The Oscilloscope - Additional AS Level Oscilloscope questions to try

Why should you maximize the volts/division scale when determining the amplitude of a wave form?

The larger the trace the smaller the error in measuring its height. If you can measure the trace height to within 2mm then the voltage error you record will be smaller if their are less volts per division on the screen.

What effect does changing the volts/division scale from 1 volt per division to 5 volts per division have on the display of the incoming wave on the screen? (Does the wave amplitude increase, decrease or remain the same?)

Think about this logically - if the trace represents 10V then it would be 10 divisions high.... changing to 5V/div would make it only 2cm high...

The wave amplitude would decrease.

What effect does changing the volts/division scale from 1 volt per division to 5 volts per division have on the incoming wave itself?

No change at all - you are altering the way you display the signal - not the signal itself.

One cycle of a waveform occupies five divisions of an oscilloscope screen. The timebase dial is set to 1 ms/div. Calculate the frequency.

T = 5ms

f = ¹/T

= ¹/(5 x 10^-3)

= 200 Hz

One cycle of a waveform occupies two divisions of an oscilloscope screen. The timebase dial is set to 5 ms/div. Calculate the frequency.

T = 10ms

f = ¹/T

= ¹/(10 x 10^-3)

= 100 Hz

A waveform occupies six divisions of an oscilloscope screen when the timebase is switched off. The voltage gain is set to 0.5V/div. Calculate the peak to peak current being tested if the resistance of the circuit is known to be 10 ohm.

Peak to peak voltage = 6 x 0.5 = 3V

V= IR

I = ^V/_R = 3/10 = 0.3 A

A waveform occupies two divisions of an oscilloscope screen when the timebase is switched off. The voltage gain is set to 10 mV/div. Calculate the peak current being tested if the resistance of the circuit is known to be 1 kΩ

Peak to Peak Voltage = 20 mV

Peak voltage = 10 mV

V_o = I_oR

I_o = ^V/_R = 10 x 10^-3/10³ = 1.0 x 10^-6A

I_o= 1.0 μA

A waveform occupies six divisions of an oscilloscope screen when the timebase is switched off. The voltage gain is set to 0.25V/div. Calculate the rms current being tested if the resistance of the circuit is known to be 19 ohm.

Peak to peak voltage = 6 x 0.25 = 1.5V

V_o = I_oR

I_o = ^V/_R = ^1.5/₁₉ A

Now, I_RMS = I_o/√2

= ^1.5/(19 x√2)

= 0.056A

A waveform occupies three divisions of an oscilloscope screen when the timebase is switched off. The voltage gain is set to 20 mV/div. Calculate the rms current being tested if the resistance of the circuit is known to be 1.4 kΩ.

Peak to peak voltage = 60 mV

V_RMS = V_o/√2 = 60/√2

V_RMS = I_RMSR

I_RMS = V_RMS /_R

= (60/√2) /(1.4 x 10³)

= 0.030A

Sketch an oscilloscope screen showing a 200 Hz frequency, 4.0 V rms trace. Include all calculations you have done to work out what you should draw, and don't forget to indicate clearly what the timebase and voltage gain settings are!

f = 200 Hz

T = 1/f = 1/200 = 5.0 x 10^-3s = 5.0 ms

If the timebase was set at 1.0 ms/div one full wave would be across 5 divisions.

V _RMS = 4.0 V

Vo = √2V _RMS

V (Peak to Peak) = 2√2V _RMS = 2√2 x 4 = 11.3 V

If the gain was set to 2V/div the trace would span 5.65 divisions