I can't tell if this is homework, or not. But if you actually are trying to figure out how to design a bridge rectifier system that yields a desired output and ripple specification, and you have complete freedom to build a transformer for it rather than being forced to buy one from a list of readily available units, then the process is something like this:
Specifications:
- \$V_{OUT_{REGULATED}}=5\:\textrm{V}\$ (the final regulated voltage)
- \$V_{DROP_{REGULATOR}}=1.5\:\textrm{V}\$ (the minimum overhead used by the linear regulator)
- \$I_{MAX}=1\:\textrm{A}\$ (the maximum load current to be supported)
- \$V_{RIPPLE}= 500\:\textrm{mV}\$ (the ripple voltage at the input of the linear regulator)
- \$f=60\:\textrm{Hz}\$ (the operating frequency of the mains supply to the transformer)
From that you can work out the rest. For example, let's use the above values to get:
$$\begin{align*}
t_{charge} = \frac{\textrm{sin}^{-1}\left(\frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}}{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}}\right)}{4 \pi f}&\approx 1.6\:\textrm{ms} \\
\\
C = \frac{I_{MAX}}{V_{RIPPLE}}\left(\frac{1}{2 f}-t_{charge}\right)&= 13.5\:\textrm{mF}\\
\\
I_{DIODE_{MAX}} = \frac{\sqrt{2}\cdot V_{RIPPLE}\cdot C}{t_{charge}}&\approx 6\:\textrm{A}
\end{align*}$$
At this point, even though the capacitor isn't a standard value, you can make an estimate for the voltage drop of your rectifier diode. I generally take this to be about \$800\:\textrm{mV}\$ when supplying \$1\:\textrm{A}\$, so in this case I'd guess about:
$$V_{DIODE}=800\:\textrm{mV}+120\:\textrm{mV}\cdot \textrm{ln}\frac{6\:\textrm{A}}{1\:\textrm{A}}\approx 1\:\textrm{V}$$
Now, you can estimate the transformer's RMS value as:
$$V_{RMS} = \frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}+2 V_{DIODE}}{\sqrt{2}}\approx 6.3\:\textrm{V}$$
Luckily, that happens to be a standard transformer value.
But note that this is under load. The capacitor chosen matters here.
I'm going to stop at this point and let you clarify your question some more. Perhaps the above process isn't what you wanted, at all. But it gives you an idea of some of the factors involved. Perhaps someone else understands you better and can provide a better response. But for now, I'm still not sure what you are doing.
I don't see anything unusual here.
You are trying to measure the dc voltage across a diode while giving the bridge an ac input. What you are actually measuring is some kind of average voltage across the diode, alternating between the forward and reverse bias cases.
If your transformer provides 5V rms then you would have a peak voltage of about 7V with no load. After the bridge rectifier a peak ac voltage with no load of about 5.5V would be reasonable. However, as soon as you start to draw current the average voltage, the dc voltage, will drop. That is simply the nature of the beast.
According to www.electronics-tutorials.ws the equivalent dc voltage from a full-wave rectified ac voltage is just 0.9 times the rms value.
Best Answer
Note: I am assuming no filter capacitor is connected.
Well, it sort-of makes sense but you won't get a good reading without a resistive load.
The average voltage of a (full wave rectified) sine wave is 0.9 times the RMS voltage, or about 19.8V in your case (minus a volt or so because of two diodes with little loading), so maybe 19V.
However, with no loading the output voltage of a transformer will be higher than the rated (full current) load, typically by 10-20% for a smallish transformer.
There's another 'however'- the input to your voltmeter will actually have a small capacitor across it and it will act to filter the voltage (since the back leakage of the bridge will be small). So, you'll get a higher than expected voltage, but not reliably.
I suggest you try putting a resistor such as 10K across the output and measure again. I would expect a reading in the 20-22V range with a basically negligible load.