yes. You are aware that the the voltage keeps rises even after the battery is disconnected.
harry ----- Original Message ---- > From: Michel Jullian <[email protected]> > To: [email protected] > Sent: Sat, March 20, 2010 3:59:08 AM > Subject: Re: [Vo]:circuit diagram > > 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 > < > href="mailto:[email protected]";>[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 < > ymailto="mailto:[email protected]"; > href="mailto:[email protected]";>[email protected]> >> To: > > href="mailto:[email protected]";>[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: > href="mailto:[email protected]";>[email protected]" >> > href="mailto: > href="mailto:[email protected]";>[email protected]"> > 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 > > href="http://en.wikipedia.org/wiki/Boost_converter"; target=_blank > >http://en.wikipedia.org/wiki/Boost_converter >> >> >> > Harry >> >> >> >> >> >> > ----- Original Message >> ---- >>> From: Michel Jullian > < >> ymailto="mailto: > href="mailto:[email protected]";>[email protected]" >> > href="mailto: > href="mailto:[email protected]";>[email protected]"> > ymailto="mailto:[email protected]"; > href="mailto:[email protected]";>[email protected]> >>> > To: >> >> href="mailto: > href="mailto:[email protected]";>[email protected]"> > ymailto="mailto:[email protected]"; > 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: > href="mailto:[email protected]";>[email protected]"> > ymailto="mailto:[email protected]"; > href="mailto:[email protected]";>[email protected]"> >> > ymailto="mailto: > href="mailto:[email protected]";>[email protected]" >> > 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=" > href="http://www.flickr.com/gift/"; target=_blank > >http://www.flickr.com/gift/"; target=_blank >> > > href="http://www.flickr.com/gift/"; target=_blank > >http://www.flickr.com/gift/ >> >> > > > > __________________________________________________________________ > > Looking for the perfect gift? Give the gift of Flickr! > > > href="http://www.flickr.com/gift/"; target=_blank > >http://www.flickr.com/gift/ > > __________________________________________________________________ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
