In article , "craigm"
wrote:

John,

If I follow what you are saying, you are saying that the sawtooth decay of
the capacitor changes and differs significantly between negative modulation
peaks and positive modulation peaks.

If so, then that ripple is caused by the RC circuit connected to the diode
where the resistor is discharging the capacitor. If you replace the resistor
with a constant current source (to discharge the capacitor), the ripple will
be much more uniform. Or instead of tying the resistor to ground, use a much
larger resistor connected to a negative supply.

If the discharge current is constant (does not change due to the voltage on
the capacitor), then this approach leads to ripple that does not vary with
the voltage on the capacitor.

With one exception, see below, the ripple should be more or less constant
with this scheme. At one time I was enamored with this scheme, but gave
it up for reasons I no longer remember. A reason that comes to mind now,
but was not the one I can't remember, is that while the ripple may be
relatively constant with modulation, the frequency at which tangential
clipping sets in will be dependent on the carrier amplitude, since the
slope of the discharge is independent of carrier level.

I would think that this improves the linearity of the detector and this
improve distortion, however, I don't know how much of an improvement it will
make.

I'm not sure, but I would think that to a first approximation the variable
ripple voltage of the normal envelope detector doesn't add distortion or
nonlinearity, but with the current source approach, there is going to be
distortion on negative modulation peaks when the amplitude of the ripple
wave form is clamped by the diode before the next carrier cycle starts.
For example at 100% negative modulation the ripple voltage must go to zero
even with a current source, and this is a nonlinear effect, since the
ripple voltage is constant up to a certain modulation depth, at which
point it abruptly starts reducing towards zero on 100 % negative
modulation peaks.

I also believe it is a lot easier to do this in the solid state world than
in the tube world.

There are high voltages available in vacuum tube circuits which in
combination with a large resistor would make a pretty good approximation
to a current source.

Regards,

John Byrns


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