DC - DC High Voltage Boost Converter Cap values?

[jwhitmore]

Hello All, I wanted to design a DC to DC Boost converter which could boost 12V input to a high output voltage, (well 200V so a high voltage to me). I've played with a basic boost circuit just to get a feel for it. I'll attach a rough circuit diagram of what I think I'm doing.

My question is to do with the values that should be used for the Capacitors on the Input and Output. I'm going to have a high current 12V input supply so I'm going to wind my own Inductor, so I can play with different values of inductance. I'm not sure of the Inductor value because a higher value of inductance will give a higher output Voltage, but at an increased input impedance. But given I'm winding the Inductor I can play with that and see what results I get from different values of inductance.

The Caps however I'm not sure about. I guess in both cases the Caps will act as energy storage devices. In the case of the Input smoothing the load on the power supply. So the higher capacitance the better? But if I make the caps too large then the capacitor might never get a change to fully charge up.

I guess I should really bite the bullet and try and get my head around spice. That might be the answer.

[snx]

Its possible to reach 200V from 12V, but i suggest using a proper smps IC,

One would be the MC34063  which is usually cheap, i used this circuit in my project: http://www.swissnixie.com/projects/NixieTester/SN_NixieTester.png
But be aware, this circuit produces a high pitch noise from the inductor which can be annoying.

Another solution would be the MAX1771 which i used in this project: http://www.swissnixie.com/projects/SUNIX-S/SN_SUNIX_S_R1.2_WTB_SCHEMATIC.png
The components are carefully matched and are expensive, but more efficient and much less noise!

Also it depends on how much current you need. If you don't need much current, check page 27 of this datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/8580fa.pdf

Or you could use the flyback technology, but it requires a transformer which usually takes up alot of space (standart model) or needs to be manufactured specifically for you.
See page 30 of this datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/3757afe.pdf

[JackJones]

Another solution would be the MAX1771 which i used in this project: http://www.swissnixie.com/projects/SUNIX-S/SN_SUNIX_S_R1.2_WTB_SCHEMATIC.png
The components are carefully matched and are expensive, but more efficient and much less noise!

MAX1771 has a pretty sensitive feedback pin though, it might not be the easiest thing to get the layout correctly. I have heard this from other people and my testing seems to suggest the same.

Some people have made HV supplies for nixies with just a standard 555, probably not the best for efficiency, but should be easier on the layout.

I recently made a 180V supply for a nixie clock with a TPS40210, there's an excellent writeup here: https://jan.rychter.com/high-voltage-power-supply-for-nixie-tube-projects (not my blog)

I did use a MC34063 for the version 1 of the clock, and it didn't really have an audible noise with the frequency I used it at. Maybe a tiny whine, but certainly not audible when inside the case.

[jwhitmore]

Maybe I should explain what I'm actually trying to do. I'm using an Infineon Profet chip which is meant to be able to switch 40Amps at 12V (Or there abouts) I wanted to test that out by pushing it to its limit, but have nothing in the shape of an electronic load which would do that. So I had the idea of making my own high power load. Now ordinarily if you're pissing away 12*40 or about 500 Watts you'd need a heat sink to soak up that power. Those heat sinks can be quite expensive so I first decided that a kettle, (for boiling water, for making a cup of tea) is actually designed for pissing away power, and turning it into hot water. So I thought I could make a load that increased the output DC voltage and the kettle would just soak it up. That assumes that you can put DC Current into a kettle's heater element. I know it's normally an AC device but a kettle is a few quid so I'll happily blow one up proving to myself that this is the stupidest idea I've ever had.

So that was my first idea. But now I'm thinking that instead of the load testing, resulting in a cup of tea, I could forget about the kettle and push the load current into Lithium battery packs charging them up. In that scenario I'd not have to boost the voltage to a very high voltage. But I might need a number of battery packs and charge controllers to soak up 40Amps.

Yeah both of these could be really bad ideas, but a load that can sink 40Amps ain't cheap

[jwhitmore]

Yes there are inverters but I'm not just trying to drive a load. I'm trying to create an adjustable electronic load to test a design. So given the example of using a kettle, to sink the energy, if the heater coil has a resistance of 20 Ohms I would hope to be able to control the output voltage of the DC-DC Converter to control the input current to the DC-DC Converter. So for example if I want to test my design switching a current of 15Amps at 12V I'd need to output (12V*15A) Watts into the 20 Ohm load, Assuming that the DC-DC is 100% efficient, which obviously it ain't. So for 15Amps I'd need an output voltage of the square root of ( 12V * 15 A * 20Ohm) = 60Volts?

If I now wanted to increase the load to 20Amps the DC-DC would have to increase its output Voltage to the square root of (12V * 20Amps * 20 Ohms) or about 69V.
To go all the way up to 40Amps the DC Output voltage would have to be square root of (12V * 40Amps * 20 Ohms) or about 98V.

Obviously these calculations assume 100% efficiency in the DC-DC converter, which ain't the case so you'd have to measure the current actually being drawn and setting Output Voltage accordingly. But I'd hope that the bulk of the power would end up in the kettle. Not that you're going to get a cup of tea at an output voltage of 98V, given that the kettle is designed to boil water at 240V AC. But a kettle is a lot cheaper then a 500W heat sink.

[jwhitmore]

Or you could use the flyback technology, but it requires a transformer which usually takes up alot of space (standart model) or needs to be manufactured specifically for you.
See page 30 of this datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/3757afe.pdf

Thanks a million for all your suggestions. The one above looks very interesting as page 30 has an output voltage of 350V, a lot more then I had intended using. I hadn't intended using a SMPS IC as I wanted to be able to dynamically adjust the load current and hence the DC-DC Output Voltage. For that reason I was thinking of a uC. I might not have thought this through completely but the uC would have a target load current, measure the actual current and set a PWM output accordingly to match the two. So a POT could be used as an input to the uC to set the desired load current.

I'll do some more reading, before I can this idea, or have a go. Thanks again for your suggestions, very helpful.

jwhitmore

[Zero999]

Yes there are inverters but I'm not just trying to drive a load. I'm trying to create an adjustable electronic load to test a design. So given the example of using a kettle, to sink the energy, if the heater coil has a resistance of 20 Ohms I would hope to be able to control the output voltage of the DC-DC Converter to control the input current to the DC-DC Converter. So for example if I want to test my design switching a current of 15Amps at 12V I'd need to output (12V*15A) Watts into the 20 Ohm load, Assuming that the DC-DC is 100% efficient, which obviously it ain't. So for 15Amps I'd need an output voltage of the square root of ( 12V * 15 A * 20Ohm) = 60Volts?

If I now wanted to increase the load to 20Amps the DC-DC would have to increase its output Voltage to the square root of (12V * 20Amps * 20 Ohms) or about 69V.
To go all the way up to 40Amps the DC Output voltage would have to be square root of (12V * 40Amps * 20 Ohms) or about 98V.

Obviously these calculations assume 100% efficiency in the DC-DC converter, which ain't the case so you'd have to measure the current actually being drawn and setting Output Voltage accordingly. But I'd hope that the bulk of the power would end up in the kettle. Not that you're going to get a cup of tea at an output voltage of 98V, given that the kettle is designed to boil water at 240V AC. But a kettle is a lot cheaper then a 500W heat sink.

It would be easier to do it the other way round. Use a lower value resistor and a buck converter to control the power output.

[jwhitmore]

It would be easier to do it the other way round. Use a lower value resistor and a buck converter to control the power output.

I'm not sure what you mean. Perhaps I'm focused on the wrong problem here. What I'm thinking is that if I'm going to test a circuit which is meant to be able to conduct 40Amps at 12V I have to be able to connect up a load which will draw that amount of power and dissipate it, in some form or other. I assume that an actual electronic load would use a heat sink something like [1] to dissipate that energy. I was trying to use a cheaper solution where a water cooled heater element is much cheaper.

If I used a buck converter and a low value resistance I'd still need a low value resistance which could sink 12*40 Watts? This all started with me thinking it'd be great if I had an actual load which could burn up about 500W, then I went to get a cup of tea So the Russell Hobbs [2] can actually burn 3000W and only costs a fraction of [1]. I thought all I had to do was convert my 40Amps @ 12V into something I could force into [2], hence I thought Boost. This thread has pointed out flybacks and stuff which I have to research.

But using a Busk and a low value resistance I think I'd still need [1]?

[1] https://ie.farnell.com/fischer-elektronik/la-9-200-24v/heat-sink-fan-cooled-24v/dp/1222517
[2] https://www.amazon.co.uk/Russell-Hobbs-Cambridge-Kettle-20070/dp/B00CFHNI8M/ref=sr_1_6?s=kitchen-appliances&ie=UTF8&qid=1541513212&sr=1-6&keywords=kettle

[max_torque]

Yeah both of these could be really bad ideas, but a load that can sink 40Amps ain't cheap

i can think of two loads that can sink 40A and neither is what i'd call expensive!

1) 10 of 50W 12 down lighter bulbs, in parallel.  Only downside is these will get hot, so need to be mounted in some form of metal plate or case (but they only need a round hole to mount them, so that's easy)

2) a roll of car brake pipe.  Change the resistance (and hence current) by just moving the point at which you connect to that roll up and down the roll, run a small pipe from a cold tap to run water through the pipe to cool it.


I've successfully used option 2) to sink 5kW from a large 12V lead acid battery........


(your step up option is complex, expensive, potentially dangerous and very likely to go bang. You need to spend your time on your device, not on making kit that might or might not work to try to test your device!)  And using a battery is fraught with potential (pun intended) dangers, like fire from over charging to huge short circuit currents from accidentally connecting unmatched potentials together....)

[ejeffrey]

I'm not sure what you mean. Perhaps I'm focused on the wrong problem here. What I'm thinking is that if I'm going to test a circuit which is meant to be able to conduct 40Amps at 12V I have to be able to connect up a load which will draw that amount of power and dissipate it, in some form or other. I assume that an actual electronic load would use a heat sink something like [1] to dissipate that energy. I was trying to use a cheaper solution where a water cooled heater element is much cheaper.

A water cooled heating element is fine, but there is no real reason to use an electric kettle designed for 120 V / 240 V.

You can find low voltage immersion heater elements.  Here is a 12 V / 300 watt (25 amp) one that I found in 30 second of googling: https://www.hurricanewindpower.com/dc-water-heater-element-12-volt-300-watt-submersible-diversion-heater/

You could use two of these in parallel with a buck converter to control the load.

Alternately you could wire your own with some nicrhome heating wire.  You could dunk it in water or wrap it around a glass rod and just let it get really hot.  With that you could easily vary the load within a reasonable range simply by adjusting where you connect the terminals.