Notes · PhysicsUK · A-Levels

Electronics (optional)

This is one of the five optional A-Level modules (AQA Paper 3, Section B); a student studies just one option. Electronics builds sensing and signal-processing systems: semiconductor devices and sensing sub-systems, the advantages of digital over analogue signals and the conversion between them, the operational amplifier and its standard circuits, and filters that shape a signal by frequency.

4 sections~11 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·131313 / 13

Exam profile

AO1 · Describe semiconductor devices, analogue and digital signals and operational-amplifier circuits (optional module)AO2 · Calculate op-amp gains, potential-divider outputs, filter cut-off frequencies and ADC resolutionsAO3 · Analyse sensing and signal-processing circuits and evaluate the advantages of digital transmission

Operators:describecalculateexplaindeterminedesignanalyse

basic level

This is an optional A2 module examined in Paper 3 Section B; it is studied only if chosen as the school's option. It requires sensing sub-systems, analogue and digital signals and the op-amp.

higher level

The full option demands quantitative op-amp gain calculations, ADC resolution, filter cut-off frequencies and the analysis of complete sensing and processing systems.

Depth

Text

Contents · 4 sections

Electronics (optional)

Discrete semiconductor devices and sensing sub-systems#

StandardAQA 7408 3.13.1DfE GCE Physics - electronics option

A light-sensing potential divider

Fig. 1An LDR and a fixed resistor form a divider across the supply; the output tapped between them varies with light level and feeds the processing stage.

An electronic system is built from sub-systems: an input transducer that turns a physical quantity into an electrical signal, a processing stage that acts on it, and an output transducer that turns the result back into a physical effect. Thinking in these blocks - sensing, processing, output - lets a complex system be designed and understood one stage at a time.

The sensing stage usually uses a potential divider with a transducer whose resistance changes with the measured quantity: a thermistor for temperature (resistance falls as it warms), a light-dependent resistor (LDR) for light (resistance falls as it brightens), or a strain gauge for force or extension (resistance rises as it stretches). The divider converts the resistance change into a voltage the next stage can use,

V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}V_{\text{in}}

Semiconductor switches control larger loads from these small signals. The MOSFET (metal-oxide-semiconductor field-effect transistor) is a voltage-controlled switch: a small voltage on its insulated gate turns a much larger current on or off between its other terminals, drawing almost no current from the sensing circuit. The bipolar transistor is a current-controlled equivalent. Either lets the tiny signal from a divider switch a heater, motor or lamp.

The design choice in a sensing sub-system is which resistor is the transducer and what value to give the fixed resistor: that value sets the operating point - the temperature or light level at which the output reaches the switching voltage of the next stage. This is why the same three blocks can make a frost alarm, an automatic light or a strain-measuring bridge.

V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}\,V_{\text{in}}

Sensing potential divider

Converts a transducer's resistance change into a voltage.

Worked example

Output of a temperature sensor

A thermistor ( $R_1 = 8.0\,\text{k}\Omega$ cold, $2.0\,\text{k}\Omega$ hot) is in series with a fixed $4.0\,\text{k}\Omega$ resistor across a $6.0\,\text{V}$ supply, with the output across the fixed resistor. Find the output when cold and when hot.

When cold
$V_{\text{out}} = \dfrac{4.0}{8.0 + 4.0}\times6.0 = \dfrac{4.0}{12}\times6.0 = 2.0\,\text{V}$ .
When hot
$V_{\text{out}} = \dfrac{4.0}{2.0 + 4.0}\times6.0 = \dfrac{4.0}{6.0}\times6.0$ .
$V_{\text{out}} = 4.0\,\text{V}$

Result: The output rises from $2.0\,\text{V}$ when cold to $4.0\,\text{V}$ when hot - a signal the processing stage can act on.

Typical mistakes

Getting the divider ratio upside down, or forgetting which way the transducer's resistance changes with the quantity.
Assuming the sensing circuit can drive a large load directly - it needs a transistor or op-amp to switch real power.

Active revision

An LDR (resistance $12\,\text{k}\Omega$ in dark, $0.50\,\text{k}\Omega$ in bright light) is in series with a fixed $2.0\,\text{k}\Omega$ resistor across a $6.0\,\text{V}$ supply, output across the fixed resistor. Find the output in dark and in bright light.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Analogue and digital signals and conversion#

StandardAQA 7408 3.13.2DfE GCE Physics - analogue and digital signals

Sampling an analogue signal

Fig. 2An analogue signal is measured at regular instants (the dots); each sample is then coded as a binary number by the ADC. A higher sampling rate captures the waveform more faithfully.

An analogue signal varies continuously, taking any value within a range - like the voltage from a microphone. A digital signal takes only discrete values, in practice two (high and low, 1 and 0). The great advantage of digital signals is noise immunity: because a receiver need only decide whether each bit is high or low, small amounts of added noise can be removed by regenerating clean pulses, whereas noise added to an analogue signal is indistinguishable from the signal itself and accumulates.

To process a real-world analogue quantity digitally, it is converted by an analogue-to-digital converter (ADC). This happens in two steps: sampling, in which the analogue voltage is measured at regular time intervals, and quantisation, in which each sample is assigned the nearest of a fixed set of levels and coded as a binary number. A digital-to-analogue converter (DAC) does the reverse, reconstructing an analogue signal from the digital codes.

The quality of the conversion depends on two things. The number of bits sets the resolution: an

n

-bit converter has

2^n

levels, so more bits give finer voltage steps and less quantisation error. The sampling rate sets how faithfully rapid changes are captured: sampling too slowly loses detail, so the rate must be well above the highest frequency in the signal.

Digital signals underpin modern communication and computing: they can be stored, copied and transmitted without degradation, encrypted, compressed and error-checked. This is why music, images and telephone calls, all originally analogue, are digitised - the small loss at conversion is far outweighed by the robustness of the digital form.

\text{number of levels} = 2^n

ADC levels

An $n$ -bit converter has $2^n$ discrete levels.

\text{resolution} = \dfrac{\text{voltage range}}{2^n - 1}

ADC resolution

The voltage represented by one step.

Worked example

Resolution of an ADC

A 10-bit ADC converts an input range of $0$ to $10.23\,\text{V}$ . Find the number of levels and the resolution.

Number of levels
$2^{10} = 1024$ levels, giving $1023$ steps.
Resolution
$\text{resolution} = \dfrac{10.23}{1023}$ .
$\text{resolution} = \dfrac{10.23}{1023} = 0.010\,\text{V} = 10\,\text{mV}$

Result: The ADC has $1024$ levels and a resolution of $10\,\text{mV}$ per step.

Typical mistakes

Confusing the number of levels ( $2^n$ ) with the number of steps between them ( $2^n - 1$ ) when finding resolution.
Thinking digital signals are noise-free - they are noise-tolerant because they can be regenerated, not immune to noise.

Active revision

An 8-bit ADC converts voltages in the range $0$ to $5.1\,\text{V}$ . Find the number of levels and the resolution (voltage per step).

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

The operational amplifier#

AdvancedAQA 7408 3.13.3DfE GCE Physics - operational amplifiers

Inverting operational-amplifier circuit

Fig. 3The input passes through

R_{\text{in}}

to the inverting input; the feedback resistor

R_f

runs from the output back to it. The gain is

-R_f/R_{\text{in}}

The operational amplifier (op-amp) is a high-gain voltage amplifier with two inputs - inverting (

-

) and non-inverting (

+

) - that amplifies the difference between them. The ideal op-amp has infinite open-loop gain, infinite input resistance (it draws no current from the previous stage) and zero output resistance. Its enormous raw gain is tamed and made useful by negative feedback, feeding part of the output back to the inverting input.

The inverting amplifier feeds the input through a resistor

R_{\text{in}}

to the inverting input, with a feedback resistor

R_f

from the output. Negative feedback holds the inverting input at the same potential as the earthed non-inverting input (a 'virtual earth'), and because no current enters the op-amp, the current through

R_{\text{in}}

equals that through

R_f

. This gives the closed-loop gain

\dfrac{V_{\text{out}}}{V_{\text{in}}} = -\dfrac{R_f}{R_{\text{in}}}

- set entirely by the two resistors, with the minus sign showing the output is inverted.

The non-inverting amplifier feeds the signal to the non-inverting input, keeping the output in phase, with a gain

\dfrac{V_{\text{out}}}{V_{\text{in}}} = 1 + \dfrac{R_f}{R_{\text{in}}}

. A summing amplifier feeds several inputs through separate resistors into the virtual earth, giving an output proportional to the (weighted) sum of the inputs - the basis of an audio mixer and of the DAC.

Without feedback the op-amp acts as a comparator: its output swings to one supply rail or the other depending on which input is higher, since even a tiny difference is amplified enormously. This makes it ideal for switching - triggering an alarm or a heater when a sensor's voltage crosses a reference. In every case the output cannot exceed the supply rails, so a large input drives the amplifier into saturation.

\dfrac{V_{\text{out}}}{V_{\text{in}}} = -\dfrac{R_f}{R_{\text{in}}}

Inverting amplifier gain

Set by the two resistors; the output is inverted.

\dfrac{V_{\text{out}}}{V_{\text{in}}} = 1 + \dfrac{R_f}{R_{\text{in}}}

Non-inverting amplifier gain

Always at least 1; the output is in phase.

Worked example

Gain and saturation of an inverting amplifier

An inverting amplifier has $R_{\text{in}} = 5.0\,\text{k}\Omega$ and $R_f = 40\,\text{k}\Omega$ , with $\pm 9\,\text{V}$ supply rails. Find the gain, and the output for inputs of $-0.80\,\text{V}$ and $-1.5\,\text{V}$ .

Gain
$\dfrac{V_{\text{out}}}{V_{\text{in}}} = -\dfrac{R_f}{R_{\text{in}}} = -\dfrac{40}{5.0} = -8.0$ .
For -0.80 V input
$V_{\text{out}} = -8.0 \times (-0.80) = +6.4\,\text{V}$ - within the rails, so valid.
For -1.5 V input
$-8.0 \times (-1.5) = +12\,\text{V}$ , which exceeds the $+9\,\text{V}$ rail, so the output saturates.
$V_{\text{out}} = +9\,\text{V (saturated)}$

Result: The gain is $-8.0$ : the $-0.80\,\text{V}$ input gives $+6.4\,\text{V}$ , but the $-1.5\,\text{V}$ input saturates the output at $+9\,\text{V}$ .

Typical mistakes

Dropping the minus sign on the inverting-amplifier gain, or using the non-inverting formula for an inverting circuit.
Forgetting that the output saturates at the supply voltage - the gain applies only while the output stays within the rails.

Active revision

An inverting amplifier has $R_{\text{in}} = 10\,\text{k}\Omega$ and $R_f = 150\,\text{k}\Omega$ , with supply rails of $\pm 12\,\text{V}$ . Find the gain and the output for inputs of $+0.50\,\text{V}$ and $+1.5\,\text{V}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Filters and signal processing#

AdvancedAQA 7408 3.13.4DfE GCE Physics - signal processing

Response of an RC low-pass filter

Fig. 4The output falls above the cut-off frequency: at

f = f_c

the gain has dropped to

0.707

, and it continues to fall for higher frequencies.

A filter passes some frequencies and attenuates others, shaping a signal by its frequency content. The simplest is a passive filter made from a resistor and a capacitor. Because a capacitor's reactance falls as frequency rises, an RC network divides the signal differently at different frequencies. Taking the output across the capacitor gives a low-pass filter (passing low frequencies, blocking high); taking it across the resistor gives a high-pass filter (the reverse).

The boundary between passed and blocked is the cut-off (corner) frequency,

f_c = \dfrac{1}{2\pi RC}

, at which the output power has fallen to half (the output voltage to

1/\sqrt{2} \approx 0.707

of the input). It is set entirely by the resistance and capacitance, and it is the same

RC

time constant met in capacitor charging - a fast-charging network passes higher frequencies.

Filters have countless uses: removing high-frequency noise from a sensor signal (a low-pass filter), blocking a slowly drifting offset (a high-pass filter), and separating the bass and treble sent to different loudspeakers (a crossover). Combining low- and high-pass sections gives a band-pass filter that passes a chosen range.

Signal processing also includes amplification (with the op-amp), and conditioning to match a signal to the next stage. Active filters combine an op-amp with resistors and capacitors to give sharper cut-offs and gain together, and the op-amp's high input resistance stops the next stage 'loading' and distorting the filter - which is why real processing chains interleave filters and amplifiers.

f_c = \dfrac{1}{2\pi RC}

Cut-off frequency

The frequency at which the output falls to $1/\sqrt{2}$ of the input.

Worked example

Cut-off frequency of a filter

An RC low-pass filter uses $R = 10\,\text{k}\Omega$ and $C = 100\,\text{nF}$ . Find its cut-off frequency.

Apply the cut-off formula
$f_c = \dfrac{1}{2\pi RC} = \dfrac{1}{2\pi (10\times10^{3})(100\times10^{-9})}$ .
Evaluate the denominator
$2\pi \times 10^{4} \times 10^{-7} = 2\pi \times 10^{-3} = 6.28\times10^{-3}$ .
$f_c = \dfrac{1}{6.28\times10^{-3}} = 159\,\text{Hz}$

Result: The cut-off frequency is about $160\,\text{Hz}$ ; frequencies well above this are attenuated.

Typical mistakes

Confusing which output (across the capacitor or the resistor) gives a low-pass or a high-pass filter.
Forgetting the factor $2\pi$ in the cut-off frequency, or muddling $f_c$ with the time constant $RC$ .

Active revision

An RC low-pass filter uses $R = 4.7\,\text{k}\Omega$ and $C = 100\,\text{nF}$ . Calculate its cut-off frequency.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Electronics (optional)

4 sections~11 min reading time3 competenciesLevel Standard 2 · Advanced 2

Electronics (optional)

Key points

Output of a temperature sensor

Key points

Resolution of an ADC

Key points

Gain and saturation of an inverting amplifier

Key points

Cut-off frequency of a filter

Sources

Electronics (optional)

Key points

Output of a temperature sensor

Key points

Resolution of an ADC

Key points

Gain and saturation of an inverting amplifier

Key points

Cut-off frequency of a filter

Sources