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As titled, the diode in my half-wave rectifier did not rectify the AC signal. Vd1 is still AC signal. Therefore Vout ~ 0V. Why could that happen? The diode's parameters are SPICE parameters.

The strange thing is, with exactly that diode's model, for a shunt-diode rectifier, it worked!!! The results are attached. Could anyone give me a hint to solve this problem? Thank you very much.

enter image description here enter image description here

Results updated when removing DC feed

All of you were right about the DC feed. Thank you very much for pointing out the reason. But there is still one question. Why did not the diode block the negative half-cycle of Vd as it should do in the theory? As observed in my following results, in negative half-cycle Vout amplitude is smaller than that of Vd, but it's not 0. I also removed the DC block, and Vout = Vd. How can it be?

enter image description here enter image description here

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    \$\begingroup\$ @MarcusMüller You're the second person to suggest this. I got Nick's permission to migrate it there. \$\endgroup\$ – Mike Waters Jan 10 '19 at 17:28
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the diode in my half-wave rectifier did not rectify the AC signal.

I don't see this in your results.

It looks like some of the time Vd and Vd1 are equal, and other times, Vd1 is about 3 V above Vd, which is exactly what you expect for half-wave rectification.

If you zoom in the time scale on the plot to see the individual cycles, this might be more clear.

Therefore Vout ~ 0V. Why could that happen?

I think this is because of the DC Feed. This device models an effectively infinite valued (maybe it is actually modeled as 1 GH or something) inductor. In a finite time, its current won't change. This component is used to provide a short in an operating point analysis, but an open in every other kind of simulation.

Run the transient simulation for a couple years of simulation time and you might see a non-zero output at Vout.

Or change the DC feed to an inductor with a realistic value (and calculate how long you'll have to run the simulation to see a response).

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  • \$\begingroup\$ You and JRE were right about the DC feed. Please see my updated results. But does why the diode still allow the negative half-cylce of Vin to pass through it? In theory, Vout should be 0 when Vin is negative. That's why this topology is called "half-wave rectifier". \$\endgroup\$ – Minh Lam Jan 11 '19 at 2:09
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Is it possible your "DC Feed" is set to block AC and DC?

Vout (which should be a blue line) seems to be on the zero volt line. It is hard to tell, but the zero volt line looks blue instead of black.

That would fit in with the DC feed blocking everything. With no load, you would see the original signal coming through the diode. No load means no current, which means both sides of the diode have the same voltage.

This is, I think, a trick of the simulator. The voltmeters (or whatever is measuring Vd1) are ideal, and draw no current. With a real voltmeter, a tiny bit of current would flow and you would see rectification.

Either remove the DC feed, or add a resistor from Vd1 to ground.

Either of those should make your diode work properly.


After your update, I am out of ideas.

The datasheet for the diode you are using says it is optimized for 915MHz to 5.8GHz operation, but that shouldn't stop it from working at lower frequencies.

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  • \$\begingroup\$ The meter is a current probe, not a voltmeter. If it's ideal, it has 0 resistance. \$\endgroup\$ – The Photon Jan 10 '19 at 18:15
  • \$\begingroup\$ @ThePhoton: The voltage at Vd1 is being measured by some kind of voltmeter in the simulation - that's the voltmeter I'm talking about. The probe for Iout is of course a current meter that looks like a short circuit. But, it doesn't matter if the DC feed is set to block AC and DC. \$\endgroup\$ – JRE Jan 10 '19 at 18:23
  • \$\begingroup\$ OK. ADS doesn't really have the idea of a "voltmeter". It just lets you plot the results of the simulation, which include the voltages at all the nodes. \$\endgroup\$ – The Photon Jan 10 '19 at 18:28
  • \$\begingroup\$ Thank you very much for your suggestion, JRE. I've removed the DC feed, please see my update for futher discussion if any. \$\endgroup\$ – Minh Lam Jan 11 '19 at 1:29
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I think the output of diode (or a diode circuit) doesn't necessarily output the so-called "half-wave". It depends on the load resistance and capacitance.

For a rectifier circuit (like your case), if you remove the load resistance or using a quite large load resistance, you could probably see the "half-wave" rectification.

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