Current limiting in simple blocking oscillator HV generator circuit
The impedance of the voltage multiplier plays a part in limiting the current from the high voltage output.

I've had a look at the impedance of the Cockcroft-Walton voltage multiplier before, and found the following equation:

$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{2 \pi fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$

What that does is to compute your output voltage from several parameters:


\$E_{pk}\$ the peak voltage of the AC input (not the peak to peak voltage, but the simple peak voltage.)
\$I_{load}\$ - the load current.
\$f\$ - the frequency of the AC input 
\$C\$ - the capacitance of your individual capacitors (in farads.)
\$n\$ - the number of stages in your multiplier.
\$V_{f}\$ - the forward voltage of your diodes.


That applies to the commonly used half wave Cockcroft-Walton voltage multiplier.

The relationship between the number of stages and the impedance is not obvious just from looking at the circuit.

The output impedance is this part:

$$\frac {4n^3 + 3n^2 - n}{2 \pi fC} $$

The maximum output voltage of the multiplier is given by this part:

$$ E_{out} = 2nE_{pk} - 2nV_{f}$$

Figure your maximum voltage, then use the impedance to figure the voltage drop at the supply current your tube needs. Vary the capacitance, the number of stages, and the frequency to get a multiplier that delivers 1600 VDC at whatever current your tube needs.



The current drawn from the low voltage supply will be higher than the current out of the high voltage side.

I never measured it while I was doing my experiments, so I don't know for sure how much higher.

I'd expect the input current to be at least \$N \times I_{out}\$ where \$N\$ is the number of stages of your multiplier.  Since capacitors leak, the input current will be higher.



I almost expect the oscillator and the driver for the transformer to draw more than 1mA.