Which voltage?
2010/3/20, Harry Veeder <[email protected]>:
> 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
>>>>
>>>>
>>>>
>>>
>>>
>>>
>>>
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>>
>>
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>
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