# Measuring an AC-voltage at the ouput of a full bridge rectifier

I've a full bridge rectifier and I put the following function into it:

$$5\sin(100\pi t)\tag1$$

So, at the output I get:

$$\left|(5-2V_d)\sin(100\pi t)\right|\tag2$$

Where $V_d$ is the voltage over one diode.

Now, why do I measure (with a Fluke multimeter) a DC-voltage of $1.4$ volts and an AC-voltage of $0.5$ volts. Why do I measure an AC-voltage, is it not completely DC?

• The AC will be a measure of the ripple. What do you mean by a "fluxe multimeter". Do you mean "Fluke"? (Note the capital 'F'.) – Transistor Apr 23 '18 at 16:05
• @Transistor Yes sorry it is Fluke. So it will measure the RMS ripple? – Looper Apr 23 '18 at 16:07
• Is it a 'true RMS' meter? Also your equation (2) is not correct. – Spehro Pefhany Apr 23 '18 at 16:49
• @SpehroPefhany What should the right equation be? – Looper Apr 23 '18 at 17:55
• @SpehroPefhany I do not understand what you mean?! – Looper Apr 23 '18 at 18:20

$$v(t) = \begin{cases}|(5\sin(100πt)|-2V_D ,& |5\sin(100πt)| > 2V_D \\0 ,& \ |5\sin(100πt)| <= 2V_D\end{cases}$$
• That is not what I mean, haha :). I mean why does the hole function get's lowered by $2Vd$ and not only the amplitude of the sine function? – Looper Apr 23 '18 at 19:19