This very simple voltage multiplier has ideal components. This isnt an actual circuit and it doesn't represent an actual circuit.

Is there a way to suppress the 8amp current spike thru the voltage source without reducing charge 600uF capacitor charge duration?

 

Can you fit 10 Lbs of olives into a 5 Lb bag?

No.

The energy required to charge the cap has to be delivered over a period of time.
Reduce the current and you slow the rate of charge.

 

Remember that your real life signal source will have a non-zero output impedance so that in itself will provide some current limiting in real life.


Adding a tiny bit of inductance in series with the capacitor (with appropriate shunt damping resistor is needed) would reduce the glitch.

 

I would not call it a spice glitch, but it is a common simulation flaw, which is that ideal components don't reflect what we actually can build. So in an "ideal circuit" with ideal components the current is only limited by the source specification provided. Ideal components like these can only be found in the Kingdom Of Utopia, which is not a place that is accessible to real people.
So to reduce that spike you can add the effective series resistance found in real capacitors and the non-linear, non-ideal diode characteristics, and you will find the spike is smaller and wider. In fact, you should add the usual inrush current limiting resistor that is normally included in many rectifier-capacitor circuits.

 

600 uF is very much for grid Voltages. Normal average fuze will blow-up at the glance. Thus how to act - incandescant (Edisson) lamp in the series, what (if harsh need) may be shortened afterward, or alternatively variac if hardly under 100W power, or inductance (probably one from "long" luminiscent lamps), or simply resistor of dozen Ohms (may short it after few seconds by a proper relay, or thyristor device if more serious power).

 

LTspice calculated everything correctly. Only for a more correct picture is necessary to reduce the maximum step. Look at the voltage difference across the capacitor. Before connecting the load, through which the current goes on the capacitor is almost equal to the magnitude of the source voltage (~ 12V), and then the capacitor discharges. Do you want the capacitor to discharge and no current flows through it?

 

See