Think about the basics that depend on time constants and duration and the possible differences between your schematic and your actual circuit that would cause 50 to 100 mA current spikes on each edge of plot 1.

(Note to all LTspice users, please get rid of blue plots and use better labels that match the schematic !)

• After some time after t=0 the cap will be charged up with a small Vf drop on the diode 0.4 and -ve pulse = -4.6V.

What will cause spikes in the diode current? (seen by series cap current Ic.)

Answer: Junction Capacitance

What should the peak current spike be anyhow with a few pF across the diode symbol?

• almost 5V /50 = 100 mA. What are you reading in blue 40? to 80 mA? Measurement error and schematic is logically incorrect without a junction capacitance note. However, diode capacitance increase towards the threshold of conduction and decreases with Vr, so it isn't constant but if the ESR of the diode capacitance is small relative to the 50 Ohms, you should always see almost a 100 mA current spike but now spewing noise up to the GHz range.

• The second plot loads the AC coupler to remove all DC faster because the diode is a high impedance in reverse bias. Now the Cap current sees 25 Ohms for positive current and 50 Ohms for negative current, so it is more like a symmetrical square wave current and the diode capacitance is still there for the negative spikes and on top of the offset square currents, but hard to see due to the LTspice plot resolution and colours but now you see some 4 ns effects of reverse recovery time. (check if that is valid on the datasheets)

Think about the basics that depend on time constants and duration and the possible differences between your schematic and your actual circuit that would cause 50 to 100 mA current spikes on each edge of plot 1.

(Note to all LTspice users, please get rid of blue plots and use better labels that match the schematic !)

• After some time after t=0 the cap will be charged up with a small Vf drop on the diode 0.4 and -ve pulse = -4.6V.

What will cause spikes in the diode current? (seen by series cap current Ic.)

Answer: Junction Capacitance

What should the peak current spike be anyhow with a few pF across the diode symbol?

• almost 5V /50 = 100 mA. What are you reading in blue 40? to 80 mA? Measurement error and schematic is logically incorrect without a junction capacitance note. However, diode capacitance increase towards the threshold of conduction and decreases with Vr, so it isn't constant but if the ESR of the diode capacitance is small relative to the 50 Ohms, you should always see almost a 100 mA current spike but now spewing noise up to the GHz range.

• The second plot loads the AC coupler to remove all DC faster because the diode is a high impedance in reverse bias. Now the Cap current sees 25 Ohms for positive current and 50 Ohms for negative current, so it is more like a symmetrical square wave current and the diode capacitance is still there for the negative spikes and on top of the offset square currents, but hard to see due to the LTspice plot resolution and colours but now you see some 4 ns effects of reverse recovery time. (check if that is valid on the datasheets)

Although Falstad's browser JS simulator does not model diode capacitance, it does a better job with plotting results on all parts.

Think about the basics that depend on time constants and duration and the possible differences between your schematic and your actual circuit that would cause 50 to 100 mA current spikes on each edge of plot 1. (Sorry if I read your plots wrong, please get rid of blue plots and use better labels that match schematic !)

After some time after t=0 the cap will be charged up with a small Vf drop on the diode 0.4 and -ve pulse = -4.6V. What will cause spikes on the diode current?

Answer: Junction Capacitance

What should the peak current spike be anyhow with a few pF across the diode symbol?

• almost 5V /50 = 100 mA. What are you reading in blue 40? to 80 mA? Measurement error and schematic is logically incorrect without a junction capacitance note. However , diode capacitance increase towards the threshold of conduction and decreases with Vr, so it isn't constant but if the ESR of the diode capacitance is small relative to the 50 Ohms, you should always see almost a 100 mA current spike but now spewing noise in the GHz range.

Think about the basics that depend on time constants and duration and the possible differences between your schematic and your actual circuit that would cause 50 to 100 mA current spikes on each edge of plot 1.

(Note to all LTspice users, please get rid of blue plots and use better labels that match the schematic !)

• After some time after t=0 the cap will be charged up with a small Vf drop on the diode 0.4 and -ve pulse = -4.6V.

What will cause spikes in the diode current? (seen by series cap current Ic.)

Answer: Junction Capacitance

What should the peak current spike be anyhow with a few pF across the diode symbol?

• almost 5V /50 = 100 mA. What are you reading in blue 40? to 80 mA? Measurement error and schematic is logically incorrect without a junction capacitance note. However, diode capacitance increase towards the threshold of conduction and decreases with Vr, so it isn't constant but if the ESR of the diode capacitance is small relative to the 50 Ohms, you should always see almost a 100 mA current spike but now spewing noise up to the GHz range.

• The second plot loads the AC coupler to remove all DC faster because the diode is a high impedance in reverse bias. Now the Cap current sees 25 Ohms for positive current and 50 Ohms for negative current, so it is more like a symmetrical square wave current and the diode capacitance is still there for the negative spikes and on top of the offset square currents, but hard to see due to the LTspice plot resolution and colours but now you see some 4 ns effects of reverse recovery time. (check if that is valid on the datasheets)