I have built a half-wave rectifier with a 5V peak sine input, 1kHz. In theory the RMS of the output should be 5V/2 = 2.5V, but the oscilloscope only measures 2.01 V. What could be the reasons for such a difference? 0.5 V seems like a pretty significant difference given the scale.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Please add a schematic and datasheets. \$\endgroup\$
    – Matt Young
    Commented Jun 9, 2013 at 4:12
  • \$\begingroup\$ It is 1N4004 diode, a simple half-wave rectifier with a 1kilo-ohm load resistor.....schematic: + source -> diode -> load resistor -> - source. output is the node linking diode and resistor \$\endgroup\$ Commented Jun 9, 2013 at 4:16
  • \$\begingroup\$ It seems like it may have to do with the 0.6V sillicon diode voltage drop. However I don't really understand the logic behind this. What does Vpeak/2 represent exactly if we need to subtract 0.6V? \$\endgroup\$ Commented Jun 9, 2013 at 4:25
  • \$\begingroup\$ What is the value of the load resistor? \$\endgroup\$
    – Matt Young
    Commented Jun 9, 2013 at 4:27
  • \$\begingroup\$ The load resistor is 1 kilo-ohm \$\endgroup\$ Commented Jun 9, 2013 at 4:28

4 Answers 4


Let's take a look at the signal waveforms:

enter image description here

You are right there is a diode voltage drop, let's assume for all intents and purposes the diode forward voltage drop is \$0.635V\$.

To compute the RMS voltage:

$$ V_{rms} = \sqrt{\frac{1}{p} \int_0^p V(t)^2 dt} $$

where \$p\$ is the period (in this case 1ms).

What is the output voltage?

Let's assume for a second that when \$V_{IN} < V_{DIODE}\$, \$V_{OUT} = 0\$. This isn't quite true, but should get us close to the correct answer.

So our output voltage for one period is:

\begin{equation} V_{OUT} = \left\{ \begin{array}{lr} 0 & : 20\mu s < t\\ 5 \sin(1000 \cdot 2 \pi t) - 0.635 & : \text{otherwise}\\ 0 & : t > 480\mu s \end{array}\right. \end{equation}

plugging into the \$V_{rms}\$ calculation,

\begin{equation} V_{rms} = \sqrt{\frac{1}{1 ms}\int_{20\mu s}^{480 \mu s}(5 \sin(1000 \cdot 2 \pi t) - 0.635)^2 dt} \approx 2.1V \end{equation}

The minor difference in calculated values here and your measured values are due to the assumptions I made about diode behavior (constant diode voltage drop, \$V_{OUT}\$ behavior when diode isn't saturated), as well as component behavior not being ideal, nor having exactly the same characteristics as those I chose for the calculations.

Ok, what was the average voltage across the same time period?

\begin{equation} V_{avg} = \frac{1}{p} \int_0^p V(t) dt\\ V_{avg} = \frac{1}{1 ms}\int_{20\mu s}^{480 \mu s}(5 \sin(1000 \cdot 2 \pi t) - 0.635) dt \approx 1.287V \end{equation}

  • 1
    \$\begingroup\$ +1, a sincere response, good explanation for anyone to grasp. \$\endgroup\$ Commented Jun 10, 2013 at 5:33

Your diode is going to drop about 0.5 to 0.7 V when it is conducting (depending on the type of diode), so what you're measuring is about right.

A "half-wave" rectifier doesn't mean Vo(pk) = 1/2 * Vin(pk), it just means you get one half of the wave's cycle. There are always some losses when rectifying a wave form.


Using a perfect diode model (no volt-drop), a full wave rectified sinewave "heats" a load resistor the same as the inputted sinewave hence, its RMS value is also Vpk/sqrt(2).

A half wave perfect diode scenario heats the load half the amount so, square the RMS value of the sinewave, divide it by 2 and then take the square root. RMS(half wave) = Vpk/2 just like the OP says in his question.

But, the effect of the diode lowers Vpk to possibly 4.5V (light load) and the waveform only presents itself for maybe 45% of the cycle. Heating reduces to 4.5 * 0.45 = 2.025V RMS.

It's down to how much of the cycle the "modified" waveform presents itself. This is due to the diode and I've taken a guess at 45%, I've also taken a guess at it reducing the peak value by 0.5V (not unreasonable for light loads). Say the diode dropped 0.6V and the waveform presented itself for 44% percent of the time, the answer would be 1.936V.

Helloworld922 calculated it at 2.1V. The scope (presumably a digital type with RMS feature) reckoned it was 2.01V - all pretty much the same in my book. The 1N4007 has a reverse recovery time of 30us - has this been factored in? The answers are near enough to substantiate the theory.

  • \$\begingroup\$ I did not factor the recovery time into my calculations, though I do agree that 2.1V and 2.01V are essentially the same. I can see where the OP got the Vpk/2 from now. \$\endgroup\$ Commented Jun 9, 2013 at 18:38

The OP and some of the replies used Peak and RMS for different parts of the question. If the input is 5V peak the rms is 5/1.414=3.536. Across the resistor for Ideal diode it would be 2.5/1.414= 1.768.

So, I think he was meaning 5 V RMS not peak, and the thought that it should be 2.5 is correct for the Ideal diode. Now that would change some numbers in the other replies, but they are good. Keep the item measure the same for clarity.

The posters that say 2.1 and 2.01 are the same are talking estimates, not good enough for precision measurements, but ok for sanity checks.


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