SQUARE LAW DETECTORS

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 it’s not.

Typical Diode V-I curve (in amperes and volts)

I make the following assumptions:

- 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!
- Acos(w
t) <<V
_{diode }__Note that under these conditions,__*the signal is not rectified* - The V-I curve of the series combination of D
_{1}and R_{L}can be represented by a power series of the form:

i_{d} = ae + be^{2} + ce^{3} + ...

4. R_{diode} + R_{L} >> R_{s}

Polynomial representation of diode curve

ANALYSIS:

Since e = Acos(w
t), i_{d }may be represented by the polynomial

i_{ d} = a Acos(w
t) + bA^{2}cos^{2}(w
t) + cA^{3}cos^{3} (w
t) + ...

Substituting in the incredibly familiar Trig identities

cos^{2}(w
t) = (1/2)[1+ cos(2w
t)] and

cos^{3}(w
t) = (1/4)[cos(3w
t) + 3cos(w
t)]

gives us the following result:

i_{d} = aAcos(w
t) + (b/2)A^{2} [1 + cos (2w
t ) ] + (c/4)A^{3} [cos (3w
t ) + 3cos(w
t) ] + ....

= (b/2)A^{2} + [aA + (3/4) cA^{3}]cos(w
t) + (b/2)A^{2}(2w
t) + (c/4)A^{3}cos(3w
t) + ...

which is of the form

** g
_{1}A^{2} + g
_{2}cos(w
t) + g
_{3}cos(2w
t) + g
_{4}cos(3w
t) + ...** where g

Since the voltage across R_{L} (v_{o}) is what is of interest, and v_{ o }= i_{d} R_{L}

v_{ o} = R_{L} g
_{1}A^{2} + R_{L} g
_{2}cos(w
t) + R_{L} g
_{3}cos(2w
t) + R_{L} g
_{4}cos(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

V_{o} = R_{L} g
_{1}A^{2}

Remembering that A is the amplitude of the RF signal, and that Power P = (V^{2}/R)

**P = a
V _{o}**

where a is a constant.

In other words, V_{o} is directly proportional to the **power** dissipated by R