
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 of a non-inverting amplifier
V₋ = V₊ = Vin, and the current through R₁ also flows through Rf, so Vout = Vin·(1 + Rf/R₁).
Av = 1 + Rf/R1
Rf = (51 − 1)·22 = 1100 kΩ
Answer
Rf = 1100 kΩ
6.Op-amp circuits
Answer: t = 603.2 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.
|Vin|·t/(RC) = Vsat → t = Vsat·RC/|Vin| = 14·103.4/2.4 = 603.2 ms
Answer
t = 603.2 ms
7.Series & parallel resistors
Answer: IR1 = 0.4402 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 = 15 + 4 = 19 Ω
R2 ∥ 19 = 8·19/(8+19) = 5.63 Ω
Req = R1 + 5.63 = 8 + 5.63 = 13.63 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 = 6/13.63 = 0.4402 A
Answer
IR1 = 0.4402 A
8.Series & parallel resistors
Answer: Req = 19.4 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 = 25 + 15 = 40 Ω
R4 ∥ 40 = 6.667 Ω
R3 + 6.667 = 36.67 Ω
R2 ∥ 36.67 = 4.4 Ω
Req = R1 + 4.4 = 15 + 4.4 = 19.4 Ω
Answer
Req = 19.4 Ω
9.Thévenin & Norton
Answer: IN = 0.5714 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 = 6·3/(6+3) = 2 Ω
Rth = (R1 ∥ R2) + R3 = 2 + 12 = 14 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) = 24·3/(6+3) = 8 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 = 8/14 = 0.5714 A
10.Thévenin & Norton
Answer: Vth = 50.5 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 = 10·5/(10+5) = 3.333 Ω
Rth = (R1 ∥ R2) + R3 = 3.333 + 25 = 28.33 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) = 8 + 1.5·3.333 = 13 V