Subject:
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FW: Re: Eliminate high-speed derailments forever! (and possibly v oid your warranty)
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Newsgroups:
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lugnet.trains
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Date:
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Mon, 23 Aug 1999 19:14:25 GMT
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Viewed:
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1161 times
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> In lugnet.trains, Ludo Soete writes:
> > In lugnet.trains, John R. Clark writes:
> > > Okay, how's this for an idea:
> > >
> > > Most electronic stores stock a kind of universal power converter. These • have
> > > several output settings, including 9v and 7.5 volt. If you use the 7.5 • volt
> > > setting, the train never gets enough power to derail on curves. This has
> > > obvious
> > > advantages at shows, and also anytime kids are given the controls (my 2 • 1/2
> > > year
> > > old loves my trains, but sometimes he runs them a bit fast!)
> > >
> > > This reduces the number of different speeds available to only two • (instead of
> > > 6). Specifically, it seems to simply eliminate the 4 highest speeds. But • that's
> > > perfect for little ones who seem to have only two speeds anyway: • full-forward,
> > > and full-reverse.
> > >
> > > So the obvious question is: Am I damaging/shortening the life of my • transformer
> > > by feeding 1.5 volts less than it expects into it?
> > >
> > > And another question that someone with more electronics knowledge than I • have
> > > can answer: The universal converter has removable male jacks, which are
> > > designed
> > > to fit many different devices. They are also reversible, which means • that the
> > > tip can be either positive or negative. So does it matter which way it • goes
> > > into
> > > the transformer?
> > >
> > > I leave the answers to these questions as an exercise for the reader.
> > >
> > > Rick Clark
> > > jrclark@nospam.aol.com
> > > http://members.xoom.com/jrickclark
> >
> > Hi,
> >
> > can i joy the conversation?
> >
> > I just opened a LEGO TRAIN SPEED REGULATOR and drawn a shematic of it.
> > I did this once before but lost the paper.
> >
> > Now to the point:
> > The INPUT of the regulator may be an AC or a DC signal.
> > Even reversing the polarity (in case of a DC voltage) won't damage the
> > regulator.There's a full bridge rectifier on the printed cirquit board • (pcb).
> >
> > Lego put on the input of the regulator : 9 - 12V AC, this has to do
> > with the following things:
> >
> > 1)The voltage regulator is a LM317T (can deliver 1.5A output current)but • is
> > mounted on a small aluminum heathsink, too small to obtain enoug cooling • when
> > drawing 1.5A, so don't do that.
> >
> > 2)The LM317 won't blow up because he's internally protected against to • high
> > tempertures.If he gets to hot,the output voltage drops.But the pcb's • layers
> > are too thin to pull 1.5A out of the regulator.
> >
> > 3) When the input voltage is 12V AC, then the internal DC voltage is • aprox.18V
> > (AC input voltage x 1.41 = DC output voltage) this means that, when the
> train
>
> not really it will be 18V - (2 x 0.7) = 16.6V (don't forget the drop
> accros
> the diode bridge)
>
> > runs at full speed (output=9V DC) that there's a voltage of 9V DC OVER • the
> > LM317. Let say that the motor uses .5A, then is the power dissipation
> > 9V x .5A = 4.5Watt
> > If you run at half the speed (4.5V) then you will messure 18V - 4.5V = • 13.5V
> > over the LM317.
> > recalculate the dissipation gives : 13.5V x .5A = 7.25 Watt.
>
> at lower speed the motor will use less current than at full speed, and
> also the
> voltage drop that the regulator as the dissipate is not constant, between
> each
> half-cycle of the AC voltage the internal DC voltage at the input of the
> regulator is dropping from ~~ 16.6 to a lot less than that because there
> is
> only a small capacitor after the diode bridge. This capacitor is charge
> only at
> each peak of the AC half-cycle and discharge the rest of the cycle trying
> to
> keep the voltage as "flat" as possible.
>
>
>
> --_ _ _
> / \ - - -_ _ _ <-LM317 input
> / \ - - -
> / \<-AC
>
>
> Martin
>
>
> >
> > As you can see is it verry important to keep the input voltage as low as
> > possible.So 9V AC is better than 12V AC
> >
> > When you want to get a lower output voltage, then you have to open the
> > speed regulator,or the input voltage must be lower than 9V DC !!
> > The output voltage is normaly independent from the input voltage as you • can
> > see in the formulla below.
> > The output voltage can be calculated with the followig formula:
> > Vout=1.25(1+(R2/R1))+Iadj(R2) -> info: data sheet National Semiconductor
> > where :
> > R1=200 ohm
> > R2=200,400,600,800,1000,1200 ohm (depending on the position of the speed • knob)
> > Iadj(R2)= current throug R2.
> > For fast calculation omit +Iadj(R2),resulting in:
> > Vout=1.25(1+(R2/R1))
> > I've done a modification on it so that the maximum output voltage is • aprox.
> > 7.8V
> >
> > If you want to do it too,open the speed regulator.
> > Remove the yellow speed knob.Keep the green LED to the bottom side.
> > you'll see on the left side of the 'circle swith' 8 solderpins going from • top
> > to bottom (resistor array).
> > Count the pins from top to bottom.You'll end up by 8.
> > Short cirquit the pins '7' and '8' with a solder pad.
> > re-assamble the speed regulator.
> > Thats it !!!
> >
> > With this action is the speed for step 5 and 6 (full speed)equal.
> >
> > Good luck !!
> > regards,
> > Ludo
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