[Lowfer] Epson Question

Tue, 20 Apr 2004 17:45:56 -0700

Bill,

So...putting this together,

The power of an AC coupled 1 volt square wave drives into 1 Ohm is  =
(Vpp/2)^2

Where Vpp is the peak-to-peak voltage of the square wave.
Assuming 1 Ohm means we don't have to write it down (Its easy to multiply by
one in you head!).

The RMS value of the fundamental frequency component of that square wave is
= ((4/Pi) (Vpp/2*SQRT(2)))^2

And after some scratching on a piece of paper, that makes the ratio of the
power in the fundamental component to the total power in the Square wave
=8/(Pi^2), or about 81.1%.  That would mean that at best, 81.1% of a square
wave driven into an antenna resonant at the fundamental frequency would be
available to drive the antenna resistances (Ohmic + radiation). Is that
right?

Seems a little higher than I expected, but agrees with P=(E^2)/R.

on 4/20/04 2:16 PM, Bill Ashlock at [email protected] wrote:

> Johan,
> 
> Thank you for pointing out the error of my ways in the amplitude of the
> square wave fundamental freq. This makes a lot of sense in my calculations
> for the loop current when the input power is known. The measured value was
> always greater than the calculated value. I just located my college textbook
> and, sure enough, there's a 4/pi factor in front of the fourier series!
> 
> Bill A
> 
> 
>> From: "Johan Bodin" <[email protected]>
>> Reply-To: [email protected]
>> To: <[email protected]>
>> Subject: Re: [Lowfer] Epson Question
>> Date: Tue, 20 Apr 2004 07:20:26 -0000
>> 
>> Hi
>> 
>> the fundamental frequency component of a square wave
>> has an amplitude of:
>> 
>> Vsine_rms = Vsq_wave_peak_to_peak * square_root (2) / pi
>> 
>> which means
>> 
>> Vsine_rms = Vsq_wave_peak_to_peak * 0.4502
>> 
>> or
>> 
>> Vsine_peak_to_peak = Vsq_wave_peak_peak * 1.2732
>> 
>> (Assuming 50/50 % square wave duty cycle)
>> 
>> Although it may appear surprising, the Vpp of the fundamental
>> frequency component is larger than the Vpp of the square wave itself :-)
>> 
>> To run class D, you can put a series resonant circuit of sufficient
>> loaded Q, say 5 or more, between the square wave generator and
>> the load (XC = XL = 5*Rload, or more).
>> 
>> The series circuit will pass only the fundamental frequency so the
>> generator will see only negligible loading on the harmonics.
>> This is much better than using a "grounded C input" lowpass
>> filter directly at the generator output (which will cause high current
>> switching spikes = more switching loss).
>> 
>> BTW, I would guess that the Epson oscillator has a quite "weak" output
>> stage so it will probably deliver more power into a load >50 ohms.
>> 
>> 73
>> Johan SM6LKM
>> 
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