# 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 ton / toff (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 Vsat.

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?

• read Application Report SLVA252B–September 2006–Revised November 2007 Oct 29, 2016 at 20:40
• @TonyStewart.EEsince'75 Thank you, that's very helpful in general, but I don't see how it answers my question (how to choose a suitable switch) Oct 29, 2016 at 20:43
• Because the App Notes have all the details like Vce and IL which Q must meet with low power or RdsOn or Rce from I^2R.. 1.8Ohm is OK for 20mA , so you start with V-I in/out specs , max temp rise spec then calculate heatsink requirement.. i.e. show us your work Oct 29, 2016 at 20:57
• My first reaction to a boost converter going from as little as $4.75\:\textrm{V}$ to $200\:\textrm{V}$ output will be the mind-numbing duty cycle required, losses, and inductor design for something like this. I haven't read the datasheet. But that's my gut, right now. I think a different topology or perhaps do it in 2 or 3 stages.
– jonk
Oct 29, 2016 at 21:24
• @jonk And by "different topology" you mean flyback - which makes sense, but I have no idea how to design a flyback converter and am pretty sure it's beyond my skill set. Other than that, what? 5V -> 40V and then a 5x voltage multiplier? Oct 29, 2016 at 21:27

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.

In the end, this design was clearly not going to work, so I 'stepped up' to the task of trying to design a flyback converter:

This seems to be working (at least in LTspice) and all I have left to do is select an output diode.