## Half wave bridge rectifier - Sine wave Amplitude problem

Hello pros ^^,

I played around with PSpice and found this weird thing ><.

I designed this circuit:

Please note that the amplitude of the sine wave is 5V
Then, I simulate it and traced RL & Vs and the result is:

Later, I changed the amplitude to 5mV
And this is what I get:

If I delete the Vs trace and left only the RL trace:

My question is, since this is a half wave rectifier, why is there still negative parts of the sine wave shown at the RL trace??

The negative part of the RL trace isn't there when amplitude = 5V, which is correct since the circuit is a Half-wave rectifier but why when amplitude = 5mV, the negative part of the RL trace is shown??

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 Quote by deejay220989 Hello pros ^^, I played around with PSpice and found this weird thing ><. I designed this circuit: Please note that the amplitude of the sine wave is 5V Then, I simulate it and traced RL & Vs and the result is: Later, I changed the amplitude to 5mV And this is what I get: If I delete the Vs trace and left only the RL trace: My question is, since this is a half wave rectifier, why is there still negative parts of the sine wave shown at the RL trace?? The negative part of the RL trace isn't there when amplitude = 5V, which is correct since the circuit is a Half-wave rectifier but why when amplitude = 5mV, the negative part of the RL trace is shown??
Since you are using a real and not an ideal diode, the current for negative voltages is a few nanoamperes, instead of zero, so you have a small voltage drop across the 5k resistor.

 you can verify this by zooming in on the 0 voltage when you applied 5 volts.. if it has enough data points and your able to zoom in into the nanovolts you will probably see the same thing. Some current will actually pass through

Recognitions:

## Half wave bridge rectifier - Sine wave Amplitude problem

Also interesting is that the red trace is leading (occurs before) the green one.

This tells you that at least part of the leakage is due to the capacitance of the diode when it is reverse biassed.

Try making the 5K a 500 K and see if this effect becomes more obvious. In the extreme case, the red trace could lead the green one by 90 degrees.