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golfguruPosted - 15 September 2013 20:31  Show Profile  Email Poster  Edit Message  
Actually, I was wondering if there is any such thing as a square law area any way?

I thought the square law area involved the squaring of a decimal amount and ending up with less than you started with (eg. 0.5*0.5 = 0.25).

But... you are already starting with decimal amounts (ie. mA versus Volts or mV's which is the usual graph "Y" axis).

If the axes were amps and volts, then squaring mv's or mA's should give square law values over the whole range of the "usual" graph.

Drawing a graph using uA's versus mV's/uV's as the "units" would give result at a different "square law point".

ie. Is there any real square law area at all?

Depends on the units used, and, what we actually define (arbitrarily) a unit to be?
(Is an amp really a "natural" unit or do we just say it is in our scheme of things).

Rambling a bit, but you should get my drift.


Edited by - golfguru on 9/15/2013 8:34:20 PM

gzimmerPosted - 15 September 2013 21:35  Show Profile  Email Poster  Edit Message
Yes, you a right. My explanation of the square law function was just a rather clumsy attempt to make it understandable. You shouldn't take it to literally.

But notwithstanding, the Square Law function is embedded in the Diode equation. It is real, you can measure it.

It's all about the Sidebands being multiplied by the Carrier.

When the signal is strong, the Carrier is toggling the diode cleanly, which means the Sidebands are being multiplied by zero or one (eg alternative half cycles are being suppressed, which is equivalent to half-wave Rectification).

However when the carrier is weak, the diode is only partly switching which means the Sidebands are being multiplied by a fractional number (eg something less than one), so the resultant audio is much weaker.

That's where the fractional part comes in.

........ Zim

Edited by - gzimmer on 9/15/2013 9:44:19 PM

golfguruPosted - 15 September 2013 22:14  Show Profile  Email Poster  Edit Message  
So that if we could draw a diode demodulation graph (that specifically suited our definition of "detection ability") it would have a point of inflection (or knee) at a specific point?


Edited by - golfguru on 9/15/2013 10:14:27 PM

gzimmerPosted - 15 September 2013 22:16  Show Profile  Email Poster  Edit Message
It would be a sharp right angle based at zero

Which is how a Synchronous Detector works.
An efficient switch, toggled by an externally derived carrier.

............ Zim

Edited by - gzimmer on 9/15/2013 10:20:11 PM

golfguruPosted - 16 September 2013 16:38  Show Profile  Email Poster  Edit Message  
>>> It would be a sharp right angle based at zero

Man !!! ... now that *is* a knee (or maybe an elbow). :-)


Edited by - golfguru on 9/17/2013 3:14:19 AM

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