What do you mean, the inductor (10 turns of wire on a core) is connected between the positive end of the supply and one end of the switch (drain of the MOSFET) isn't it?
2010/3/20 Harry Veeder <[email protected]>: > The toroid is also wired in differently from the inductor in the wiki > diagram, but I suppose that doesn't matter either? > > > harry > > > > > ----- Original Message ---- >> From: Michel Jullian <[email protected]> >> To: [email protected] >> Sent: Fri, March 19, 2010 1:42:52 PM >> Subject: Re: [Vo]:circuit diagram >> >> The capacitor on your photo 2 is in parallel with the battery so it's > part of >> the converter's input supply. The capacitor in the operating > principles >> diagram of the wikipedia article is the converter's output > capacitor, which >> might as well not be there in steady state is there > is no load (once charged >> it just stays charged at a high voltage, and > the Boost's diode never >> conducts-- so the diode might as well not be > there either). So everything to >> the right of the switch in the boost > converter diagram could be removed in no >> load condition, that's why I > say the circuit operates like a Boost converter >> without a load. Which > explains why it steps up the input voltage, that's what >> Boost > converters do. > > Michel > > 2010/3/19 Harry Veeder < >> ymailto="mailto:[email protected]"; >> href="mailto:[email protected]";>[email protected]>: >> I'll pass >> that along. >> But the capacitor looks like it is in the wrong place to be >> a booster >> converter with or without a load. >> compare photo >> 2: >> >> >http://tinyurl.com/ycw4xm4 >> >> with operating >> principles >> >> target=_blank >http://en.wikipedia.org/wiki/Boost_converter >> >> >> Harry >> >> >> >> >> >> ----- Original Message >> ---- >>> From: Michel Jullian < >> ymailto="mailto:[email protected]"; >> href="mailto:[email protected]";>[email protected]> >>> To: >> >> href="mailto:[email protected]";>[email protected] >>> Sent: Fri, >> March 19, 2010 4:54:02 AM >>> Subject: Re: [Vo]:circuit >> diagram >>> >>> 2010/3/19 Harry Veeder < >>> >> href="mailto: >> href="mailto:[email protected]";>[email protected]"> >> ymailto="mailto:[email protected]"; >> href="mailto:[email protected]";>[email protected]>: >>> Here is >> a >>> reply from Magluvin who is also a member of >> overunity.com: >>> "This is not >>> a boost >> converter >> >> I said it was a boost converter _without a >>> >> load_. >> >>> as none of them will recharge the input >>> >> source(cap) >>> while being operated. Ive tried. >> >> This is >> because he hasn't tried removing >>> the load. If you do, in the >> >> course of one oscillation cycle, the input source >>> first >> sources >> current, and then sinks current. Note there is a >> hidden >>> component in >> the circuit which is important to >> understand where the >>> inductor's >> current flows to and from in >> this no load operation, that's >>> the >> MOSFET's output >> capacitance. The IRF640's antiparallel diode >>> is >> another >> hidden component which plays an important role, it prevents >> >> the >>> drain voltage from going below zero. >> >> >> Michel >> >>> And you wont find >>> any >>> dc/dc >> converters with magnets on the coil core. >> ;]" >>> >>> >>> >> Harry >>> >>> >>> >> >> >> >> __________________________________________________________________ >> >> Looking for the perfect gift? Give the gift of Flickr! >> >> >> href="http://www.flickr.com/gift/"; target=_blank >> >http://www.flickr.com/gift/ >> >> > > > __________________________________________________________________ > Looking for the perfect gift? Give the gift of Flickr! > > http://www.flickr.com/gift/ > >
