mosfet - MC34063 help - step-up with an external switch

I'm attempting to design a ~200V, relatively low-current (10mA expected, I'm allowing a bit more though) power supply for some Nixie tubes, and I've decided on using an MC34063 step-up converter. It is my understanding that to reach >40V some kind of external switch is necessary. By looking at specs and at several other designs like this, I have chosen the following components:

I should add that I am relatively inexperienced in this area, and I'm trying to use this as a learning opportunity.

Looking at the formulae on the MC34063 datasheet, the very first calculation for t_on / t_off (related to duty cycle) is:

V    + V  - V
 OUT    F    IN (min)
---------------------
   V         - V
    IN (min)    sat

Assuming these calculations are even valid for an external-switch configuration, I'm stuck with how to determine V_sat.

If the saturation voltage from the datasheet (1V typical, 1.3V max) adds with the gate-source threshold voltage of the IRF720 (2V min, 4V max), the worst case scenario well exceeds my minimum input voltage (4.75V, I'm allowing 5V ± 5%). This immediately smells like it's not going to work.

Upon further examination, all the designs I've been referencing have either a higher input voltage (~12V) or use another external transistor to drive the MOSFET.

Does anyone have any advice on how to choose a more suitable external switch, or how to calculate these parameters correctly for an external switch configuration?

Answer

When designing a boost converter, you can choose a switched inductor (aka flyback or switched boost) for reasonable size inductors up to 100W or a feedforward boost regulator with a stepup transformer and opto or tertiary winding feedback.

You must consider for any type that the voltage ratio is linear while the impedance ratio is squared so your source drivers must be <5% of the load Z referred back to source for high efficiency.

Voltage ratio of 200/5 = 40 implies an impedance ratio of 1600 @5V so a rated load of 200V/10mA=20K appears as 20k/1600=12 Ohm load to 5V source thus 5% of this is 600 mOhm while your part is 1.8 Ohm which dissipates 3x what I recommend for switch losses and a good size little heat is needed.

So the higher the input voltage, the lower the switch losses from input V*I. or a lower RdsOn SMD switch.

A flyback design can be considered suitable here or a feed forward step up transformer. There are many choices, but keep in mind the impedance ratios and input referred switch losses.