Electronic – what is the dc component in the output of diode half wave rectifier circuit will it be the average value or the rms value of the output waveform

Let's take a look at the signal waveforms:

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}

There is never more than a single diode voltage drop across either diode (disregarding the small voltage across the meter itself for sake of the discussion). In this configuration he diodes will not break down due to reverse overvoltage as for each half period one diode will be conducting.