129 Cathode Ray Osc

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Number 129

Cathode Ray Oscilloscope A Cathode Ray Oscilloscope (CRO) is a useful tool for investigating alternating potential difference. The following points are covered in this Factsheet: • finding the period and frequency of an alternating signal; • choosing the correct CRO sensitivity; • calculating peak and rms potential difference; • Lissajous figures and beats are explained, different wave-forms discussed and AC power calculations demonstrated.

Fig 1. Dual beam oscilloscope

potential difference

Time sensitivity per division

time

Potential difference sensitivity per division

After the p.d. is applied, the beam moves 3cm upwards. What is the p.d. of the cell? The sensitivity of the y-plates can be varied. In figure 2, the sensitivity is set to 1V cm-1. This tells us the p.d. of the cell is 3V.

Inside a CRO a beam of electrons passes through an evacuated glass tube and hits a fluorescent screen. Applying an alternating potential difference produces a wave-form on the screen. This wave-form can be used to find the period and frequency of the signal as well as the peak and average potential difference. Two pairs of deflecting metal plates inside the oscilloscope are crucial.

How would a sinusoidally alternating potential difference appear on the screen? The p.d. of the y-plates would therefore vary sinusoidally and the trace is given in Fig 3.

The y-plates The y-plates can deflect the electron beam up or down, in the ydirection.

Fig 3. Aleternating potential difference (no x-deflection)

Fig 2. Applying a constant p.d. to an oscilloscope

OCR Screen

CRO Screen

+ Y plate Beam of electrons - Y plate

Alternating potential difference

deflected trace

The sensitivity of the y-plates in this case is 0.020 Vcm-1. What is the maximum observed potential difference according to this trace? The trace reaches approximately 3.1cm from the origin, giving a maximum peak to trough p.d. of 0.062 V.

orginal trace

External cell Peak potential = difference

A constant p.d. is applied to the y-plates of an oscilloscope. The original, undeflected beam is shown in the centre of the screen.

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peak height oscilloscope p.d. × in cm sensitivity

Physics Factsheet

129. Cathode Ray Oscilloscope The x-plates

How precisely can the peak p.d. for each signal be determined? Assume oscilloscope readings can be made to a precision of 0.1cm. The lower p.d. trace is approximately 0.5cm but could be between 0.4 and 0.6cm; 20% uncertainty. The higher p.d. trace is around 3.0 cm but could be between 2.9 and 3.1cm or 3% uncertainty.

The second pair of plates can deflect the beam in the x-direction. These are normally used to show time, as they can scan the beam from left to right across the screen at varying rates. This is not very useful with a constant p.d, giving a horizontal straight line.

Exam Hint: More precise potential differences can be measured from an oscilloscope by choosing a sensitivity such that the trace has a large range in the y-direction.

Deflection in the x-direction is useful to study alternating potential difference: In Fig 4 the same alternating potential difference is shown, with an x-deflection timebase.

Two inputs are shown in Fig 6.

Fig 4. Alternating potential difference (with x-deflection)

Fig 6. Oscilloscope sensitivity

CRO Screen

CRO Screen alternating potential difference

If the x-sensitivity is 0.02s cm-1, what is the period of this signal? The wavelength is 7cm, giving a period of 0.14s. What is the frequency of this signal? Frequency = 1 / 0.14s = 7.14Hz.

The signals have been deflected up and down to make it easier to measure the period. How precisely can we calculate the periods? Again assume we can make measurements to a precision of 0.1cm.

Frequency = 1/ Period One full wavelength on the lower trace spans around 1.5cm, although this could be between 1.4 and 1.6cm. The period of the lower signal can be determined with an uncertainty of 7%. One full wavelength on the upper trace spans around 4.2cm, although this could be between 4.1 and 4.3 cm. The upper signal can be determined to a precision of 2%.

Two separate traces are shown in Fig 5.

CRO Screen

Choose an oscilloscope timebase scale so one full wave has the widest possible range in the x-direction.

Fig 7. Time sensitivity (Time/div) Example question: What time per division should be used to precisely determine the period of an alternating signal that has a frequency of around 1600Hz? Answer: The period is approximately 1/1600Hz = 6.25×10-4s. Most oscilloscopes have a 10cm wide screen. We need to select the time/division so that at least one full wavelength appears on the screen. Each cm should represent at least 6.25×10-5s, or 63µs. The 50µs setting would not show a complete wave, so we choose the 0.1ms setting.

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Physics Factsheet

129. Cathode Ray Oscilloscope Phase difference: Lissajous figures

Take two identical alternating inputs; apply one to the x-plates and one to the y-plates. The resulting single trace displays the relationship between the two signals.

Fig 8. Lissajous figures

x-deflection y-deflection overall trace

x-deflection y-deflection overall trace

Two signals in phase

x-deflection y-deflection overall trace

Two signals 900 out of phase

Two signals 1800 out of phase

When the two inputs are in phase, there would be a large amplitude peak. However, the two inputs would cancel when they are 180o degrees out of phase with each other.

Lissajous figures represent phase difference between two signals: In phase = / 180o out of phase = \ 90o degree out of phase, different amplitudes = 0 90o out of phase, identical amplitudes =

Fig 10. Beats: combination of two inputs with similar frequency

Frequency difference: Beats In Fig 9, inputs 1 and 2 have similar frequencies. To compare the two inputs, you could use a dual beam oscilloscope. The resulting combined trace is shown in the figure. Input 1 moves slowly out of phase from input 2 and then moves slowly back into phase.

waves in phase, large resulting amplitude

Fig 9. Two signals with similar frequency

waves out of phase, small resulting amplitude

Different waveforms

Input 1

Look at the three traces in Fig 11. Can you think of a way of altering the original signal to produce the second trace?

Fig 11. Different wave-forms Orginal signal rectification Input 2

Half wave rectification Input 1 and 2

Full wave rectification

How would this look if both inputs were applied to a single channel, with just a single trace?

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129. Cathode Ray Oscilloscope A diode is used to allow current in one direction only. A network of Practice Questions four diodes is used to produce the third trace. 1. Draw a 20Hz, 30Vrms signal on an oscilloscope screen, choosing appropriate p.d. and time scales.

Fig 12. How to achieve half and full-wave rectification

2. Describe and explain the Lissajous figure caused by two signals with identical amplitude and frequency that are 90o out of phase with each other. Alternating input

Half wave rectifier circuit

3. What time/division setting would you choose to precisely determine the period of a 2000Hz signal?

load

load

4. What V/division setting would you choose to precisely determine the peak p.d. of a signal where V0 is about 1.3mV?

Full wave rectifier circuit

5. Describe the action of the x-plates for timebase mode. 6. Explain, using a diagram, the principle of beats. Alternating input

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Alternating input

7. What is the power of a circuit where V0 = 130V and R=1kΩ?

load

8. With what precision can you determine the peak of a 10V signal where V/div = 10V/cm and 2V/cm? positive potential difference

Answers

Negative potential difference

1. V0 = Vrms × √ 2 = 30 × √ 2 =42.4V Frequency = 20Hz Period = 1 / Frequency = 1/20 = 0.05s Peak and average potential difference P.d. scale: 42.4V over 5cm of screen so p.d. sensitivity of over The relationship between power, current and potential difference is 8.5V/div is required: 10V/div P=IV. We can use V/R=I to substitute for I in the power equation, Time scale: period = 0.05s giving P=V2/R. One full wave across 10cm screen, time/div of 0.005s or 5ms is required. But potential difference continually varies with an alternating input: which p.d. do we use? We use an average p.d., known as Vrms. 2. In text. The relationship between the peak p.d. V0 and this average p.d. is 3. 2000Hz Frequency Vrms = V0/√ 2. Period = 1/2000 = 5×10-4s or 500µS. Time/ div = 50µS The relationship between peak,V0, and average, Vrms, √ 2. potential difference is given by Vrms = V0/√ 4. V0 = 1.3mV V/div around 1.3mV/5cm = 0.26mV/div is required Most oscilloscopes can achieve around 1mV/div sensitivity Example question: If Vrms in the UK mains supply is 240V, what is the peak p.d? 5. In text 240V × √2 = 339V. Answer: Vrms × √2 = V0 6. In text 7. P=Vrms2/R Vrms = V0/√ 2 = 130/√ 2 = 91.9V 91.9V2/1000Ω = 8.45W 8. 10V at 10V/div results in a 1cm peak 1cm ± 0.1cm = 10% precision 10V at 2V/div results in a 5cm peak 5cm ± 0.1cm = 2% precision

Acknowledgements: This Physics Factsheet was researched and written by J Carter The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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