IN THE IN SYLLABUS

Noise, interference, shielding and grounding

Every measurement system has a floor set by random noise generated inside its own components (thermal, shot and flicker noise) and by interference coupled in from outside through conductive, capacitive and inductive paths. GATE IN questions ask you to compute rms noise voltages and currents, add uncorrelated sources correctly, use noise bandwidth, SNR and noise figure, and pick the right cure (shielding, twisting, guarding or grounding) for a given coupling mechanism.

2 previous-year questions · 2 solved · 6 practice questions

Noise and interference: sources and effects

Noise is a random signal generated inside a circuit by the physics of its own components: thermal agitation of charge carriers, the discrete nature of charge crossing a junction, and trapping and release of carriers at defects. Its instantaneous value cannot be predicted, only its statistics (mean-square value, spectrum). It can be reduced but never eliminated, and it sets the smallest signal a system can resolve.

Interference is an unwanted signal coupled in from another circuit or an outside source: 50 Hz mains wiring and its harmonics, motors and relays, switching regulators, digital clocks, radio transmitters, electrostatic discharge. It has an identifiable source, so in principle it can be removed by breaking the coupling path.

Every interference problem has three parts: a source, a coupling channel and a receptor (the victim). The channel is one of four kinds: conductive (a shared wire or common impedance), capacitive (electric field), inductive (magnetic field), or radiated (an electromagnetic wave, once dimensions are comparable with the wavelength). Suppression can act on any of the three parts: quieten the source, break the channel, or make the receptor less sensitive (lower impedance, differential input, narrower bandwidth).

	Noise (intrinsic)	Interference (extrinsic)
Origin	Inside the components: thermal, shot, flicker, burst, avalanche	Outside the circuit, or another part of it: mains, motors, relays, switching converters, digital logic, RF
Nature	Random; described by its rms value and power spectral density	Usually deterministic or periodic (hum at 50 Hz and harmonics, clock spikes)
Spectrum	Broadband (white) or rising at low frequency (1/f)	Concentrated at the source frequencies and their harmonics
Can it be removed?	No, only minimised (lower $R$ , $T$ or bandwidth; low-noise devices)	Yes, in principle, by attacking the source, the coupling path or the receptor
Typical cures	Component choice, bandwidth limiting, filtering, averaging	Shielding, twisting, guarding, grounding, filtering, isolation

Effects on electronic circuits

Lower signal-to-noise ratio: the smallest resolvable signal rises.
Measurement error in instrumentation: hum or drift superimposed on a slow sensor signal (thermocouple, strain gauge, pH electrode).
DC offsets: RF interference rectified by the nonlinear input stage of an op-amp or ADC shifts the DC output.
False triggering of comparators, counters and digital logic by spikes on signal or ground lines.
Timing jitter, and fewer effective bits from an ADC.

Other noise-like error sources met in instrumentation

Thermoelectric (thermocouple) EMFs at junctions of dissimilar metals, such as solder joints and connector pins at different temperatures.
Triboelectric noise: charge generated by friction between a cable's dielectric and its conductors when the cable flexes.
Electrochemical EMFs and leakage currents from contamination (flux residue, moisture) on a circuit board.
Microphonic and piezoelectric pickup from mechanical vibration of cables, capacitors and connectors.

Noise generated by components: thermal, shot and flicker

Thermal (Johnson) noise comes from the random thermal motion of charge carriers in any resistance above absolute zero. It needs no current: an open-circuited resistor on a shelf generates it. Its spectrum is flat (white) up to frequencies of the order of $k T / h$ (about 6 THz at 300 K), far beyond electronic frequencies; its amplitude distribution is Gaussian; and it depends only on $R$ , $T$ and the bandwidth $B$ , not on what the resistor is made of. Only the resistive (dissipative) part of an impedance generates thermal noise; an ideal capacitor or inductor generates none.

THERMAL NOISE VOLTAGE (THEVENIN MODEL)

\overline{v_{n}^{2}} = 4 k T R B, V_{n, rms} = 4 k T R B

Open-circuit noise of resistance $R$ at absolute temperature $T$ over noise bandwidth $B$ ; $k = 1.38 \times 1 0^{- 23} J/K$ . Model: a noiseless $R$ in series with this voltage source. One-sided spectral density $4 k T R$ in $V^{2} / Hz$ .

THERMAL NOISE CURRENT (NORTON MODEL)

\overline{i_{n}^{2}} = \frac{4 k T B}{R} = 4 k T GB

Short-circuit noise current: a noiseless $R$ in parallel with this current source. Use it for resistors in parallel or for a node driven by a current source.

THERMAL NOISE OF ANY PASSIVE NETWORK (NYQUIST)

\overline{v_{n}^{2}} = 4 k T \int_{f_{1}}^{f_{2}} Re [Z (f)] df

$Z (f)$ = impedance between two terminals of a passive network of resistors, capacitors and inductors, all at temperature $T$ and in thermal equilibrium. Only $Re [Z]$ contributes. It follows that a network of resistors at one temperature has exactly the noise of its Thevenin resistance.

AVAILABLE NOISE POWER

P_{n} = k T B

The maximum noise power a resistor can deliver, into a matched noiseless load $R_{L} = R$ (load voltage $4 k T R B /2$ ). It does not depend on $R$ . At $T_{0} = 290$ K, $k T$ corresponds to $- 174$ dBm/Hz.

NOISE ON A CAPACITOR (KT/C NOISE)

\overline{v_{n}^{2}} = \frac{k T}{C}

Total mean-square noise across $C$ , integrated over all frequencies, when $C$ is shunted by any resistor $R$ at temperature $T$ : the RC filter's noise bandwidth $1/ (4 R C)$ cancels $R$ . More generally it follows from equipartition, $\frac{1}{2} C \overline{v_{n}^{2}} = \frac{1}{2} k T$ , for a capacitor in any passive network in thermal equilibrium. It sets the noise floor of sample-and-hold and switched-capacitor circuits.

Shot noise arises because current is a stream of discrete charges crossing a potential barrier (a pn junction, the base-emitter junction of a BJT, a photodiode) at random, independent instants. It is present only when a DC current flows, it is white and Gaussian, and it does not depend directly on temperature. Current through an ordinary ohmic conductor, such as a metal-film resistor, shows negligible shot noise, because the motion of its carriers is correlated rather than a stream of independent barrier crossings.

SHOT NOISE CURRENT

\overline{i_{s h}^{2}} = 2 q I_{D C} B, I_{s h, rms} = 2 q I_{D C} B

$q = 1.6 \times 1 0^{- 19}$ C and $I_{D C}$ = DC current crossing the barrier. In a BJT the base and collector currents carry their own shot noise, $2 q I_{B} B$ and $2 q I_{C} B$ . Valid at frequencies well below the reciprocal of the carrier transit time.

Flicker (1/f) noise has a power spectral density that rises as frequency falls, roughly as $1/ f$ . It is linked to carriers being trapped and released at crystal defects and surface states, so it is large in MOSFETs and carbon-composition resistors and small in metal-film and wirewound resistors. In a device it grows with DC current, $\overline{i^{2}} = K_{1} I^{a} Δ f / f^{b}$ with $a$ roughly between 0.5 and 2 and $b \approx 1$ . For $b = 1$ its power in a band from $f_{L}$ to $f_{H}$ depends only on the ratio $f_{H} / f_{L}$ , so every decade of frequency carries the same flicker-noise power. The corner frequency $f_{c}$ is where the 1/f density equals the white density; below $f_{c}$ flicker noise dominates.

FLICKER NOISE IN A BAND

S (f) = \frac{K}{f} \Rightarrow \overline{v_{n}^{2}} = \int_{f_{L}}^{f_{H}} \frac{K}{f} df = K ln \frac{f _{H}}{f _{L}}

$K$ in $V^{2}$ (or $A^{2}$ for current noise). Equal power per decade: each decade adds $K ln 10 \approx 2.303 K$ .

WHITE PLUS 1/F NOISE WITH CORNER FREQUENCY

S (f) = e_{w}^{2} (1 + \frac{f _{c}}{f}) \Rightarrow \overline{v_{n}^{2}} = e_{w}^{2} [(f_{H} - f_{L}) + f_{c} ln \frac{f _{H}}{f _{L}}]

$e_{w}$ = white-noise density (for example in $nV / Hz$ ), $f_{c}$ = 1/f corner frequency, brick-wall band from $f_{L}$ to $f_{H}$ . The same form applies to current noise.

Two more intrinsic mechanisms are worth recognising. Burst (popcorn) noise is random switching of the output between two or more levels, caused by defects in a junction; its spectrum is flat at low frequency and falls as $1/ f^{2}$ above a characteristic frequency. Avalanche noise is produced in a reverse-biased junction in breakdown and is large, which is why a breakdown-diode voltage reference is usually followed by an RC filter or bypass capacitor.

	Thermal	Shot	Flicker (1/f)
Physical cause	Random thermal motion of carriers in a resistance	Discrete charges crossing a potential barrier at random	Carrier trapping at defects and surface states
Needs DC current?	No	Yes	Yes in devices; in resistors it is the current-dependent excess noise
Spectrum	White	White	PSD $\propto 1/ f$ ; equal power per decade
Mean-square value	$4 k T R B$ (voltage)	$2 q I_{D C} B$ (current)	$K ln (f_{H} / f_{L})$
Depends on	$R$ , $T$ , $B$	$I_{D C}$ , $B$	$I_{D C}$ , device technology, $f_{H} / f_{L}$
Where it dominates	Resistors: source and feedback resistances	Diodes, BJTs, photodiodes	Low frequencies (below $f_{c}$ ); MOSFETs, carbon resistors

BJT INPUT NOISE SOURCES (WHITE PART)

e_{n}^{2} \approx 4 k T (r_{b} + \frac{1}{2 g _{m}}), i_{n}^{2} \approx 2 q I_{B}

Densities per hertz ( $V^{2} / Hz$ , $A^{2} / Hz$ ). Hybrid-pi model with $g_{m} = I_{C} / V_{T}$ , $V_{T} = k T / q$ (about 25.9 mV at 300 K); $r_{b}$ = base spreading resistance, whose thermal noise gives $4 k T r_{b}$ . The collector shot noise $2 q I_{C}$ , referred to the input, gives $2 q I_{C} / g_{m}^{2} = 2 k T / g_{m}$ , and the base shot noise gives $i_{n}$ . Flicker noise and the correlation between $e_{n}$ and $i_{n}$ are neglected. Raising $I_{C}$ lowers $e_{n}$ but raises $i_{n}$ .

MOSFET CHANNEL THERMAL NOISE (SATURATION)

\overline{i_{d}^{2}} = 4 k T γ g_{m} B, e_{n}^{2} = \frac{i _{d}^{2}}{g _{m}^{2} B} = \frac{4 k T γ}{g _{m}}

$γ = 2/3$ for a long-channel MOSFET in saturation (larger in short-channel devices); $e_{n}^{2}$ is the gate-referred voltage density in $V^{2} / Hz$ . The gate current noise is negligible at low frequency, which is why MOS and JFET inputs suit high source resistances. Flicker noise, inversely proportional to gate area, adds a term that rises as $1/ f$ .

WORKED EXAMPLE · WORKED EXAMPLE: WHITE PLUS FLICKER NOISE

An amplifier's input-referred voltage noise has a white density of $10 nV / Hz$ and a 1/f corner frequency of 100 Hz. Find the total rms input noise over a brick-wall band from 0.1 Hz to 10 kHz, and how much of it is flicker noise.

Solution

Density: $S (f) = e_{w}^{2} (1 + f_{c} / f)$ with $e_{w}^{2} = (1 0^{- 8})^{2} = 1 0^{- 16} V^{2} / Hz$ and $f_{c} = 100$ Hz.
White part: $e_{w}^{2} (f_{H} - f_{L}) = 1 0^{- 16} \times (1 0^{4} - 0.1) \approx 1.000 \times 1 0^{- 12} V^{2}$ .
Flicker part: $e_{w}^{2} f_{c} ln (f_{H} / f_{L}) = 1 0^{- 16} \times 100 \times ln (1 0^{5}) = 1 0^{- 14} \times 11.513 = 1.151 \times 1 0^{- 13} V^{2}$ .
The two parts are uncorrelated, so add mean squares: $\overline{v_{n}^{2}} = 1.000 \times 1 0^{- 12} + 0.115 \times 1 0^{- 12} = 1.115 \times 1 0^{- 12} V^{2}$ .
rms: $V_{n} = 1.115 \times 1 0^{- 12} = 1.056 μ V$ . Flicker alone is $1.151 \times 1 0^{- 13} = 0.339 μ V$ rms, shared equally by the five decades: $0.152 μ V$ rms per decade, and $0.152 5 = 0.339$ .

Answer: Total noise is about $1.06 μ V$ rms, of which flicker noise is about $0.34 μ V$ rms. Extending the band down to 0.01 Hz would add one more equal decade of flicker noise.

EXAM TIP

Thermal noise depends only on $R$ , $T$ and $B$ . A carbon-composition resistor and a metal-film resistor of the same value have identical thermal noise; the carbon one is noisier only through its current-dependent excess (flicker) noise.

EXAM TIP

Handy number: at 300 K, $4 k T R$ is about $4.07 nV / Hz$ for 1 k $Ω$ and scales as $R$ : about $12.9 nV / Hz$ for 10 k $Ω$ and $128.7 nV / Hz$ for 1 M $Ω$ .

EXAM TIP

Scaling traps: rms noise $\propto B$ , so four times the bandwidth doubles the rms noise. Thermal noise $\propto T$ and is independent of current; shot noise $\propto I_{D C}$ and does not depend directly on $T$ .

EXAM TIP

A forward-biased diode with small-signal resistance $r_{d} = V_{T} / I_{D}$ (where $V_{T} = k T / q$ ) has a shot-noise voltage $\overline{v^{2}} = 2 q I_{D} B r_{d}^{2} = 2 k T r_{d} B$ , half the mean-square thermal noise of a physical resistor of value $r_{d}$ . $r_{d}$ is not a physical resistor, so do not add $4 k T r_{d} B$ on top.

Adding noise sources, noise bandwidth, SNR and noise figure

Independent (uncorrelated) noise sources add in power, that is, in mean-square value, never in rms value. Refer every source to the same point (usually the input, called input-referred) and to the same kind (all voltages in series, or all currents in parallel), then square, add and take the square root. Only fully correlated sources, such as the same noise reaching a node by two paths, add as amplitudes.

UNCORRELATED SOURCES

V_{n, tot} = V_{n 1}^{2} + V_{n 2}^{2} + \dots + V_{nk}^{2}

Rms values of independent sources in the same bandwidth. It follows that resistors at the same temperature have, in series, the noise of $R_{1} + R_{2}$ and, in parallel, the noise of $R_{1} ∥ R_{2}$ . Series resistors at different temperatures give $\overline{v_{n}^{2}} = 4 k B (R_{1} T_{1} + R_{2} T_{2})$ .

PARTIALLY CORRELATED SOURCES

\overline{v^{2}} = \overline{v_{1}^{2}} + \overline{v_{2}^{2}} + 2 C \overline{v_{1}^{2}} \overline{v_{2}^{2}}, - 1 \leq C \leq 1

$C$ = correlation coefficient. $C = 0$ : powers add; $C = 1$ : rms values add; $C = - 1$ : rms values subtract.

Real filters have no brick-wall edges, so noise above the 3-dB frequency still gets through. The equivalent noise bandwidth $B_{N}$ is the width of an ideal rectangular filter, with the same peak power gain, that passes the same total white-noise power. Multiply a white density by $B_{N}$ , not by the 3-dB bandwidth.

EQUIVALENT NOISE BANDWIDTH

B_{N} = \frac{1}{∣ H _{m a x} ∣ ^{2}} \int_{0}^{\infty} ∣ H (f) ∣^{2} df

$H (f)$ = voltage transfer function and $∣ H_{m a x} ∣$ its peak magnitude (the DC gain, for a low-pass filter). For a white input density $e_{w}$ , the output noise is $∣ H_{m a x} ∣ e_{w} B_{N}$ .

FIRST-ORDER (SINGLE RC) LOW-PASS

B_{N} = \frac{π}{2} f_{3 dB} = \frac{1}{4 R C}, f_{3 dB} = \frac{1}{2 π R C}

About 1.57 times the 3-dB bandwidth. For an $n$ -th order Butterworth low-pass, $B_{N} = f_{3 dB} \frac{π /2 n}{sin ( π /2 n )}$ : about $1.11 f_{3 dB}$ for $n = 2$ , approaching $f_{3 dB}$ as $n$ grows. A single-tuned (parallel RLC) band-pass has the same ratio as the single RC: $B_{N} = (π /2) B_{3 dB}$ , with $B_{3 dB} = f_{0} / Q$ .

SIGNAL-TO-NOISE RATIO

SNR = \frac{P _{s}}{P _{n}}, SNR_{dB} = 10 lo g_{10} \frac{P _{s}}{P _{n}} = 20 lo g_{10} \frac{V _{s, rms}}{V _{n, rms}}

The voltage form needs both voltages across the same resistance. Use the rms signal, not the peak.

NOISE FACTOR, NOISE FIGURE AND NOISE TEMPERATURE

F = \frac{SNR _{in}}{SNR _{out}} = 1 + \frac{T _{e}}{T _{0}}, N F = 10 lo g_{10} F dB

Defined with the source resistance at the standard temperature $T_{0} = 290$ K. $F \geq 1$ ( $N F \geq 0$ dB), and a noiseless amplifier has $F = 1$ . Equivalently, $F$ = total output noise power divided by the output noise power due to the source alone. $T_{e} = (F - 1) T_{0}$ is the equivalent input noise temperature. In decibels, $N F = SNR_{in,dB} - SNR_{out,dB}$ .

FRIIS FORMULA FOR CASCADED STAGES

F = F_{1} + \frac{F _{2} - 1}{G _{1}} + \frac{F _{3} - 1}{G _{1} G _{2}} + \dots

Noise factors $F_{i}$ and available power gains $G_{i}$ as linear ratios, not dB. When $G_{1}$ is large the first stage dominates, so a low-noise, high-gain stage goes first. Same form for noise temperature: $T_{e} = T_{1} + T_{2} / G_{1} + \dots$

MATCHED LOSSY PASSIVE NETWORK (CABLE, ATTENUATOR)

F = 1 + (L - 1) \frac{T}{T _{0}}, T_{e} = (L - 1) T, F = L when T = T_{0}

$L = 1/ G \geq 1$ = power loss of the matched two-port as a linear ratio, $T$ = its physical temperature. At $T_{0}$ its noise figure in dB equals its loss in dB. Placed before the first amplifier (noise factor $F_{2}$ ), Friis gives $F = L + (F_{2} - 1) L = L F_{2}$ : the loss in dB adds directly to the system noise figure.

An amplifier's internal noise is modelled at its input by a voltage-noise source $e_{n}$ in series and a current-noise source $i_{n}$ in parallel (densities in $nV / Hz$ and $pA / Hz$ ). The current noise flows through the source resistance $R_{s}$ and becomes a voltage $i_{n} R_{s}$ . Treating $e_{n}$ , $i_{n}$ and the source's thermal noise as uncorrelated gives the results below.

INPUT-REFERRED NOISE DENSITY WITH SOURCE RESISTANCE R_S

e_{ni}^{2} = 4 k T R_{s} + e_{n}^{2} + i_{n}^{2} R_{s}^{2} (V^{2} / Hz)

Multiply by $B_{N}$ for the mean-square noise and by the gain for the output noise. With low $R_{s}$ , $e_{n}$ dominates (a bipolar-input op-amp usually has the lower $e_{n}$ ); with high $R_{s}$ , $i_{n} R_{s}$ dominates (a JFET or CMOS input usually has the lower $i_{n}$ ). Add the feedback resistors' noise separately if it is not negligible.

NOISE FACTOR OF AN AMPLIFIER AND OPTIMUM SOURCE RESISTANCE

F = 1 + \frac{e _{n}^{2} + i _{n}^{2} R _{s}^{2}}{4 k T R _{s}}, R_{s, opt} = \frac{e _{n}}{i _{n}}, F_{m i n} = 1 + \frac{e _{n} i _{n}}{2 k T}

$T$ = source temperature ( $T_{0} = 290$ K for the standard noise figure). $F$ is smallest when $R_{s} = e_{n} / i_{n}$ (for uncorrelated $e_{n}$ and $i_{n}$ ). Minimum noise factor is not minimum total noise: $e_{ni}^{2}$ always grows with $R_{s}$ , so never add series resistance to a source to reach $R_{s, opt}$ ; a transformer can present $R_{s, opt}$ to the amplifier without adding resistance.

SNR IMPROVEMENT BY AVERAGING

SNR_{N} = N SNR_{1} (rms-voltage ratio)

Averaging $N$ synchronised records of a repetitive signal whose noise is uncorrelated between records: the signal adds as $N$ , the noise rms as $N$ . As a power ratio the SNR improves by $N$ , that is, by $10 lo g_{10} N$ dB.

WORKED EXAMPLE · WORKED EXAMPLE: AMPLIFIER NOISE AND NOISE FIGURE

A signal source of internal resistance $R_{s} = 10 k Ω$ at 300 K drives a non-inverting amplifier of voltage gain 100. The amplifier's input noise is white: $e_{n} = 4 nV / Hz$ and $i_{n} = 1 pA / Hz$ . Its response is single-pole low-pass with $f_{3 dB} = 20$ kHz. Neglect feedback-resistor noise and take the amplifier's input resistance as infinite. Find (a) the rms output noise voltage and (b) the noise figure, taking the source at 300 K as the reference. Use $k = 1.38 \times 1 0^{- 23}$ J/K.

Solution

Source thermal density: $4 k T R_{s} = 1.656 \times 1 0^{- 20} \times 1 0^{4} = 1.656 \times 1 0^{- 16} V^{2} / Hz$ (that is, $12.87 nV / Hz$ ).
Voltage noise: $e_{n}^{2} = (4 \times 1 0^{- 9})^{2} = 0.16 \times 1 0^{- 16} V^{2} / Hz$ .
Current noise through $R_{s}$ : $i_{n} R_{s} = 1 0^{- 12} \times 1 0^{4} = 10 nV / Hz$ , so $(i_{n} R_{s})^{2} = 1.0 \times 1 0^{- 16} V^{2} / Hz$ .
The three are uncorrelated, so add: $e_{ni}^{2} = (1.656 + 0.16 + 1.0) \times 1 0^{- 16} = 2.816 \times 1 0^{- 16} V^{2} / Hz$ , i.e. $e_{ni} = 16.78 nV / Hz$ .
Noise bandwidth of a single pole: $B_{N} = (π /2) (20 kHz) = 31.42 kHz$ , so $B_{N} = 177.2 Hz$ .
Input-referred rms: $16.78 nV \times 177.2 = 2.974 μ V$ . Output: $100 \times 2.974 μ V = 297.4 μ V$ rms.
Noise factor: $F = e_{ni}^{2} / (4 k T R_{s}) = 2.816/1.656 = 1.700$ , so $N F = 10 lo g_{10} 1.700 = 2.31$ dB.

Answer: (a) About $297 μ V$ rms at the output (using $f_{3 dB}$ in place of $B_{N}$ would wrongly give $237 μ V$ ). (b) $N F \approx 2.3$ dB.

EXAM TIP

Never add rms noise values directly: $3 μ V$ and $4 μ V$ of independent noise give $5 μ V$ , not $7 μ V$ .

EXAM TIP

Friis and noise-figure formulas use linear ratios. Convert $N F$ and gain from dB first ( $F = 1 0^{N F /10}$ ), work in ratios, and convert back at the end.

Interference coupling: conductive, capacitive and inductive

Conductive (common-impedance) coupling happens when two circuits share a conductor, most often a ground return or a supply line. The current of one circuit, flowing through the shared impedance $Z_{c} = R_{c} + j ω L_{c}$ , develops a voltage that appears in series with the other circuit's signal. Noise is also conducted directly along power and signal leads from a noisy source, for example switching-regulator ripple on a supply rail.

COMMON-IMPEDANCE COUPLING

V_{N} = I_{2} Z_{c} = I_{2} (R_{c} + j ω L_{c})

$I_{2}$ = current of the aggressor circuit in the shared conductor. At high frequency or with fast edges the inductive term $L_{c} d I_{2} / d t$ usually dominates.

Capacitive (electric-field) coupling acts through the stray capacitance $C_{12}$ between a conductor at a changing voltage $V_{1}$ and the victim conductor, whose resistance to ground is $R$ and whose own capacitance to ground is $C_{2 G}$ . The coupling injects a noise current $C_{12} d V_{1} / d t$ into the victim node, so the resulting noise voltage is proportional to the victim's impedance to ground.

CAPACITIVE COUPLING

V_{N} = \frac{j ω R C _{12}}{1 + j ω R ( C _{12} + C _{2 G} )} V_{1} \approx j ω R C_{12} V_{1} for ω R (C_{12} + C_{2 G}) ≪ 1

For a low-impedance victim the noise is proportional to $ω$ , $R$ and $C_{12}$ , and leads $V_{1}$ by $9 0^{\circ}$ . Equivalent to a current source $I_{N} = j ω C_{12} V_{1}$ from the victim node to ground.

CAPACITIVE COUPLING INTO A HIGH-IMPEDANCE VICTIM

V_{N} \to \frac{C _{12}}{C _{12} + C _{2 G}} V_{1} for ω R (C_{12} + C_{2 G}) ≫ 1

A plain capacitive divider, independent of frequency and of $R$ : the worst case, met by floating or very-high-impedance inputs.

Inductive (magnetic-field) coupling happens when a changing current $I_{1}$ in one circuit sets up a magnetic flux that links the loop formed by the victim's signal and return conductors. By Faraday's law this induces a noise voltage in series with the victim loop, proportional to the loop area and to $d I_{1} / d t$ , and independent of the victim circuit's impedance.

INDUCTIVE COUPLING

V_{N} = j ω M I_{1}, ∣ V_{N} ∣ = ω B A cos θ

$M$ = mutual inductance; $B$ = sinusoidal flux density (an rms $B$ gives an rms $V_{N}$ ), taken as uniform over the loop area $A$ ; $θ$ = angle between $B$ and the normal to the loop. Sign: $V_{N} = - d Φ/ d t$ (Lenz's law); usually only the magnitude matters.

FIELD OF A LONG STRAIGHT CONDUCTOR

B = \frac{μ _{0} I}{2 π r}, μ_{0} = 4 π \times 1 0^{- 7} H/m

Flux density at distance $r$ from an isolated long wire. The field of a closely spaced go-and-return pair falls much faster (roughly as $1/ r^{2}$ at distances large compared with the spacing), which is why supply and return conductors should run together.

	Conductive (common impedance)	Capacitive (electric field)	Inductive (magnetic field)
Driven by	Current in a shared conductor	Voltage change $d V_{1} / d t$ on the source conductor	Current change $d I_{1} / d t$ in the source loop
Equivalent circuit in the victim	Voltage $I_{2} Z_{c}$ in series with the signal	Current source $C_{12} d V_{1} / d t$ from the victim node to ground	Voltage source $j ω M I_{1}$ in series with the victim loop
Effect of victim impedance	Set by the shared impedance and current, not by the victim's load	Noise voltage rises with the victim's impedance to ground	Independent of the victim's impedance
Geometry that matters	Length of the shared conductor	Separation (through $C_{12}$ )	Loop area, separation, orientation
Main cures	Separate returns, single-point ground, low-impedance ground plane, decoupling, filtering, isolation	Grounded electrostatic shield, separation, lower victim impedance, shorter exposed leads	Smaller loop area, twisted pairs, separation, perpendicular orientation, magnetic shielding, smaller source loop

WORKED EXAMPLE · WORKED EXAMPLE: CAPACITIVE VERSUS INDUCTIVE PICKUP

A victim signal wire runs alongside a 230 V (rms), 50 Hz mains conductor. The stray capacitance between them is 5 pF and the victim's own capacitance to ground is 50 pF. (a) Find the capacitively coupled noise voltage when the victim circuit's resistance to ground is 10 k $Ω$ , and when it is 1 M $Ω$ . (b) Separately, a 50 Hz magnetic field of 2 $μ$ T (rms), uniform and normal to the plane of the victim loop, links a loop area of 20 cm $^{2}$ . Find the induced noise voltage.

Solution

Check the approximation: $ω = 2 π \times 50 = 314.2$ rad/s. For $R = 1 M Ω$ , $ω R (C_{12} + C_{2 G}) = 314.2 \times 1 0^{6} \times 55 \times 1 0^{- 12} = 0.0173 ≪ 1$ , so $∣ V_{N} ∣ \approx ω R C_{12} V_{1}$ holds for both values of $R$ .
$R = 10 k Ω$ : $∣ V_{N} ∣ = 314.2 \times 1 0^{4} \times 5 \times 1 0^{- 12} \times 230 = 3.61 mV$ rms.
$R = 1 M Ω$ : 100 times larger, $∣ V_{N} ∣ = 361 mV$ rms. Capacitive pickup scales with the victim's impedance.
Inductive: $∣ V_{N} ∣ = ω B A = 314.2 \times (2 \times 1 0^{- 6}) \times (20 \times 1 0^{- 4}) = 1.26 μ V$ rms, whatever the victim's resistance.
Cures: a grounded electrostatic shield (or a lower victim impedance) for part (a); a smaller loop area, by twisting the signal and return wires, for part (b).

Answer: (a) 3.61 mV rms and 361 mV rms; (b) 1.26 $μ$ V rms.

EXAM TIP

To tell electric from magnetic pickup on a cable, measure the noise across the impedance at one end while lowering the impedance at the other end. If the measured noise falls, the coupling is capacitive (the injected current now sees a lower impedance); if it rises, it is inductive (a larger share of the series-induced voltage appears across the unchanged end).

EXAM TIP

Capacitive coupling is a voltage-driven problem that hurts high-impedance circuits; inductive coupling is a current-driven problem that hurts low-impedance loops. A non-magnetic shield grounded at one end cures the first but does almost nothing for the second at low frequency; a twisted pair cures the second but not the first.

Shielding, twisted pairs, guard rings and grounding

An electrostatic shield is a conducting enclosure (foil, braid, metal box) around the victim that intercepts the electric field lines from the source. The field now ends on the shield, and the coupling capacitance from source to victim falls almost to zero. The shield works only when it is connected to the victim circuit's reference (ground or common). A floating shield simply couples the source to the victim through two capacitors in series and can make matters worse. Any length of victim conductor sticking out beyond the shield, and any gap in the shield, still couples.

Magnetic fields are much harder to shield. At low frequencies (50 Hz and its harmonics) a non-magnetic conductor such as copper or aluminium does almost nothing, because its skin depth is several millimetres; a high-permeability material (mu-metal, silicon steel) is needed to divert the flux around the victim. At high frequencies a good conductor shields well: the field induces eddy currents whose own field opposes it, and the wave is absorbed within a few skin depths. In a coaxial cable grounded at both ends, above about five times the shield cut-off frequency $f_{s h} = R_{s h} / (2 π L_{s h})$ (shield resistance and inductance), the return current flows on the shield rather than through ground; the go and return currents are then coaxial, the loop area is tiny, and magnetic coupling in both directions is small.

SKIN DEPTH

δ = \frac{1}{π f μ σ}

Depth at which a field falls to $1/ e$ (37 %) inside a conductor of permeability $μ$ and conductivity $σ$ . For copper: about 9.3 mm at 50 Hz and about 66 $μ$ m at 1 MHz.

ABSORPTION LOSS OF A SHIELD

A = 8.686 \frac{t}{δ} dB

Shield thickness $t$ : each skin depth of material adds 8.686 dB of attenuation. Total shielding effectiveness $S E = A + R + B$ , where $R$ is the reflection loss and $B$ a correction for thin shields.

	Electrostatic shield	Magnetic shield, low frequency	Magnetic shield, high frequency
Material	Any good conductor: copper or aluminium foil, braid	High-permeability material: mu-metal, silicon steel	Good conductor, several skin depths thick
Mechanism	Terminates electric field lines on a surface held at the reference potential	Low-reluctance path that diverts flux around the victim	Eddy currents oppose the field; absorption and reflection
Must it be grounded?	Yes, to the circuit reference	No; grounding does not affect its magnetic action	Not for absorption; it is grounded anyway so that it also shields electrostatically
Protects against	Capacitive coupling	Inductive coupling at mains frequencies	Inductive coupling and radiated fields

Twisted pair. Twisting the signal and return conductors divides the loop into many small loops of alternating orientation. A uniform external field induces equal and opposite voltages in adjacent twists, which cancel, so the net induced voltage is small; the tighter the twist, the better. This works only when the whole return current flows in the return conductor, that is, when the circuit is grounded at one end only; if it is grounded at both ends, part of the return current flows through the ground loop, whose area twisting does not reduce. Twisting also reduces the field the pair itself radiates. It gives little protection against capacitive coupling, so a shield is added for that (shielded twisted pair). Driving the pair from a balanced source into a differential receiver also cancels capacitive pickup that reaches both wires equally.

Guard ring. On a board, a high-impedance node (an electrometer or photodiode amplifier input) collects leakage current across the board surface from nearby traces at other potentials. A guard ring is a copper track surrounding the node, held at (nearly) the node's own potential by a low-impedance point in the circuit: the inverting-input potential for a non-inverting amplifier or follower, or ground for an inverting amplifier whose inverting input sits at virtual ground. With no voltage across the gap, no leakage flows into the node; leakage from other traces is intercepted by the guard and returned to a low-impedance point. A driven shield on a cable applies the same idea to cable leakage and capacitance.

A ground is the reference against which voltages are measured and the path by which currents return. Real ground conductors have resistance and inductance, so two ground points generally differ in potential whenever current flows. Keep three purposes apart: safety ground (the enclosure bonded to earth so that a fault cannot leave it live), signal ground (the reference for low-level signals) and power or noisy ground (returns of motors, relays, digital logic and high-current stages). Separate returns for these, joined at a single point, avoid common-impedance coupling between them. Never remove a safety earth to cure noise.

Scheme	How it is wired	Strength	Weakness and typical use
Single-point, series (daisy chain)	Circuits connected one after another along one ground wire to the ground point	Simplest wiring	All circuits' currents share the wire, giving the worst common-impedance coupling; avoid for sensitive circuits
Single-point, parallel (star)	Each circuit has its own ground wire to one common point	No shared return impedance; best at low frequency	Many long wires whose inductance makes it poor at high frequency; rule of thumb: use below about 1 MHz
Multipoint	Each circuit connects by the shortest path to a nearby low-impedance ground plane	Lowest ground impedance (inductance) at high frequency	Forms ground loops that trouble low-frequency circuits; rule of thumb: use above about 10 MHz. Between 1 and 10 MHz, single-point grounding still works if the longest ground lead is shorter than about one-twentieth of a wavelength
Hybrid	Single-point at low frequency, multipoint at high frequency (for example through capacitors)	Suits systems with both low-level analog and fast signals	Needs careful design

Ground loops. When a signal circuit is grounded at two points, say a grounded sensor at one end and a grounded amplifier at the other, the signal return and the ground path form a closed loop. Any potential difference $V_{G}$ between the two ground points, and any magnetic flux linking the loop, drives a current around it, and part of the resulting voltage appears in series with the signal, typically as 50 Hz hum. Ground loops matter most at low frequency and with low-level signals.

Ways to break or tolerate a ground loop

Ground the signal circuit at one point only: float the sensor, or float the amplifier's signal common, while keeping the safety earth on the enclosure.
Use a differential or instrumentation amplifier: $V_{G}$ becomes a common-mode voltage that the CMRR rejects, leaving an input-referred error of about $V_{G} / CMRR$ .
Break the conductive path with an isolation transformer (AC signals only), an optocoupler or an isolation amplifier (DC signals too).
Put a common-mode choke (balun) in the cable, which presents a high impedance to the loop current while leaving the differential signal untouched.
Ground a low-frequency cable shield at one end only (where the signal circuit is grounded), so the shield is not part of a loop.

COMMON-MODE REJECTION

V_{o} = A_{d} V_{d} + A_{c m} V_{c m}, CMRR = \frac{A _{d}}{A _{c m}}, input-referred error = \frac{V _{c m}}{CMRR}

$V_{d}$ = differential signal, $V_{c m}$ = common-mode voltage (for example a ground-potential difference). $CMRR_{dB} = 20 lo g_{10} CMRR$ . An imbalance in source resistances converts part of $V_{c m}$ into a differential error, so the real rejection can be worse than the datasheet CMRR.

Normal-mode and common-mode interference. Normal-mode (series-mode, differential-mode) interference appears in series with the signal, between the two input leads, so the input stage cannot tell it apart from the signal: it has to be removed by filtering or by the conversion method. Common-mode interference appears equally on both input leads with respect to ground (a ground-potential difference is the usual source) and is rejected by a differential input according to its CMRR. Unequal lead or source impedances convert part of a common-mode voltage into a normal-mode one.

NORMAL-MODE REJECTION OF AN INTEGRATING ADC

NMRR = \frac{π f T}{sin ( π f T )}

An integrating (for example dual-slope) ADC averages its input over the integration time $T$ , so a normal-mode sinusoid at frequency $f$ is attenuated by this factor. It is ideally infinite when $T$ is a whole number of periods of the interference: $T = 20$ ms, or any multiple of it, rejects 50 Hz and all its harmonics. $NMRR_{dB} = 20 lo g_{10} NMRR$ .

Cable-shield grounding rules

Low frequency (roughly below 1 MHz, cable short compared with a wavelength): ground the shield at one point only, at the end where the signal circuit is grounded, so the shield does not form a ground loop.
High frequency, or a cable longer than about one-twentieth of a wavelength: ground the shield at both ends, or at many points, so it stays at ground potential along its length.
In a low-level circuit, prefer a shielded twisted pair to using the shield as the signal return, so that shield current stays out of the signal path.
Terminate the shield with a full 360-degree connection at the connector rather than a long pigtail wire, whose inductance spoils high-frequency shielding.

EXAM TIP

An electrostatic shield must be connected to the circuit reference to work; a magnetic shield at 50 Hz must be a high-permeability material. Copper foil grounded at one end is an excellent electrostatic shield and almost useless against 50 Hz magnetic fields.

EXAM TIP

Grounding a low-frequency cable shield at both ends creates a ground loop: it adds a path for ground-loop current and does not remove hum. To remove hum, ground the shield at one end and break or reject the signal circuit's own ground loop.

Component selection, board realisation and testing

Selecting a component means reading its datasheet against the job. The part must meet the electrical requirement with margin over the whole temperature range and its lifetime, and its non-ideal behaviour (tolerance, drift, noise, parasitics) must not dominate the error budget. Derate: operate parts well below their maximum voltage, current, power and temperature ratings, following the manufacturer's derating curve.

Component	Key specifications	Selection points
Resistor	Value and tolerance (E-series), power rating, temperature coefficient (ppm/°C), voltage rating, excess-noise index, parasitic L and C	Metal-film for precision and low noise; carbon-composition has high excess (1/f) noise; wirewound is precise and quiet but inductive; matched networks where ratios matter
Capacitor	Capacitance and tolerance, working voltage, dielectric (temperature and voltage coefficients), ESR, leakage, self-resonant frequency, dielectric absorption	C0G/NP0 ceramic for stable timing and filter values; X7R for decoupling; electrolytic for bulk storage (polarised, leaky, higher ESR); polypropylene film for sample-and-hold and integrators (low dielectric absorption)
Op-amp or instrumentation amplifier	Offset voltage and its drift, bias current, $e_{n}$ and $i_{n}$ , gain-bandwidth product, slew rate, CMRR, PSRR, supply range, output swing, input common-mode range	Bipolar input for low source impedance (low $e_{n}$ ); JFET or CMOS input for high source impedance (low $I_{B}$ and $i_{n}$ ); chopper-stabilised (zero-drift) for DC precision below the 1/f corner
Diode or transistor	Maximum voltage, current and power, leakage, switching speed, gain, thermal resistance	Ratings with margin; junction temperature checked against its maximum
All parts	Temperature grade: commercial 0 to 70 °C, industrial −40 to 85 °C, military −55 to 125 °C; package	Match the grade to the environment

SLEW-RATE REQUIREMENT (FULL-POWER BANDWIDTH)

SR \geq 2 π f_{m a x} V_{p}

A sinusoidal op-amp output of peak $V_{p}$ at frequency $f_{m a x}$ is undistorted only if the slew rate meets this; equivalently the full-power bandwidth is $f_{F P} = SR / (2 π V_{p})$ .

JUNCTION TEMPERATURE

T_{J} = T_{A} + P_{D} θ_{J A}

$T_{A}$ = ambient temperature, $P_{D}$ = dissipated power, $θ_{J A}$ = junction-to-ambient thermal resistance in °C/W. $T_{J}$ must stay below the rated maximum with margin.

RESISTANCE OF A PCB TRACE

R = \frac{ρ L}{w t}

Length $L$ , width $w$ , copper thickness $t$ ; copper $ρ \approx 1.7 \times 1 0^{- 8} Ω m$ , and 1 oz/ft $^{2}$ copper is about 35 $μ$ m thick. A shared trace is a common impedance: a few milliohms carrying amperes of load current put millivolts into a low-level ground.

Circuit board realisation, in order

Schematic capture of the verified design, with every part's footprint and rating fixed from its datasheet.
Placement: group by function (analog, digital, power); keep the analog front end away from switching regulators, clocks and relays; put decoupling capacitors next to each IC supply pin.
Ground and power: give every circuit a low-impedance return. Either use one unbroken ground plane and place the analog and digital parts in separate regions so their return currents do not cross, or use separate analog and digital grounds joined at one point, usually at the mixed-signal converter. Never route a signal across a gap in its reference plane. Distribute supplies in a star, with bulk and local decoupling.
Routing: keep sensitive traces short and away from fast edges; route each signal over an unbroken plane so its return current flows directly beneath it (minimum loop area); run differential pairs together; surround high-impedance nodes with guard rings.
Design-rule check (clearances, trace widths for the current, creepage for high voltage), then fabrication data (Gerber and drill files).
Fabrication, assembly and soldering, then cleaning flux residue from high-impedance areas.

Testing a built board, in order

Bare-board test for opens and shorts (flying probe or bed of nails) before assembly.
Visual or automated optical inspection of the assembled board: part orientation and values, solder bridges, dry joints.
Power-off checks: resistance from each supply rail to ground, to catch shorts before power is applied.
First power-up from a current-limited supply; check each rail's voltage and ripple and the supply current.
DC checks: bias points, offsets and reference voltages against the design values.
Functional (AC) tests with a signal generator and oscilloscope: gain, bandwidth, linearity, distortion; then calibration.
Noise and interference checks: output noise with the input shorted and with a known source resistance, hum with the real cables and grounding; temperature and burn-in tests where required.

EXAM TIP

Estimating rms noise from an oscilloscope: for Gaussian noise, the peak-to-peak value exceeded only about 0.1 % of the time is about 6.6 times the rms value ( $\pm 3.3 σ$ ).

EXAM TIP

Put a small ceramic decoupling capacitor (typically 0.1 $μ$ F) as close as possible to each IC supply pin, with a short path to ground, and a bulk electrolytic where the supply enters the board. A decoupling capacitor far from the pin sits behind the trace inductance and loses its effect at high frequency.

PREVIOUS-YEAR QUESTIONS

2 questions from the IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE IN 2024 · GATE IN 2018

GATE 2024 IN Q.56

GATE 2024 · INQ.56 · 2 marks · NAT

A shielded cable with $C_{stray} = 20 pF$ and $R_{wire} = 10 Ω$ is used to connect the inductive sensors as shown in the figure. The RMS value of $V_{out}$ is ____ V (rounded off to two decimal places).

Note: Assume all components are ideal, and sensors are not magnetically coupled.

GATE 2024 IN question 56: Two variable inductors (inductive sensors, each drawn with a diagonal arrow) are in series. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

GATE 2024 IN Q.56: answer and solution

GATE 2018 IN Q.52

GATE 2018 · INQ.52 · 2 marks · NAT

Assuming ideal opamp, the RMS voltage (in mV) in the output $V_{o}$ only due to the 230 V, 50 Hz interference is (up to one decimal place) ______.

GATE 2018 IN question 52: An inverting amplifier. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

GATE 2018 IN Q.52: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · NAT

The rms thermal noise voltage across an open-circuited 1 M $Ω$ resistor at 300 K, measured over a noise bandwidth of 10 kHz, is ________ $μ$ V (rounded off to two decimal places). Take $k = 1.38 \times 1 0^{- 23}$ J/K.

Answer and solution

Answer: $12.87$ (accepted range $\pm 0.1$ )

$4 k T = 4 \times 1.38 \times 1 0^{- 23} \times 300 = 1.656 \times 1 0^{- 20}$ J.
$\overline{v_{n}^{2}} = 4 k T R B = 1.656 \times 1 0^{- 20} \times 1 0^{6} \times 1 0^{4} = 1.656 \times 1 0^{- 10} V^{2}$ .
$V_{n} = 1.656 \times 1 0^{- 10} = 1.287 \times 1 0^{- 5}$ V $= 12.87 μ$ V.

PRACTICE 22 marks · NAT

A reverse-biased photodiode, modelled as an ideal (infinite-impedance) current source carrying a DC photocurrent of 10 $μ$ A, is connected across a 100 k $Ω$ load resistor at 300 K. The voltage across the resistor is read by a noiseless amplifier of infinite input impedance over a noise bandwidth of 10 kHz. Considering only the shot noise of the photocurrent and the thermal noise of the load resistor, the total rms noise voltage across the resistor is ________ $μ$ V (rounded off to two decimal places). Take $q = 1.6 \times 1 0^{- 19}$ C and $k = 1.38 \times 1 0^{- 23}$ J/K.

Answer and solution

Answer: $18.35$ (accepted range $\pm 0.2$ )

Shot noise current: $\overline{i_{s h}^{2}} = 2 q I B = 2 \times 1.6 \times 1 0^{- 19} \times 1 0^{- 5} \times 1 0^{4} = 3.2 \times 1 0^{- 20} A^{2}$ .
Thermal noise current of the resistor (Norton form): $\overline{i_{t h}^{2}} = 4 k T B / R = 1.656 \times 1 0^{- 20} \times 1 0^{4} /1 0^{5} = 0.1656 \times 1 0^{- 20} A^{2}$ .
Both currents flow into the same node and are uncorrelated: $\overline{i_{n}^{2}} = 3.3656 \times 1 0^{- 20} A^{2}$ , so $i_{n} = 1.8345 \times 1 0^{- 10}$ A rms.
The only impedance at the node is $R$ (ideal current source, infinite amplifier input impedance): $V_{n} = i_{n} R = 1.8345 \times 1 0^{- 10} \times 1 0^{5} = 18.35 μ$ V.
Check: shot part $17.89 μ$ V and thermal part $4.07 μ$ V; $17.8 9^{2} + 4.0 7^{2} = 18.35 μ$ V.

PRACTICE 31 mark · MCQ

A grounded-junction thermocouple at a furnace is connected through a 40 m shielded twisted-pair cable to a single-ended amplifier whose signal common is tied to the panel ground. A 50 Hz potential difference exists between the furnace ground and the panel ground, and hum appears at the amplifier output. Which one of the following does NOT break or reject this ground loop?

(A)
Replacing the amplifier with an isolation amplifier whose input side floats
(B)
Replacing the amplifier with a high-CMRR differential (instrumentation) amplifier
(C)
Replacing the thermocouple with an insulated-junction (ungrounded) type
(D)
Connecting the cable shield to ground at both the furnace end and the panel end

Answer and solution

Answer: (D)

A breaks the loop: a floating isolation-amplifier input leaves no conductive path between the two grounds.
B rejects it: the ground-potential difference becomes a common-mode voltage, rejected by the amplifier's CMRR.
C breaks the loop: an insulated junction removes the ground connection at the sensor end.
D adds a second conductive loop through the shield and the two grounds; at 50 Hz it neither breaks nor rejects the ground loop. The answer is D.

PRACTICE 42 marks · NAT

A receiver front end consists of a low-noise amplifier with noise figure 2 dB and available power gain 20 dB, followed directly by a mixer with noise figure 10 dB. All noise figures are referred to $T_{0} = 290$ K. The noise figure of the cascade is ________ dB (rounded off to two decimal places).

Answer and solution

Answer: $2.24$ (accepted range $\pm 0.02$ )

Convert to linear ratios: $F_{1} = 1 0^{2/10} = 1.585$ , $G_{1} = 1 0^{20/10} = 100$ , $F_{2} = 1 0^{10/10} = 10$ .
Friis: $F = F_{1} + \frac{F _{2} - 1}{G _{1}} = 1.585 + \frac{9}{100} = 1.675$ .
$N F = 10 lo g_{10} 1.675 = 2.24$ dB.
Check of the principle: with an LNA gain of only 10 dB, $F = 1.585 + 0.9 = 2.485$ and $N F = 3.95$ dB, so the first stage's gain decides how much the mixer's noise matters.

PRACTICE 51 mark · MCQ

A resistor of unknown value $R$ , at 300 K, is connected across an ideal 10 pF capacitor, and nothing else is connected. The rms noise voltage across the capacitor, taken over all frequencies, is closest to (take $k = 1.38 \times 1 0^{- 23}$ J/K)

(A)
$2.03 μ V$
(B)
$20.3 μ V$
(C)
$40.7 μ V$
(D)
It cannot be found without the value of $R$

Answer and solution

Answer: (B)

The resistor's thermal noise, density $4 k T R$ , is filtered by the RC network, whose noise bandwidth is $1/ (4 R C)$ .
$\overline{v_{n}^{2}} = 4 k T R \times \frac{1}{4 R C} = \frac{k T}{C}$ , independent of $R$ , so D is wrong.
$\overline{v_{n}^{2}} = \frac{1.38 \times 1 0 ^{- 23} \times 300}{10 \times 1 0 ^{- 12}} = 4.14 \times 1 0^{- 10} V^{2}$ , so $V_{n} = 20.3 μ V$ .
Option C comes from wrongly using $4 k T / C$ ; option A is off by a factor of 10.

PRACTICE 62 marks · MSQ

A low-level signal circuit (the victim) runs parallel to a 50 Hz mains cable; all dimensions are small compared with the wavelength, and the victim circuit is grounded at one end only. $C_{12}$ is the stray capacitance between the mains conductor and the victim conductor, $C_{2 G}$ the victim conductor's capacitance to ground, and $R$ the victim circuit's resistance to ground. Which of the following statements is/are correct?

(A)
When $ω R (C_{12} + C_{2 G}) ≪ 1$ , the capacitively coupled noise voltage increases in proportion to $R$ .
(B)
The magnetically coupled noise appears as a voltage source in series with the victim loop and, in a field uniform over the loop, is proportional to the loop area.
(C)
A solid copper tube placed around the victim conductor and grounded at one end only substantially reduces the 50 Hz magnetically coupled noise.
(D)
Twisting the victim's signal and return conductors together reduces the magnetically coupled noise.

Answer and solution

Answer: (A), (B), (D)

A is correct: for $ω R (C_{12} + C_{2 G}) ≪ 1$ , $V_{N} \approx j ω R C_{12} V_{1}$ , which is proportional to $R$ .
B is correct: Faraday's law gives $∣ V_{N} ∣ = ω B A cos θ$ in series with the loop, proportional to its area for a uniform field.
C is wrong: a shield grounded at one end only carries no current along its length to cancel the field, and the skin depth of copper at 50 Hz is about 9.3 mm, so it gives almost no magnetic shielding. It is an electrostatic shield only.
D is correct: with the victim grounded at one end only, the whole return current flows in the return conductor, and adjacent twists form loops of opposite orientation whose induced voltages cancel. (Had the victim been grounded at both ends, part of the return current would flow through the ground loop, which twisting does not shrink.)

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE IN

Sources and effects of noise and interference in electronic circuits; conductively, capacitively, inductively coupled interference; noise generated by electronic components, thermal noise, shot noise, flicker noise; shielding and grounding. Basic principles of component selection based on specifications, circuit board realization and testing

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