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in my electronics lab at university we were studying diodes and I stumbled upon a behaviour that I cannot understand nor explain (I'm I physics undergrad, not and EE).

The circuit we were analysing is a simple single diode rectifier (a 1N4148 signal diode) with a 100 ohm load in parallel with a 330uF electrolytic cap and a 50Hz input sine wave. I noticed that the input signal was clipped. To eliminate the doubt that the problem was to be caused by something other than the circuit (signal generator, oscilloscope etc) I simulated it in LTSpice and got the same result, whitch is the one in the image. clipped signal

Why is the circuit doing this? If I measure directly the input signal shouldn't I see a 50Hz sine wave?

If it helps the circuit is this: enter image description here

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If I measure directly the input signal shouldn't I see a 50Hz sine wave?

If you were measuring the input signal at source then it would be a pure sinewave but you have output resistance set for your generator of a value 50 ohms and this will allow the output voltage of the source to become distorted.

You can prove this by setting Rser to 0 and using an external 50 ohm resistor in series with V1 - now V1 will be pure and undistorted.

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  • \$\begingroup\$ So the problem is the internal resistente of the signal generator! 50ohm is the expected input impedance of the signal generator that we were using (whitch from what I now is a pretty standard value). But the question remains, why is it doing that? I cannot wrap my head around it \$\endgroup\$ – Davide Morgante May 24 '18 at 18:17
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    \$\begingroup\$ It does "that" because when the sinewave voltage rises to an input level that exceeds the forward voltage drop of the diode plus the residual voltage on the capacitor, current is suddenly starting to be taken and that increase in current causes a sudden increase in volt drop across the series resistor. Hence the input voltage peak (while current is charging the capacitor) is somewhat clipped. \$\endgroup\$ – Andy aka May 24 '18 at 18:26

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