Please correct me if I am wrong, but is one way to transfer the energy 
efficiently accomplished by creating a tuned LC circuit on the primary side 
of the transformer to resonate at the frequency of the driver circuit (100 
KHz), and match the impedance at say 50 ohms to 50 ohms?  Is that tuned 
circuit (on the primary winding) not the load?  If this is not correct can 
someone who knows explain why?

----- Original Message ----- 
From: "Apptech" <apptech@paradise.net.nz>
To: "Microcontroller discussion list - Public." <piclist@mit.edu>
Sent: Saturday, August 09, 2008 9:50 AM
Subject: Re: EE


>> Nice way of computing... A square signal bumped with a
>> positive glich
>> at the end of the rising edge will increase the current
>> with 10-20% at
>> the same secondary load.
>> Does the primary impedance seen by the driver will be
>> different ?
>
> My figure is approximate. Reality will want to do its own
> thing.
>
> My key point was that transformer impedance is not generally
> seen but load impedance, whatever the value may be.
>
>
>        Russell.
>
>
>>> > Where did you get 16 ohms from?
>
>>> R = V/I.
>
>>> The driver sees the load reflected via the transformer.
>>> If
>>> 230 VAC causes 14 amps to flow then the AC is seeing
>>> 230/14
>>> ~= 16.
>
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