[Premium-Rx]

Mon Jul 7 16:29:50 EDT 2003

Hi All,

My Friend Bob and I were looking at AM detectors a while ago and he came up
with this one.

I thought I would share it with everyone. A lot better than the common high
Z detectors.   fc 

  An Improved Precision Full-wave AM Detector

by Rob Schenck, K2CU
 <http://www.amwindow.org/tech/htm/[email protected]> K2CU at arrl.net

 Am RulZ! <http://www.amwindow.org/tech/gif/alowdisdet/fig1.gif> 
Detector Schematic

Theory of Operation

This circuit is a variation of the classic "Precision Half Wave Rectifier"
as described in many op-amp application notes as shown below:

  <http://www.amwindow.org/tech/gif/alowdisdet/fig2.gif> 

Its operation is based upon basic principles of operational amplifier
circuits. An ideal op-amp has extremely high input impedance, extremely low
output impedance, and extremely high gain. Note the operative word
"extremely". The non inverting input (+) of the op-amp is connected to
ground. Feedback is provided by the two paths R2+D1 and D2. The op-amp will
drive its output pin to a voltage such that the inverting input (-) is at
ground voltage. There will be a slight variation (a few millivolts) from 0
volts as a result of what is known as input offset voltage, a property of
real op-amps. The inverting input is often called the summing junction, as
the sum of the currents in that node will be zero. Since virtually no
current flows into the inverting input (-) of the op-amp itself, any current
coming in through the input resistor R1 will be met by an equal and opposite
current coming through either of the two paths of R2+D1 or D2.

The I.F. input signal will produce an input current of:

Iin = Vin/R1

During the negative half of the input AC waveform, the op-amp output will
drive positive such that the current through R2+D1 will be equal and
opposite the input current, or:

Iin = Vin/R1 = -ID1 = -IR2

The voltage developed across R2, and hence the output voltage as the left
side of R2 is at ground potential, will be:

V = Vout = -IR2 * R2 = -Vin/R1 * R2

By setting R1 = R2, the output voltage will be;

Vout = Vin

Similarly, the positive going input cycle produces a feedback current
through D2. Any non linear voltages developed across the diodes is not
relevant as the output is derived from the feedback resistor only.

A full wave detector was desired for a 50 KHz IF application so that the
ripple frequency out of the detector would be 100 KHz, and more easily
filtered. In the full wave detector version of the circuit, a second
feedback resistor is added to the D2 diode path to produce a voltage of the
other half of the input waveform. A virtual ground at +6 Volts was created
with R814/R815/C812, so that the circuit could operate from a single supply.
Again, any nonlinear voltages developed across the diodes are not relevant.
The two rectified half waves are combined in the next two op-amp stages. The
two outputs of the rectifier on R811 and R813, are connected directly to the
non-inverting inputs of the op-amp stages so that there will be minimal
loading to these outputs. U6A provides a gain of two to the signal voltage
on R813. U6B provides a gain of two to the voltage on R811, and inverts and
adds the output of U6A. The result is a full wave rectified signal at the
output of U6B. Simple low pass filtering will remove the carrier energy
which is at twice the ripple frequency.

The ability of this circuit to accurately track the input waveform is
dependant on the speed of the op-amp and the switching speed of the diodes.
For the circuit to track the input, the op-amp must be considerably faster
than the input frequency. The LM318 with its 15 MHz bandwidth was selected
for this reason and because it is readily available and stable. This circuit
was built up and set up for comparison with the stock envelope detector in a
Drake R4-C with a 50 KHz I.F. A HP8640 signal generator was used to provide
a 1Khz modulated carrier at 3.885 MHz which was input to the antenna
connector. The input level was set to S9 (30 uV). The detectors were
compared at modulation levels of 80% and 99+ %. A Link Inc. , PC based
oscilloscope was used to record and display the resulting waveforms.

R4-C Stock Detector at 99+% 

  <http://www.amwindow.org/tech/gif/alowdisdet/fig3.gif> 

R4-C Stock Detector at 80%

   <http://www.amwindow.org/tech/gif/alowdisdet/fig4.gif> 

The distortion is visible at the crest of the wave, where the input signal
is at the modulation trough in this inverting detector. The distortion at
80% is less visible than at 99+%, but still audible.

Applying the 99+% modulated signal and using the precision rectifier results
in the following waveform:

Precision Detector 99+% Modulation

   <http://www.amwindow.org/tech/gif/alowdisdet/fig5.gif> 

This waveform displays virtually no distortion not only visually, but also
audibly. A pure tone is heard coming from the speaker. Actual on air
comparative listening tests of 75 meter AM signals at levels from just over
the noise to well over S9 were quite conclusive. Low level signals could be
more easily heard through the noise, and high level signals we noticeably
clearer.

Comments: Purists may want to use LM318 op-amps for all three stages, but I
found no significant improvement when doing same at 50 KHz. I wanted to pick
parts that were readily available and in particular in DIP packages so that
any builder would be able to build up the circuit on a perf board,
preferably the type with a ground plane on one side. This circuit is so
stable at 50 KHz, that it was actually built and tested using a prototyping
board with no ground plane at all. Operation at 455 KHz will definitely
require a ground plane in the construction. At 455 KHz IF frequency and
higher, the combining circuitry requires the use of three LM318's or
equivalent.

The Math Behind The Detection Process

There have been several articles written discussing the production of
harmonic distortion by diode envelope detectors. Almost all of these start
with the exponential diode equation and then examine the cross modulation
products derived from the series expansion of the exponential function.

e^X = 1 + X + 1/2 X^2+ 1/6X^3 + 1/24 X^4 +......

The simplistic AM modulated signal is:

V(t) = cos(Wc) t + m/2 cos(Wc - Wm) t + m/2 cos( Wc + Wm) t , where m is
modulation index

= Carrier + LSB + USB

The term of significance in the eX expansion series is the square term as it
provides the cross product of the carrier with each sideband, as in
synchronous detection. Unfortunately, it also provides the cross product of
the two sidebands which produces second harmonic distortion of the
modulating signal.

Consider : (a + b + c)^2 = a^2 + ab + ac + b^2 + ba + bc + c^2 + ca + cb =
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

Where;

cos(Wc) t = a

m/2 cos(Wc - Wm) t = b

m/2 cos(Wc + Wm) t = c

The three squared terms produce signals at twice the carrier frequency by
the trigonometric double angle formulas:

cos2(W) = 1/2 ( cos (2W) + 1)

The 2ab and 2ac terms multiply each sideband with the carrirer:

2 (cos(Wc)t) (m/2cos(Wc-Wm)t) = m/2 [cos (2Wc Wm)t + cos (Wm)t]

2 (cos(Wc)t) (m/2 cos(Wc+Wm)t) = m/2 [cos (2Wc +Wm)t + cos (Wm)t]

Which results in double sideband AM at twice the carrier (2Wc) plus the
desired, demodulated output m cos (Wm )t.

These detectors are often referred to as square law detectors for this
reason.

The problem is the 2bc term which is the cross modulation of the two
sidebands themselves. It results in:

2bc = 2 (m/2 cos(Wc-Wm)t) (m/2cos(Wc+Wm)t)

= m/4 [ cos(2Wc)t + cos (2Wm)t]

The m/4 cos (2Wm)t term represents some 25% second harmonic distortion,
which is much more than what is usually experienced in the real world.

In the real world, diode detectors work into a load resistor which often has
a parallel "filter" capacitor. If the diode were an ideal device, it would
only allow current to flow in the load resistor during precisely one half of
the input waveform. The resultant voltage developed across the resistor
would then be precisely the half wave rectified voltage of the input signal.
The reality is that the current/voltage relation in a diode is a non linear,
exponential function as has been discussed. This means that the output
voltage on that resistor would be the half waveform of the input signal
minus the voltage across the diode.

Vout = Vin - Vdiode

It is the voltage across the diode that is the non-linear portion of the
output signal, which becomes dominant at low input levels, such as the
modulation trough of an AM signal. There are techniques that have been
employed to reduce the diode component of the output waveform. One is to
drive the detector into a high impedance load resistor. With the current
very low, the voltage produced on the diode will be low. This approach is
limited by the generally ignored reverse current of the diode. At high
impedance levels, the rectifying function of the diode becomes compromised
by the reverse leakage current, resulting in other performance problems.

The precision rectifier uses the features of op-amp feedback design to
eliminate the diode error from the rectification process, resulting in a
"perfect" rectifier. Well, why does a perfect rectifier work as a detector
anyway? Most diode detector discussions talk about the square law function
as discussed above to explain the demodulation process. The received carrier
multiplies against the received sidebands. This is where the error comes in
the thinking that it also should produce an objectionable level of second
harmonic distortion. Let's go back to the pure half wave rectified signal
that the precision detector produces and the diode with the load resistor
tries to emulate. What is the process that takes an input signal at some
frequency, Wc, and outputs only the positive half of the waveform. It is
just as if the signal were multiplied by one during the positive half cycle,
and zero during the negative half cycle. The type of waveform to do this
would be a square wave, with a values of one or zero, and at the same
frequency and phase as the incoming carrier. The mathematical representation
of a square wave is what is known as a Fourier Series expansion. For our
sequence of one, zero, one, zero... at a frequency of Wc the function is:

f(X) = 1/2 + 2/Pi cos Wc t + 2/3Pi cos 3Wc t + 2/5Pi cos 5Wc t + ......

When this is multiplied times the incoming signal, we get a result that is
the half wave rectified waveform. Our interest is in the fundamental term,
cos Wc t. The higher frequency harmonic terms and the DC term will produce
results that are all at or above the incoming IF frequency. So the product
term of interest becomes:

Vout(t) = {A cos(Wc) t + 1/2 AB cos(Wc - Wm) t + 1/2 AB cos( Wc + Wm) t} *
2/Pi cos Wc t

The demodulated sideband terms are:

Vout(ty) = AB/2Pi * cos(Wm) t + AB/2Pi * cos(Wm) t , or

Vout(ty) = AB/Pi * cos(Wm) t

There are no intermodulation terms to contend with. In fact, functionaly,
this is identical to synchronous detection. The negative half wave signal is
derived in a similar fashion. The two signals, when combined in the summing
circuit produce an output with twice the ripple frequency to be filtered by
an output low pass filter. This is of value when the IF frequency is only 50
KHz, as in the Drake R4-C and others. For receivers with a 455 KHz or 500
KHz IF, replace the two LCMC660 op amps with a pair of LM318's.

Here's a  <http://www.amwindow.org/tech/htm/alowdisdet.pdf> PDF version of
this article, suitable for printing.

 <http://www.amwindow.org/tech/htm/gen.htm> Back to the General Technical
Page

8 March  <mailto:webmaster at amwindow.org> 2002

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