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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?

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    \$\begingroup\$ 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'.) \$\endgroup\$ – Transistor Apr 23 '18 at 16:05
  • \$\begingroup\$ @Transistor Yes sorry it is Fluke. So it will measure the RMS ripple? \$\endgroup\$ – Looper Apr 23 '18 at 16:07
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    \$\begingroup\$ Is it a 'true RMS' meter? Also your equation (2) is not correct. \$\endgroup\$ – Spehro Pefhany Apr 23 '18 at 16:49
  • \$\begingroup\$ @SpehroPefhany What should the right equation be? \$\endgroup\$ – Looper Apr 23 '18 at 17:55
  • \$\begingroup\$ @SpehroPefhany I do not understand what you mean?! \$\endgroup\$ – Looper Apr 23 '18 at 18:20
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Equation (2) should be:

$$v(t) = \begin{cases}|(5\sin(100πt)|-2V_D ,& |5\sin(100πt)| > 2V_D \\0 ,& \ |5\sin(100πt)| <= 2V_D\end{cases}$$

The exact reading of a voltage that is not a pure sine wave (zero average voltage) on the AC range of a multimeter depends on whether the multimeter is AC or DC coupled and whether it is true-RMS or average-reading, corrected to RMS.

Example (resistive load of 1K):

enter image description here

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  • \$\begingroup\$ Where does that formula comes from? \$\endgroup\$ – Looper Apr 23 '18 at 19:07
  • \$\begingroup\$ @Looper Comes from my brain. Is there a problem with it? \$\endgroup\$ – Spehro Pefhany Apr 23 '18 at 19:18
  • \$\begingroup\$ 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? \$\endgroup\$ – Looper Apr 23 '18 at 19:19
  • \$\begingroup\$ The input voltage to the bridge has to exceed 2 diode drops or the output stays at zero. So there are flat sections with voltage zero between the rectified half sine waves (assuming a resistive load to deal with diode leakage and capacitance). \$\endgroup\$ – Spehro Pefhany Apr 23 '18 at 19:25

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