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I am learning how to work with rectifiers and I built a half-wave rectifier to test voltages in simulation.

When I add a probe at the output of a diode and a voltmeter across that diode, the voltages are different. It seems fair but I can't explain why the voltages are different.

What is (in technical terms) the output voltage of the diode (VF1) (it is a rectified signal now) and the voltage across the diode (VM1)?

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  • \$\begingroup\$ Well, what do you know about diodes to help explain the circuit? What is your current understanding of the circuit, so we can start from there? \$\endgroup\$
    – Justme
    Sep 14, 2020 at 7:43
  • \$\begingroup\$ Diode dont pass negative half-cycle of AC and so positive half-cycle of AC is pass through diode in forward-biased mode. So as expected, we can see the signal with no negative component beside output of diode. And also we have capacitor in the circuit. Capacitor acts like open circuit when they meet DC ( if it is fully-charged) and it also acts passing AC. So AC component of input signal goes through the ground. Because of DC component meet high reactance, current pass through output. Please correct me if I have some technical faults or missing sides in my explanation. \$\endgroup\$
    – sapphire
    Sep 14, 2020 at 9:33

1 Answer 1

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VF1 shows a voltage related to ground (VF1 - Vground), VM1 shows another voltage not directly related to ground (VF1 - VF2), which is the diode's output voltage minus its input voltage.

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  • \$\begingroup\$ So we can say that VF1 is output signal of diode and also half-wave rectifier not VM1, am I right \$\endgroup\$
    – sapphire
    Sep 14, 2020 at 10:42
  • \$\begingroup\$ VF1=VM2 Voltage related to ground. \$\endgroup\$
    – csabahu
    Sep 14, 2020 at 13:10

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