step up converter

May 28, 2007 8 Replies

Hello,



I am building a step converter, the circuit works and is based on the following web site:

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My circuit boost from 5v to 150v, bit it takes about 45 seconds to reach the maximum voltage with the proper duty cycle. I found out by using a bigger coil I get a faster result. Is there a way to reach the



150v within 10 seconds or less? or is this how it usually works and we cannot do anything about it..

Thanks



ken



Use a push-pull converter configuration for a much faster response time.

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time.http://schmidt-walter.fbe.fh-darmstadt.de/snt/snt_eng/snteng4a.pdf

Okk thank you

ken

site:

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Strewth!. I didn't think these things could get that horribly complicated. Any chance of you posting somewhere your actual circuit?.

site:

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Figure 1 on the link above.

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with a 2n3055 and a 3300UF 350v elec. cap, no resistor at the load. That simple. The coil I wound myself with 300 feet of magnetic wire.

ken

site:

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above.http://services.eng.uts.edu.au/~venkat/pe_html/ch07s3/ch07s3p1.htm

(at some point you'll need to know your inductor value)

You've complete control over how fast the capacitor charges up. At 150V your capacitor stores a whopping 40 Joules of energy. A one second recharge time could be actioned by storing 40J in the inductor and switching across once per second (an impractical 0.3H and 17amps). Or storing 4mJ and switching across 10000 times a second. A 10 second recharge would get you down to a more reasonable couple of amps. Joules energy stored in the inductor is equal to 1/2 x L x current^2. (it's the inductor 'peak' 'current at the point where the transistor is switched OFF, that counts.)

Basically it's how many amps you can get through the inductor when the transistor is ON and the number of charging/switching cycles you can put up with each second. More power implies a lower inductor value as ... Amps per second rise through inductor = Voltage across inductor / L. I.e low L values allow a high current (hence energy) build up in a short time. Somewhere there is a happy balance. You've got to look for the sweet spot.

With the 300' of wire and old 2N3055 transistor, your lazy charge problem could be down to voltage loss across the ON transistor and high wire resistance. You've only 5V to play with and at the needed couple of amps or so, an odd volt and a few ohms means the inductor sees only a fraction of the full voltage and the design is lost.

Hi,

I did tried different inductor values, I actually get better performance using a higher value coil. In one case i used a 12v supply battery and a 220uF inductor (Dn4516- ND at digikey). I get increment just about the same as before. With a high value inductor, the current might be less, but the potential peak is higher.

Ken

On 31 May, 19:51, lerameur wrote: [...]

The lower the coil resistance the better. A high R value can quite easily dominate the circuit performance. Dependant on quality you can have a higher inductance but lower R. Commercial inductor makers strive constantly to reduce their R values. I can't believe what I've just gone and done but have downloaded and actually read a 2N3055 datasheet. :). Looks like you'll be losing 1V across it when the inductor is charging but more to the point, you'll need to be driving the base with something like 1/2 amp, which is a biblical quantity of current . Would suggest it's far, far better to get hold of a cheap power FET instead. You'll get that volt back and have far simpler switching needs.

Hi, I have been driving the 2N3055 with a microcontroller, its far from driving 0.5 amp and it works. I am also looking into driving a high speed step up transformer. I cannot even find that at digikey. Is it me or does somebody knows a part number at digikey ?

ken

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