Examkin

Question

Difficulty: MediumOscilloscopes

A technician is evaluating an AC ripple voltage on a DC power supply using an oscilloscope. The vertical sensitivity control (Volts/Div) is set to 50 mV/div50\text{ mV/div}. If the displayed waveform occupies a total of 6 vertical divisions6\text{ vertical divisions} from its lowest trough to its highest crest, what is the peak voltage (VpkV_{\text{pk}}) of the ripple?

  1. A
    300 mV300\text{ mV}
  2. 150 mV150\text{ mV}Answer
  3. C
    25 mV25\text{ mV}
  4. D
    1.5 V1.5\text{ V}

Answer

The peak voltage of the ripple is 150 mV150\text{ mV}.
Multiplying the total vertical deflection (6 divisions6\text{ divisions}) by the vertical sensitivity (50 mV/div50\text{ mV/div}) yields a peak-to-peak voltage of 300 mV300\text{ mV}. Peak voltage (VpkV_{\text{pk}}) measures the amplitude from the reference centerline to the positive crest, which equals half of the peak-to-peak voltage: 300 mV2=150 mV\frac{300\text{ mV}}{2} = 150\text{ mV}.

Step-by-Step Solution

1
Calculate the peak-to-peak voltage (Vp-pV_{\text{p-p}})
Vp-p=6 divisions×50 mV/div=300 mVV_{\text{p-p}} = 6\text{ divisions} \times 50\text{ mV/div} = 300\text{ mV}
The total vertical span from crest to trough represents the full peak-to-peak voltage.
2
Calculate the peak voltage (VpkV_{\text{pk}})
Vpk=Vp-p2=300 mV2=150 mVV_{\text{pk}} = \frac{V_{\text{p-p}}}{2} = \frac{300\text{ mV}}{2} = 150\text{ mV}
Peak voltage is the amplitude measured from the center baseline to either peak, which is half of the peak-to-peak value.

Key Concept

Determining peak voltage from peak-to-peak vertical graticule measurements on an oscilloscope.
Estimated Time:1m 0s
Rate this question

Topics

Oscilloscopes

More questions from this topic