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boost converter


scuba14c
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I have been trying to build a boost converter, but am having problems. If you aren't familiar with these, they use an inductor, diode, a semiconducting switch, and a 555 timer. The voltage builds up in the inductor, and is pumped into the capacitors. The diode prevents it from flowing out. I can get it to charge to 45V, and then it won't go any farther. With a 12V input, it does work, but should be able to charge to a much higher voltage. The load on the capacitor doesn't matter either. I have used 100uF and 3600uF banks and both stop at 45V. I have a 100uH inductor and a fast recovery diode. This is the schematic I built it off of...

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I have changed the duty cycle, but it also changed the frequency. The duty cycle is anywhere from 66.6-99.1%. Because the frequency changed, I can't tell if the duty cycle or the frequency makes the most difference. The timer that works the best is 2186Hz @ 66.66% duty. Is it better to have a larger voltage driving the chip; right now it is a 9V battery. I am also not using the 1200V 20A IGBT, it only charged to 25V or so. I have replaced it with (2) 400V 8A NPN transistors in parallel to reduce the heating of the switch. The NPNs work the best of any switch I have tried.

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Hi scuba14c

apart from the HV switch, the inductor design is about the most important part of what you're trying to do, can I suggest you have a look at http://www.semiconductors.philips.com/acrobat_download/applicationnotes/AN96052_1.pdf
and particularly the section on inductor design

Best of Luck

Ed

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Not sure where to find it. I actually gave up on the IGBT and am using two 400V 8A NPN transistors in parallel because they gave me a higher voltage. If you tell me where it would be, I'll give it to you. I think the switch may be the problem after trying to alter the duty cycles on the timers.

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Guest Alun

scuba14c,
What voltage output do you require, and at what power level?

There are lots of ICs avliable to help you, here are some links:
Google results
http://www.linear.com/pc/productDetail.do?navId=H0,C1,C1003,C1042,C1031,C1061,P2410
http://www.amplicon.co.uk/dr-prod3.cfm/subsecid/10090/secid/6/groupId/12114.htm
http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1545
http://www.chipcatalog.com/National/LM1577.htm
http://www.sparkfun.com/datasheets/IC/MC34063A.pdf
http://www.datasheetcatalog.com/datasheets_pdf/A/D/P/3/ADP3000.shtml

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The capacitor bank I have been using to this point is 330V 1900uF. When I charged this with a photoflash charger, it took 4 minutes. The new bank is 350V @ 56700uF or 700V @ 14400uF. I would like it to be able to charge to 700V, but keep the charge time down to about 30 seconds. The schematic for the boost converter said it would charge a smaller bank in 3 seconds.

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Guest Alun

So basically you want a high voltage at a high energy level, my advice to you would be to use a transformer rather than an inductor, a mains transformer with a 6V secondary (connected to the output transistor) and 240V primary connected to the capacitor bank via a high voltage rectifier would be perfect. You might want to stop the voltage from going too high and destroying the capacitors, you can add an auto switch off circuit for this.

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Guest Alun

A variac is no better than using the mains directly.

I didn't mean directly using the mains anyway. ::)

I meant using a mains transformer as a step up transformer.

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Guest Alun

Pulsed DC will do the job just as well as AC, you can get a couple of kV from a mains transformer as it will bahave as a flyback transformer.

You know how just using the back EMF from an indutor can give you a high voltage, a transformer is simply two inductors coupled to each other. The output in voltage is the ratio of primary turns to secondary, for example a 240V mains transformer with a 12V has a turns ratio of 20:1 so 240V at the primary becomes 12V at the secondary. If you use it the other way round and put 12V into what would normally be the secondary you'll get 240V at the primary (but you'd be using it as the secondary)

The flyback effect..
As I explained before the back EMF in a normal inductor when given pulses of DC current can be 100s of volts, the same is true with the mains transformer. Say for example if the mains 12V coil on a transformer gives a back EMF of 100V and the turns ratio is 1:20 then that 100V will be multiplied by 20 and give you 2000V - this is how ignition coils and TV fly back transformers work. You will need to limit the voltage to the maximum voltage rating of you components but this is easy enough to do with a comparator circuit.

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How do I get the pulsed DC, a 555 timer with low frequency? From what you have said, this seems like it would be better than the boost converter. Will any type of transformer work? I have one sitting in the basement from a dehumidifier. How do you make the comparator? I've heard of them, but couldn't find a diagram of one online.

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Couldn't figure out what happened, thanks for reposting it.

I've got a couple more questions about the pulsed DC and transformer. If it is pulsed DC, do I get DC or AC output? If it is AC, would I use an inverter to change it to DC? How much current could I expect to get from it? If it is powered with batteries, the current would have to be lower than the battery, right?

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Hi scuba14c

pulsed DC usually refers to switching on and off the primary of a transformer which is supplied by a DC voltage such that when there is a change of current (switch on or off) that rate of change of current becomes available at the secondary.  The secondary is electrically separated from the primary and the on/off action induces a voltage which alternates, it's not a sine wave however, it's nearer a triangle wave. 

If it is AC, would I use an inverter to change it to DC?


No need,  use diodes to rectify the AC waveform to DC.

How much current could I expect to get from it?


This is the $64,000 question, the answer is that the amount of energy that you put into the primary circuit will be what you can expect from the secondary less any losses. ie you can't get out more than you put in!

Suppose that the most power that you can extract from a battery is 1 Watt (purely ficticious and just for argument) and the battery has a voltage of 10v, here I=P/V  = 100mA.  This means that if you apply this voltage and current to a perfect DC-DC converter you would be able to get the same power out, because you have a transformer you can increase the output voltage as much as you like. Suppose you wanted 1000v output, no problem, but the energy balance has to be maintained and you'll find that 100 times more voltage means that you have 100 times less current available. 

Look at the Joules per second level and suddenly it gets clearer

Best of Luck

Ed
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Guest Alun

No need,  use diodes to rectify the AC waveform to DC.


While you don't need to convert the AC to DC, you do need a diode to stop the capacitor discharging though the secondary.
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