Frederick Glenn


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:

  1. The diode has a forward bias applied so it is operating somewhere in the "knee" of the curve. In this case, say about 500 uA. This operating point is still very non-linear!
  2. Acos(w t) <<Vdiode Note that under these conditions, the signal is not rectified
  3. The V-I curve of the series combination of D1 and RL can be represented by a power series of the form:

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 .