Notes · ElectronicsUK · A-Levels

AC circuits and passive filters

Alternating signals behave quite differently from steady voltages once capacitors and inductors are involved. This chapter defines the AC quantities and root-mean-square value, develops the frequency-dependent reactance of capacitors and inductors, and uses them to design and analyse RC low-pass, high-pass and band-pass filters, including the cut-off frequency, the decibel and the roll-off.

5 sections~19 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

T·0333 / 16

Exam profile

AO1 · Define AC quantities (peak, peak-to-peak, rms, frequency, period) and describe capacitor and inductor reactance and passive filters.AO2 · Calculate rms values, reactance, cut-off frequency, gain in decibels and filter response.AO3 · Design a low-pass, high-pass or band-pass filter to a frequency specification and evaluate its response.

Operators:definecalculatedeterminedesignsketchanalyseevaluate

basic level

At AS the emphasis is on AC quantities, rms, the reactance formulae and the single-pole RC low-pass and high-pass cut-off frequency.

higher level

The full A-Level treats the decibel, roll-off and the combination of stages into a band-pass filter, with quantitative response design.

Depth

Text

Contents · 5 sections

AC circuits and passive filters

AC signals and the rms value#

FoundationWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (AC signals)

Sinusoidal voltage: peak and rms

Fig. 1One cycle of a

5\,\text{V}

peak sinusoid. The peak is marked; the horizontal line at

3.54\,\text{V}

is the rms value,

V_p/\sqrt{2}

An alternating signal repeatedly reverses direction. A sinusoid is described by its peak voltage

V_p

(the maximum from zero), its peak-to-peak voltage

V_{pp} = 2V_p

(the full swing from the most negative to the most positive point), its period

T

(the time for one complete cycle) and its frequency

f = 1/T

(cycles per second, in hertz). These are the basic descriptors you read off an oscilloscope trace, and every AC calculation starts from them.

The root-mean-square (rms) value is the single most useful AC quantity because it captures the heating effect: the rms value of an alternating current is the steady direct current that would dissipate the same average power in a resistor. For a sinusoid the rms value is the peak divided by the square root of two,

V_{rms} = V_p/\sqrt{2} \approx 0.707\,V_p

. This is why UK mains is quoted as

230\,\text{V}

rms even though its peak is about

325\,\text{V}

The average power delivered to a resistor by an AC signal is found from the rms values exactly as for DC:

P = V_{rms}^2/R = I_{rms}^2 R

. Using peak values here would overstate the power by a factor of two, which is a common and costly error. Whenever a question asks for power or for an equivalent heating value, convert to rms first; whenever it asks what a scope shows, work in peak or peak-to-peak.

Frequency and period are reciprocals, so a

1\,\text{kHz}

tone has a period of

1\,\text{ms}

, and a

2\,\text{ms}

period corresponds to

500\,\text{Hz}

. These conversions underpin everything that follows: reactance depends on frequency, filters are specified by frequency, and communication systems are described by the range of frequencies they carry. Fluency in moving between

T

and

f

, and between peak and rms, is assumed throughout the analogue units.

It is worth being clear that rms is defined for any waveform, not just a sinusoid — but the convenient factor

1/\sqrt{2}

applies only to a sine. A square wave that spends equal time at

+V_p

and

-V_p

has an rms value of

V_p

, because its magnitude is always

V_p

. Recognising that the rms-to-peak relationship depends on the waveform shape is a mark of genuine understanding and appears in the more searching questions.

V_{rms} = \dfrac{V_p}{\sqrt{2}}, \quad V_{pp} = 2V_p

Sinusoid values

rms is 0.707 of peak for a sine; peak-to-peak is twice the peak.

f = \dfrac{1}{T}, \quad P = \dfrac{V_{rms}^2}{R}

Frequency and power

Frequency is the reciprocal of period; power uses rms values.

Worked example

Reading a sinusoid and finding power

An oscilloscope shows a sinusoid of peak-to-peak voltage $6.0\,\text{V}$ and period $0.25\,\text{ms}$ . Find the peak voltage, rms voltage, frequency, and the average power it delivers to a $50\,\Omega$ resistor.

Peak and rms
$V_p = V_{pp}/2 = 3.0\,\text{V}$ ; $V_{rms} = V_p/\sqrt{2} = 3.0/1.414 = 2.12\,\text{V}$ .
Frequency
$f = 1/T = 1/(0.25\times 10^{-3}) = 4000\,\text{Hz} = 4.0\,\text{kHz}$ .
Power
$P = V_{rms}^2/R = (2.12)^2/50 = 4.50/50 = 0.090\,\text{W}$ .
$P = \dfrac{V_{rms}^2}{R} = \dfrac{2.12^2}{50} = 90\,\text{mW}$

Result: $V_p = 3.0\,\text{V}$ , $V_{rms} = 2.12\,\text{V}$ , $f = 4.0\,\text{kHz}$ , and the resistor dissipates about $90\,\text{mW}$ .

Typical mistakes

Using peak instead of rms when calculating average power, overstating it by a factor of two.
Applying $V_p/\sqrt{2}$ to a non-sinusoidal waveform, where the rms-to-peak factor is different.

Active revision

A sinusoidal signal displayed on an oscilloscope has a peak-to-peak voltage of $8.0\,\text{V}$ and a period of $0.50\,\text{ms}$ . Find its peak voltage, rms voltage and frequency, and the average power it would deliver to a $100\,\Omega$ resistor.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Reactance and impedance#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (reactance)

Capacitor and inductor reactance versus frequency

Fig. 2Capacitor reactance

X_C

(falling) and inductor reactance

X_L

(rising) for a

100\,\text{nF}

capacitor and a

0.25\,\text{H}

inductor; they are equal near

1\,\text{kHz}

(the resonance condition).

A capacitor and an inductor oppose alternating current, but their opposition depends on frequency, unlike a resistor. This frequency-dependent opposition is called reactance, measured in ohms. A capacitor's reactance is

X_C = 1/(2\pi f C)

: it is large at low frequency (a capacitor blocks DC, where

f=0

makes

X_C

infinite) and small at high frequency (a capacitor passes high frequencies easily). This single behaviour — high reactance at low frequency, low at high — is the mechanism behind every capacitive filter.

An inductor behaves in exactly the opposite way. Its reactance is

X_L = 2\pi f L

: small at low frequency (an inductor passes DC freely, since

f=0

gives

X_L=0

) and large at high frequency (an inductor blocks rapid changes). Capacitors and inductors are therefore complementary; a capacitor favours high frequencies and an inductor favours low ones, and pairing them lets a designer select a band of frequencies.

Unlike resistance, reactance stores energy rather than dissipating it, and the current through a reactance is a quarter-cycle out of phase with the voltage across it. For the purposes of this course the essential quantitative skill is to calculate reactance at a given frequency and to combine a resistor and a capacitor into an impedance. In a series RC circuit the impedance is

Z = \sqrt{R^2 + X_C^2}

, the resistance and reactance combining at right angles because of the phase difference.

The frequency at which the reactance of a capacitor equals a given resistance is the pivot around which a filter is designed. When

X_C = R

, the impedance is shared equally between the resistor and the capacitor, and the output of an RC network is

1/\sqrt{2}

(about 0.707) of the input. This is exactly the condition that defines the cut-off frequency of a filter, developed in the next sections; recognising that

X_C = R

at cut-off ties the reactance formula directly to filter design.

Because reactance varies with frequency while resistance does not, a network of resistors and capacitors treats different frequencies differently — and that is the whole point. A steady (DC) level, a low-frequency hum and a high-frequency hiss all see different impedances in the same circuit, so they can be separated. Keeping clear which way each component's reactance moves with frequency is the key to predicting, without any calculation, whether a given network passes or blocks a signal.

X_C = \dfrac{1}{2\pi f C}, \quad X_L = 2\pi f L

Reactance

Capacitor reactance falls with frequency; inductor reactance rises.

Z = \sqrt{R^2 + X_C^2}

Series RC impedance

Resistance and reactance combine at right angles because of the 90 degree phase difference.

Worked example

Capacitor reactance and the crossover frequency

A $100\,\text{nF}$ capacitor is used in an AC circuit. Find its reactance at $2.0\,\text{kHz}$ , and the frequency at which its reactance is $1.0\,\text{k}\Omega$ .

Reactance at 2 kHz
$X_C = 1/(2\pi f C) = 1/(2\pi \times 2000 \times 100\times 10^{-9}) = 1/(1.257\times 10^{-3}) = 796\,\Omega$ .
$X_C = \dfrac{1}{2\pi (2000)(100\times 10^{-9})} = 796\,\Omega$
Frequency for 1 k-ohm
Set $X_C = 1000\,\Omega$ and solve for $f$ : $f = 1/(2\pi C X_C) = 1/(2\pi \times 100\times 10^{-9} \times 1000) = 1592\,\text{Hz}$ .

Result: $X_C = 796\,\Omega$ at $2.0\,\text{kHz}$ ; its reactance equals $1.0\,\text{k}\Omega$ at about $1.59\,\text{kHz}$ .

Typical mistakes

Adding resistance and reactance directly instead of combining them as $Z = \sqrt{R^2 + X_C^2}$ .
Getting the frequency dependence the wrong way round — $X_C$ falls with frequency, $X_L$ rises.

Active revision

A $100\,\text{nF}$ capacitor is used at $2.0\,\text{kHz}$ . Calculate its reactance, and find the frequency at which its reactance equals $1.0\,\text{k}\Omega$ .

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The RC low-pass filter#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (low-pass filter)

RC low-pass filter circuit

Fig. 3Series resistor, shunt capacitor: the output across

C

is large at low frequency (high

X_C

) and small at high frequency.

A low-pass filter passes low-frequency signals and attenuates high-frequency ones. The simplest is a resistor in series with the input and a capacitor to ground, with the output taken across the capacitor. The resistor and capacitor form a frequency-dependent potential divider: at low frequency the capacitor's reactance is huge, so almost all of the input appears across it at the output; at high frequency the reactance is tiny, so the output is small. The network therefore favours low frequencies.

The boundary between 'passed' and 'attenuated' is the cut-off frequency

f_c = 1/(2\pi R C)

, the frequency at which

X_C = R

. At the cut-off the output has fallen to

1/\sqrt{2}

(about 70.7%) of the input, which corresponds to a power ratio of one half and a gain of

-3\,\text{dB}

. Below

f_c

the filter passes the signal with little loss; above

f_c

the output falls steadily, at a rate of

-20\,\text{dB}

per decade for this single-stage (first-order) filter.

Designing a low-pass filter to a specification is a two-value problem: choose

R

and

C

so that

1/(2\pi R C)

equals the required cut-off. Because only the product

RC

matters, there is freedom to choose one component and calculate the other; a sensible choice keeps

R

in the kilohm range so the source is not overloaded and the capacitor is a convenient value. The time constant

\tau = RC

links directly to the timing chapter: a low-pass filter and an RC timing circuit are the same network viewed in the frequency and the time domains respectively.

The low-pass filter is everywhere in real systems: it smooths the output of a rectifier and of a digital-to-analogue converter, it removes high-frequency noise and hiss from an audio signal, and it defines the bandwidth of a sensor channel. Wherever a slow, wanted signal must be separated from fast, unwanted noise, a low-pass filter is the tool. Recognising it by topology — series resistance, shunt capacitance, output across the capacitor — is a basic exam skill.

A single RC stage rolls off gently, at only

-20\,\text{dB}

per decade, so it does not sharply separate wanted from unwanted frequencies. Where a steeper cut is needed, several stages are cascaded (buffered by op-amps so they do not load one another), each adding another

-20\,\text{dB}

per decade. Understanding that the order of the filter sets the steepness of the roll-off is the bridge to the more sophisticated active filters and to the audio tone-control circuits later in the course.

f_c = \dfrac{1}{2\pi R C}

Cut-off frequency

The frequency at which X_C = R and the output is 0.707 of the input.

\dfrac{V_{out}}{V_{in}} = \dfrac{1}{\sqrt{2}} = 0.707 \ \text{at } f_c\ (-3\,\text{dB})

Cut-off condition

Half power point, a gain of minus three decibels.

Low-pass frequency response

Fig. 4The response is flat below

f_c

, is

-3\,\text{dB}

f_c

(marked, at

\log_{10}(f/f_c)=0

), and falls at

-20\,\text{dB}

/decade above it.

Worked example

Designing a low-pass filter

Design an RC low-pass filter with a cut-off frequency of $1.0\,\text{kHz}$ using a $10\,\text{k}\Omega$ resistor. Find the capacitor, and the output relative to the input at $1.0\,\text{kHz}$ and at $10\,\text{kHz}$ .

Capacitor
Rearrange $f_c = 1/(2\pi R C)$ : $C = 1/(2\pi R f_c) = 1/(2\pi \times 10000 \times 1000) = 1.59\times 10^{-8}\,\text{F} = 15.9\,\text{nF}$ (use $16\,\text{nF}$ ).
$C = \dfrac{1}{2\pi R f_c} = \dfrac{1}{2\pi (10^4)(10^3)} = 15.9\,\text{nF}$
At the cut-off
At $1.0\,\text{kHz}$ the frequency equals $f_c$ , so the output is $0.707$ of the input, i.e. $-3\,\text{dB}$ .
A decade above
At $10\,\text{kHz}$ the frequency is one decade above $f_c$ , so the output has fallen a further $-20\,\text{dB}$ , to about $-23\,\text{dB}$ (roughly $7\%$ of the input).

Result: A $16\,\text{nF}$ capacitor gives $f_c \approx 1.0\,\text{kHz}$ ; the output is $0.707$ of the input at cut-off and about $-23\,\text{dB}$ a decade higher.

Typical mistakes

Taking the output across the resistor (that gives a high-pass) instead of across the capacitor.
Forgetting the factor $2\pi$ in the cut-off frequency formula.

Active revision

Design an RC low-pass filter with a cut-off frequency of $1.0\,\text{kHz}$ using a $10\,\text{k}\Omega$ resistor. State the capacitor value and the output level, relative to the input, at the cut-off frequency.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The RC high-pass filter#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (high-pass filter)

RC high-pass filter circuit

Fig. 5Series capacitor, shunt resistor: the output across

R

is small at low frequency (high

X_C

) and large at high frequency — and the series capacitor blocks DC.

A high-pass filter passes high frequencies and attenuates low ones. It is built from the same two components as the low-pass but interchanged: the capacitor is now in series with the input and the resistor connects the output node to ground, with the output taken across the resistor. At low frequency the capacitor's large reactance drops most of the input, leaving little at the output; at high frequency the reactance is small, so the signal passes to the output almost unattenuated. The topology alone tells you which way it filters.

The cut-off frequency is again

f_c = 1/(2\pi R C)

, the point where

X_C = R

and the output has fallen to

0.707

of the input (

-3\,\text{dB}

). Above

f_c

the filter passes the signal; below

f_c

the output falls at

-20\,\text{dB}

per decade. Notice that the cut-off formula is identical to the low-pass — what differs is which side of

f_c

is passed, and that follows entirely from whether the output is taken across the resistor or the capacitor.

The most common use of a high-pass filter is as a coupling or DC-blocking capacitor. A series capacitor blocks any steady (DC) level while passing the AC signal above the cut-off, which is exactly what is needed to couple one amplifier stage to the next without disturbing their separate bias voltages. Choosing the coupling capacitor so that the cut-off sits below the lowest wanted signal frequency ensures the whole audio band passes while the DC is removed.

High-pass filters also remove unwanted low-frequency content: mains hum, the rumble of a turntable, or a slow drift in a sensor signal. In each case the design task is identical to the low-pass — pick

R

and

C

for the wanted cut-off — but the interpretation is reversed. Being able to state, from the circuit alone, whether a given RC network is a high-pass or a low-pass, and to place its cut-off, is a routine but essential exam skill.

Because both filters share the same cut-off formula and the same first-order

-20\,\text{dB}

per decade slope, they are natural partners: cascade a high-pass and a low-pass with the high-pass cut-off below the low-pass cut-off, and only the band of frequencies between the two cut-offs passes. This band-pass combination, developed in the next section, is the basis of tuning, tone control and channel selection, and shows how two simple first-order blocks combine into a more selective system.

f_c = \dfrac{1}{2\pi R C}

Cut-off frequency

Same formula as the low-pass; above f_c the high-pass passes the signal.

High-pass frequency response

Fig. 6The response rises at

+20\,\text{dB}

/decade below

f_c

, is

-3\,\text{dB}

f_c

(marked), then flattens in the pass band above it.

Worked example

High-pass cut-off and hum rejection

A coupling high-pass filter uses $C = 100\,\text{nF}$ and $R = 1.6\,\text{k}\Omega$ . Find the cut-off frequency and state, with reasoning, whether a $50\,\text{Hz}$ hum and a $2\,\text{kHz}$ tone are passed.

Cut-off
$f_c = 1/(2\pi R C) = 1/(2\pi \times 1600 \times 100\times 10^{-9}) = 1/(1.005\times 10^{-3}) = 995\,\text{Hz} \approx 1.0\,\text{kHz}$ .
$f_c = \dfrac{1}{2\pi (1600)(100\times 10^{-9})} \approx 1.0\,\text{kHz}$
50 Hz hum
At $50\,\text{Hz}$ , well below $f_c$ , the output is heavily attenuated — the hum is largely rejected (about $-26\,\text{dB}$ , since $50\,\text{Hz}$ is roughly $1.3$ decades below cut-off).
2 kHz tone
At $2\,\text{kHz}$ , above $f_c$ , the tone is in the pass band and passes with little loss.

Result: $f_c \approx 1.0\,\text{kHz}$ ; the $50\,\text{Hz}$ hum is strongly attenuated while the $2\,\text{kHz}$ tone passes — a useful hum-rejecting coupling filter.

Typical mistakes

Confusing the high-pass and low-pass topologies — the output is across the resistor for a high-pass.
Choosing a coupling capacitor whose cut-off is above part of the wanted band, so low notes are lost.

Active revision

A high-pass filter uses a $100\,\text{nF}$ capacitor and a $1.6\,\text{k}\Omega$ resistor. Find the cut-off frequency and state whether a $50\,\text{Hz}$ mains hum is passed or attenuated.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Decibels, roll-off and the band-pass filter#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (decibels, band-pass)

Band-pass filter response

Fig. 7A band-pass response from a high-pass at

100\,\text{Hz}

cascaded with a low-pass at

10\,\text{kHz}

: flat between the two

-3\,\text{dB}

cut-offs (marked), falling outside them.

Gains and losses in electronics span such huge ranges that they are usually expressed logarithmically, in decibels. For a voltage ratio the gain in decibels is

G = 20\log_{10}(V_{out}/V_{in})

; for a power ratio it is

10\log_{10}(P_{out}/P_{in})

. The factor of 20 for voltage arises because power depends on the square of voltage. A gain of

20\,\text{dB}

means a tenfold voltage increase,

40\,\text{dB}

a hundredfold, and

-3\,\text{dB}

a fall to

0.707

(half power) — the value that defines a filter's cut-off.

The decibel makes cascaded stages easy: because logarithms add, the overall gain of stages in series is simply the sum of their individual decibel gains. An amplifier of

+26\,\text{dB}

followed by an attenuator of

-6\,\text{dB}

gives an overall

+20\,\text{dB}

, a tenfold voltage gain, without ever multiplying the raw ratios. This additive property is why the decibel is the natural currency of a signal chain and the gain budget of an audio system.

The steepness with which a filter cuts is its roll-off, quoted in decibels per decade (a tenfold change in frequency) or per octave (a doubling). A first-order RC filter rolls off at

-20\,\text{dB}

per decade, equivalently

-6\,\text{dB}

per octave. Cascading

n

first-order stages gives an

n

-th order filter that rolls off at

-20n\,\text{dB}

per decade — a sharper transition between pass and stop bands, at the cost of more components. Reading a roll-off rate off a response graph and relating it to the filter order is a standard A2 skill.

A band-pass filter passes a band of frequencies and rejects those above and below it. The simplest way to make one is to cascade a high-pass filter (whose cut-off

f_L

sets the lower edge of the band) with a low-pass filter (whose cut-off

f_H

sets the upper edge). Between

f_L

and

f_H

both filters pass the signal, giving a flat pass band; outside this band one or other filter attenuates it. The width of the band is

f_H - f_L

, and its centre is where the response peaks.

Band-pass filters are the heart of tuning and channel selection: a radio receiver uses one to pick out a single station's band of frequencies from the crowded spectrum, and an audio tone control shapes the sound by boosting or cutting chosen bands. The design brings together everything in this chapter — reactance sets each cut-off, the decibel measures the response, and the roll-off decides how cleanly the wanted band is separated from its neighbours — which is why it is a favourite subject for extended, evaluate-and-design exam questions.

G_{dB} = 20\log_{10}\!\left(\dfrac{V_{out}}{V_{in}}\right)

Voltage gain in decibels

Factor of 20 for a voltage ratio; use 10 for a power ratio.

\text{Band-pass: } f_L < f < f_H, \quad \text{width} = f_H - f_L

Band-pass filter

High-pass sets the lower edge, low-pass the upper edge.

Worked example

Gain in decibels and a telephone band-pass

An amplifier raises a signal from $20\,\text{mV}$ to $2.0\,\text{V}$ . Express its gain in decibels. It is followed by a band-pass made from a high-pass at $300\,\text{Hz}$ and a low-pass at $3.4\,\text{kHz}$ . State the pass band and the width.

Gain ratio
$V_{out}/V_{in} = 2.0/0.020 = 100$ .
In decibels
$G = 20\log_{10}(100) = 20 \times 2 = 40\,\text{dB}$ .
$G = 20\log_{10}(100) = 40\,\text{dB}$
Pass band
The high-pass sets the lower edge at $300\,\text{Hz}$ and the low-pass the upper edge at $3.4\,\text{kHz}$ , so the pass band is $300\,\text{Hz}$ to $3.4\,\text{kHz}$ — the standard telephone speech band — of width $3400 - 300 = 3.1\,\text{kHz}$ .

Result: The amplifier gain is $40\,\text{dB}$ ; the band-pass passes $300\,\text{Hz}$ to $3.4\,\text{kHz}$ , a $3.1\,\text{kHz}$ -wide band.

Typical mistakes

Using the factor 10 instead of 20 for a voltage ratio in decibels (or vice versa for power).
Setting the high-pass cut-off above the low-pass cut-off, which leaves no pass band at all (a band-stop).

Active revision

An amplifier increases a signal from $20\,\text{mV}$ to $2.0\,\text{V}$ . Express its gain in decibels. A band-pass filter is then made from a high-pass at $300\,\text{Hz}$ and a low-pass at $3.4\,\text{kHz}$ : state the pass band.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

AC circuits and passive filters

Key points

Reading a sinusoid and finding power

Key points

Capacitor reactance and the crossover frequency

Key points

Designing a low-pass filter

Key points

High-pass cut-off and hum rejection

Key points

Gain in decibels and a telephone band-pass

Sources