I have this reading of a oscilloscope for a half wave rectifier:

Pk-Pk voltage is 17.6V.

The channel is set to 5V division. I asked my dad who is a electric engineer and he told me that Vp = 17.6/2 = 8.8V because the oscilloscope shows the Pk-Pk voltage of the unrectified sine wave. However I have my doubts about what he said because if I count "boxes" from top to bottom until I reach 0V I can count 3.something ->Vp = 3.something x5 = 17.6V. Any help?

  • \$\begingroup\$ What help you need? The oscilloscope measures peak-to-peak of the waveform it sees, and it sees the half-wave rectified waveform. \$\endgroup\$
    – Justme
    Apr 3, 2022 at 15:09
  • \$\begingroup\$ Waht is the AC voltage at the input to the rectifier. The peak voltage of an AC voltage is about 1.4 times the RMS voltage, and we normally specify AC as the RMS voltage. \$\endgroup\$ Apr 3, 2022 at 15:34

3 Answers 3


Normally the "Vpp" function on a scope displays, as you'd expect, the peak to peak voltage of whatever waveform is on screen, which is the difference between maximum voltage and minimum voltage of your waveform.

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Vpp is the red arrow in each plot.

In case of waveform 1 Vpp is the difference between peak positive and peak negative voltage.

For waveform 2, although voltage is always positive, you can still define Vpp as the difference between maximum and minimum.

Waveform 3 is an example where Vpp could be measured wrong, if the peaks are sharp and brief, and the scope doesn't have a high enough sample rate, it could soften the edge and miss the actual peak value. So if it looks like that, and you want an accurate value, a bit of double-checking is recommended.

In your case, waveform 4, since the minimum is close to zero, Vpp is equal to peak voltage.

What your dad says implies that the scope would know that the waveform is a halfwave rectified sine, and it would give you the Vpp of the original sine. But the scope doesn't know that. The algorithm is always the same no matter what's on screen: max(voltage)-min(voltage).


Your oscilloscope shows the 0V level with the arrow on the left.

The waveform is unipolar and has peak voltage +17.6V (which is also equal to the peak-to-peak voltage in this particular case).

Try switching the input to AC-coupled and see the difference.


We need to be clear whether we are talking about the total signal, or just the AC component of it. In various circumstances, either might be more useful.

The peak voltage of the AC component of the reading is half the peak to peak voltage. That is, the AC component swings 8.8 V above and below the mid-level of the AC component.

Another way to describe this particular signal is that its positive peak is at +17.6 V, and its negative peak is 0 V.

When dealing with rectifiers, and the power supply filters that often come after them, we often want to know about the residual ripple that is put onto the circuit's power supply lines. That tends to be expressed as the peak, or the peak to peak, of the AC component only.


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