In article , "craigm"
wrote:

"John Byrns" wrote in message
...

Patrick, this is a joke right? This is the really simple stuff, this
isn't one of the more complex and subtle points that we can all get
wrong. I am going to try and keep things simple by just sticking to this
one point. It is easily demonstrated that your claim is wrong with
respect to increasing carrier levels. Your circuit discharges the peak
hold capacitor with what is a reasonable approximation of a current
source, which means that the discharge is at a fixed rate of volts/sec.
Assuming a given fixed modulating frequency, and depth of modulation, the
maximum slope of the modulation that must be tracked by the voltage on the
peak hold capacitor is proportional to the average carrier amplitude.
That implies that if the carrier level is increased by say 6 dB, then the
slope in volts/sec that must be tracked increases by a factor of two,
while the discharge slope of your constant current circuit remains fixed,
ultimately leading to tangential clipping at some carrier amplitude. On
the other hand, while the traditional RC circuit has its problems, it is
not affected by the average level of the carrier that is feed to it. If
the average carrier increases by 6 dB, the peak modulation slope that must
be tracked increases by a factor of two as before, but since the discharge
current is not fixed, and varies in proportion to the carrier level, the
discharge slope also increases by a factor of two, and there will be no
additional tangential clipping with the traditional circuit when the
average carrier level is increased.

The bottom line is that the traditional circuit can handle any carrier
level no matter how large, without an increase in tangential clipping,
while the tangential clipping in your circuit, with a fixed discharge
rate, increases as the carrier level increases above the design point,
hence a poor dynamic range.

Now specifically what is wrong with what I have just said, where is my
error? This is the simple part of the problem, it is not even the complex
stuff where we all go wrong from time to time, yet you don't seem to be
able to grasp it.


John,

You seem to be limiting your considerations to the 'tangential clipping' and
not to other distortions that will occur.

On the contrary I alluded to the other problems in my post above where I
said "while the traditional RC circuit has its problems". The problem is
that Patrick has a very thick head, so I am trying to keep things simple
so he might get the point.

With the simple RC circuit the
decay of the signal differs from positive to negative peaks in the
modulation. This imparts an assymetry to the recovered signal. You are
trading one type of distortion for another.

Yes I think I discussed this in another recent post, although I didn't
call it asymmetry, I called it a problem with the negative peaks, at least
I hope I actually posted that bit, and that it didn't get edited out, I
will have to check back in the message archive. In the interest of full
disclosure, I will also own up to having talked in a previous message like
the negative peak problem didn't exist in the traditional circuit, this
seems to be the tack most text books take, concentrating on the point of
maximum slope. It may be justifiable, as just like the tangential
clipping in the high slope area, the negative peak problem is related to
both modulating frequency and modulation depth, creating a problem only
with heavy modulation at high frequencies. For the reader who may be
confused by this negative peak asymmetry problem, it should be noted that
this is not the same clipping phenomenon caused by a poor AC/DC load
ratio, and that the negative peak clipping caused by a poor AC/DC load
ratio is not frequency dependent. All in all I don't believe the
asymmetry you are talking about is a serious problem in practice, but that
is an individual matter of judgment.

Using a constant current to
drain the capacitor provides a more linear output and one where slew rate
limiting can be easily computed as a function of frequency and amplitude.

If you look closely at the operation ot the constant current "drain", you
will see that it too has a distortion problem at the negative peaks, and
it affects all modulating frequencies, unlike the traditional circuit
where the asymmetry tends to disappear at lower modulating frequencies.

Using a resistor to drain the capacitor provides an output where slew rate
limiting is more a function of frequency and less of amplitude.

I wouldn't say that.

However, if you know the maximum amplitude and modulating frequency of the
signal you are trying to detect, then for either detector one can determine
the proper component values for the desired result. These are the tradeoffs
that go into every design.

Quite true.

I suggest that those who are interested and/or following this discussion
would be well served by doing some modeling of the two proposals and
consider the results and how they are affected by changes in the input
signal. For a simple approach you could consider an ideal diode and signal
source, a capacitor and either a current source or a resistor. Try various
amplitudes and modulation levels.

Both circuit approaches work within their limitations. The question is 'what
are the limitations?'.

I don't think Patrick will approve this sort of activity.

Suggestion: Consider a triangle wave for the modulation source. The math is
a lot easier.

Hint: Linear is good.

Once you understand the limitations of each circuit topology, then you can
understand how it interacts with the rest of the radio, or what requirements
each places on the rest of the radio.

Have fun, I'll be watching,

This is all way too complex, I was trying to keep things simple for
Patrick, and get him to try on one small bite sized piece at a time. The
first piece is how the modulating frequency and depth of modulation affect
the tangential clipping in the high slope part of the wave form with the
two circuits.

Once that is grasped, then he can move on to the negative peak asymmetries
both circuits have, but that is considerably harder to understand.


Regards,

John Byrns


Surf my web pages at, http://users.rcn.com/jbyrns/