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I'd like to double the 150–220VDC 50mA supply from this Nixie power circuit, for use with electrowetting, but I'm not sure of the best method. Most boost converters I've seen require a low input voltage, and I haven't seen any other circuits that do DC-DC voltage multiplication (though I am almost completely new to electronics).

This seems like a super-simple question, and I'm sure it has a simple answer. Are there just higher-rated components out there I should use? Anything that isn't a big box that requires a quote?

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    \$\begingroup\$ What's powering the \$150\:\textrm{V}_\textrm{DC}\$ supply? Is it your own design? Or COTS? Do you have any other handy sources to work with? Or is this it? And what kind of current compliance do you expect from the \$300\:\textrm{V}_\textrm{DC}\$ supply? (I'm assuming you may not need a precision voltage supply here... but if I'm wrong about that, then say so.) \$\endgroup\$ – jonk Jul 13 '17 at 3:54
  • \$\begingroup\$ I added a link to the supply, which should include a schematic and EAGLE files. I don't have any other sources handy. I was considering perhaps ditching the Nixie and going from 120VAC wall to 240 DC, or some higher multiple, and then using a buck converter. And actually, within certain bounds, precision would be nice. I just need it not to fluctuate too much (it doesn't have to be exact, but I'll be measuring it) \$\endgroup\$ – Jack Lynch Jul 13 '17 at 4:01
  • \$\begingroup\$ @jonk I guess another option would be to employ a smaller boost converter from 5V to 30V, then use that 30V as the supply to the linked schematic, which would give me closer to the range I want. \$\endgroup\$ – Jack Lynch Jul 13 '17 at 4:03
  • \$\begingroup\$ If you have AC at hand, it's reasonable and more vintage to use a straight AC→DC source or doubler. \$\endgroup\$ – Janka Jul 13 '17 at 4:30
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    \$\begingroup\$ If you told us what's the intended application, we could be able to help you better. You still don't mention what's the max current output you need from your HV source and what's its intended load. \$\endgroup\$ – Lorenzo Donati supports Monica Jul 13 '17 at 5:05
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This seems like a super-simple question, and I'm sure it has a simple answer. Are there just higher-rated components out there I should use? Anything that isn't a big box that requires a quote?

There are a few flyback boost designs on the internet such as this one from Linear technology: -

enter image description here

The output voltage can be chosen by setting the FB (feedback) resistor ratio from the output. Read the data sheet.

Offerings from TI are also likely to be similar.

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The boost converter can be controlled in several ways but the simplest one is what is called direct duty ratio control or voltage-mode control. The idea is to close the power SW (the MOSFET) and magnetize the inductor for a time duration labelled \$t_{on}\$ and leave it off for a time duration \$t_{off}\$. Then a new switching cycle takes place, leading to a switching period \$T_{sw}=t_{on}+t_{off}\$. We define the duty ratio \$D\$ by \$D=\frac{t_{on}}{T_{sw}}\$. If you determine the dc transfer function of this converter operating in the continuous conduction mode or CCM (the inductor current never returns to 0 A within a switching cycle), you find the following relationship: \$\frac{V_{out}}{V_{in}}=\frac{1}{1-D}\$. You can see that, in theory, having a duty ratio approaching 1 or 100% should lead an infinite output voltage. In reality, when you close the power switch, there are several losses in the current path such as the inductor equivalent series resistance (ESR) \$r_L\$ and the MOSFET \$r_{DS(on)}\$. They limit the maximum inductor during the on-time: even if you leave the MOSFET on for a long time, the inductor current is clamped. At turn off, you see the diode forward drop \$V_f\$ in series with the inductor which introduces another loss. As such, the dc transfer function complicates when you include these losses. If you plot the new transfer function linking \$V_{out}\$ to \$D\$ then you can observe a so-called latch-up phenomenon:

enter image description here

Despite an increase of the duty ratio, the output voltage drops and you can no longer boost the input voltage. That is the reason why practically-realisable boost coefficients are usually around 4-5 while 15 are less common. If you want to go higher than that, perhaps a flyback transfer would be better suited as the transfer featuring a turns ratio \$N\$ changes the dc transfer function to \$\frac{V_{out}}{V_{in}}=\frac{ND}{1-D}\$.

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  • \$\begingroup\$ The circuit author claims he can go from 12..15V up to 180V@2kΩ. See the tests he's done with his circuit. \$\endgroup\$ – Janka Jul 13 '17 at 6:51
  • \$\begingroup\$ This is indeed a neat circuit and the efficiency is pretty decent despite the large ratio he has adopted. As I said, as long as \$V_{in}\$ is high enough to impose the right inductor current despite the ohmic losses at turn-on, this is fine. Here, \$V_{in}=12\;V\$ so quite a bit of headroom. The latch-up phenomenon becomes pregnant at low input voltages, when the voltage drop on the various elements is no longer negligible. Thank you for pointing this out. \$\endgroup\$ – Verbal Kint Jul 13 '17 at 7:18
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Please understand the topology of this converter first:

schematic

simulate this circuit – Schematic created using CircuitLab

The regulator IC is controlling the switch. First it's closed, we give the inductor a kick with the input current. Then we open the switch and the current has no choice but running through the diode, into the capacitor and output.

Okay, but where does the 1:15 voltage lift comes from? That one is only possible because the impedance of the output circuit is much higher than that of the switch. To maintain the same current, the inductor has to increase the voltage across its terminals, which adds up to U_in. Well, and that pretty much is it.

I think you now understand why you can't simply place another converter at the output. Because then, the impedance of your output circuit drops, and so does the voltage lift of the first stage.

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  • \$\begingroup\$ Interesting. Thanks! So perhaps I should instead multiply the wall AC and then use a buck converter to step it down \$\endgroup\$ – Jack Lynch Jul 13 '17 at 5:41

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