SQUARE LAW DETECTORS
Some people say square law detectors are "incredibly linear". Not quite so.
Consider the following circuit and the typical diode curve in the forward bias region. This looks like a half-wave rectifier circuit, but itís not.
Typical Diode V-I curve (in amperes and volts)
I make the following assumptions:
id = ae + be2 + ce3 + ...
4. Rdiode + RL >> Rs
Polynomial representation of diode curve
Since e = Acos(w t), id may be represented by the polynomial
i d = a Acos(w t) + bA2cos2(w t) + cA3cos3 (w t) + ...
Substituting in the incredibly familiar Trig identities
cos2(w t) = (1/2)[1+ cos(2w t)] and
cos3(w t) = (1/4)[cos(3w t) + 3cos(w t)]
gives us the following result:
id = aAcos(w t) + (b/2)A2 [1 + cos (2w t ) ] + (c/4)A3 [cos (3w t ) + 3cos(w t) ] + ....
= (b/2)A2 + [aA + (3/4) cA3]cos(w t) + (b/2)A2(2w t) + (c/4)A3cos(3w t) + ...
which is of the form
g 1A2 + g 2cos(w t) + g 3cos(2w t) + g 4cos(3w t) + ... where g n are constants
Since the voltage across RL (vo) is what is of interest, and v o = id RL
v o = RL g 1A2 + RL g 2cos(w t) + RL g 3cos(2w t) + RL g 4cos(3w t) + ...
This represents a DC term + all the harmonics of cos(w t). By passing v o through a low pass filter we get the DC output voltage
Vo = RL g 1A2
Remembering that A is the amplitude of the RF signal, and that Power P = (V2/R)
P = a Vo
where a is a constant.
In other words, Vo is directly proportional to the power dissipated by Rs .