Many detectors using diodes have an active bias voltage/current applied.
An Autek WM-1 peak reading watt meter is a great circuit of said.

cheers
skipp
http://sonic.ucdavis.edu

: Clifton T. Sharp Jr. wrote:
: I was going through my bookmarks when I came upon one about diodes
: (http://uweb.superlink.net/bhtongue/7diodeCv/7diodeCv.html). In his
: third paragraph, the author asserts:

: For RF diode detectors to work, one needs a device that has a
: non-linear V/I curve. In other words, the slope of the V/I curve
: must change as a function of applied Voltage. The slope must be
: steeper (or shallower) at higher voltages and shallower (or steeper)
: at lower voltages than at the quiescent operating point.

: If there's any truth to this, someone please enlighten me. I've always
: been of the belief that an envelope detector diode would be most perfect
: if the diode was a perfect switch, i.e. zero attoamps reverse current
: and perfectly linear forward current (as though the diode was a wire
: during the forward conduction period). I don't see how a changing slope
: during forward conduction could do anything other than distort the
: demodulated waveform, especially on tiny signals.

: --
: The function of an asshole is to emit quantities of crap. Spammers do
: a very good job of that. However, I do object to my inbox being a
: spammer's toilet bowl. -- Walter Dnes

Hi,

For RF diode detectors to work, one needs a device that has a
non-linear V/I curve. In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

If there's any truth to this, someone please enlighten me. I've always
been of the belief that an envelope detector diode would be most perfect
if the diode was a perfect switch, i.e. zero attoamps reverse current
and perfectly linear forward current (as though the diode was a wire
during the forward conduction period). I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

You believe correctly! A diode detector is nothing much more than a
rectifier and should be characterised accordingly. Curvature of the
forward characteristic has to be lived with and is not something generally
to be desired. There are exceptions to this where the law of the initial
curvature is employed in certain instruments but not usually in
demodulating an AM signal.

However, with very weak signals a small degree of forward biassing is
sometimes employed to obtain the greatest recovered audio. This comes
about because the positive peaks are driven onto the greatest slope while
the negative ones, which would normally distract from the charge on the
load capacitor, are placed on the flatter part below the bias point and so
not recovered so well.


Cheers - Joe

Hi,

For RF diode detectors to work, one needs a device that has a
non-linear V/I curve. In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

If there's any truth to this, someone please enlighten me. I've always
been of the belief that an envelope detector diode would be most perfect
if the diode was a perfect switch, i.e. zero attoamps reverse current
and perfectly linear forward current (as though the diode was a wire
during the forward conduction period). I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

You believe correctly! A diode detector is nothing much more than a
rectifier and should be characterised accordingly. Curvature of the
forward characteristic has to be lived with and is not something generally
to be desired. There are exceptions to this where the law of the initial
curvature is employed in certain instruments but not usually in
demodulating an AM signal.

However, with very weak signals a small degree of forward biassing is
sometimes employed to obtain the greatest recovered audio. This comes
about because the positive peaks are driven onto the greatest slope while
the negative ones, which would normally distract from the charge on the
load capacitor, are placed on the flatter part below the bias point and so
not recovered so well.


Cheers - Joe

"Clifton T. Sharp Jr." wrote in message
...
I was going through my bookmarks when I came upon one about diodes
(http://uweb.superlink.net/bhtongue/7diodeCv/7diodeCv.html). In his
third paragraph, the author asserts:

For RF diode detectors to work, one needs a device that has a
non-linear V/I curve. In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

If there's any truth to this, someone please enlighten me. I've always
been of the belief that an envelope detector diode would be most perfect
if the diode was a perfect switch, i.e. zero attoamps reverse current
and perfectly linear forward current (as though the diode was a wire
during the forward conduction period). I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

Dude, you and they are talking about the exact same thing.

Think about the V-I curve on an ideal diode. Observe that, if the diode is
reverse-biased, no conduction takes place. In principle, you can crank the
reverse voltage up as high as you like, and STILL no conduction takes place.
(In practice, you run into avalanche and zener breakdown.) If the diode is
forward-biased, the diode conducts like crazy. There is a CORNER in the
curve, i.e. a change in the slope of the V/I curve, at NOMINALLY V=0V.
(Actually, V=0.7V for silicon, V=0.3V for germanium, V=1.7V for red LEDs,
about 2.2V for yellow, about 2.4V for green, and so on and so forth...)

John R. Strohm wrote:
"Clifton T. Sharp Jr." wrote...
(http://uweb.superlink.net/bhtongue/7diodeCv/7diodeCv.html).

In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

Think about the V-I curve on an ideal diode. Observe that, if the diode is
reverse-biased, no conduction takes place. In principle, you can crank the
reverse voltage up as high as you like, and STILL no conduction takes place.
(In practice, you run into avalanche and zener breakdown.) If the diode is
forward-biased, the diode conducts like crazy. There is a CORNER in the
curve, i.e. a change in the slope of the V/I curve, at NOMINALLY V=0V.
(Actually, V=0.7V for silicon, V=0.3V for germanium, V=1.7V for red LEDs,
about 2.2V for yellow, about 2.4V for green, and so on and so forth...)

I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

....while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

--
The function of an asshole is to emit quantities of crap. Spammers do
a very good job of that. However, I do object to my inbox being a
spammer's toilet bowl. -- Walter Dnes

You're right. The author is wrong.

Roy Lewallen, W7EL

Clifton T. Sharp Jr. wrote:


I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

...while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

You're right. The author is wrong.

Roy Lewallen, W7EL

Clifton T. Sharp Jr. wrote:


I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

...while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

"Clifton T. Sharp Jr." wrote in message
...

I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

...while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Am I missing the contradiction? Say you have a quiescent operating point of
0V. An ideal diode would have an infinately high slope above that point and
a 0 slope below that point. Or vice-versa. That seems to fit the given
statement.

Real world diodes have differing ideal q-points and variable slopes. And
most of the other articles seem to be aimed at getting the most out of such
diodes.


Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

I think your perfect diode analysis is correct. I think the only
nonlinearity the author requires is around a q-point. I don't get the
impression the author requires a diode to have a variable slope
characteristic at other voltages.


He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

The author's last revision was on 07/2/2003, so he must still have an
interest. I suppose you could send him an e-mail. The address is the first
item in article #1.

--
Frank Dresser

"Clifton T. Sharp Jr." wrote in message
...

I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

...while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Am I missing the contradiction? Say you have a quiescent operating point of
0V. An ideal diode would have an infinately high slope above that point and
a 0 slope below that point. Or vice-versa. That seems to fit the given
statement.

Real world diodes have differing ideal q-points and variable slopes. And
most of the other articles seem to be aimed at getting the most out of such
diodes.


Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

I think your perfect diode analysis is correct. I think the only
nonlinearity the author requires is around a q-point. I don't get the
impression the author requires a diode to have a variable slope
characteristic at other voltages.


He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

The author's last revision was on 07/2/2003, so he must still have an
interest. I suppose you could send him an e-mail. The address is the first
item in article #1.

--
Frank Dresser

"Clifton T. Sharp Jr." wrote in message ...
John R. Strohm wrote:
"Clifton T. Sharp Jr." wrote...
(http://uweb.superlink.net/bhtongue/7diodeCv/7diodeCv.html).

In other words, the slope of the V/I curve
must change as a function of applied Voltage. The slope must be
steeper (or shallower) at higher voltages and shallower (or steeper)
at lower voltages than at the quiescent operating point.

I don't see how a changing slope
during forward conduction could do anything other than distort the
demodulated waveform, especially on tiny signals.

Think about the V-I curve on an ideal diode. Observe that, if the diode is
reverse-biased, no conduction takes place. In principle, you can crank the
reverse voltage up as high as you like, and STILL no conduction takes place.
(In practice, you run into avalanche and zener breakdown.) If the diode is
forward-biased, the diode conducts like crazy. There is a CORNER in the
curve, i.e. a change in the slope of the V/I curve, at NOMINALLY V=0V.
(Actually, V=0.7V for silicon, V=0.3V for germanium, V=1.7V for red LEDs,
about 2.2V for yellow, about 2.4V for green, and so on and so forth...)

I'm not sure I was clear on my question. The only nonlinearity I want in
my detector diode is the noncontinuous function

| Vin when Vin = 0.000000 micronanofemtoattovolts
Vout = |
| 0 when Vin 0.000000 micronanofemtoattovolts

...while the author seems to be saying that the nonlinearity which exists
just above the barrier voltage is essential to detection. He does say,

"The slope must be steeper (or shallower) at higher voltages and
shallower (or steeper) at lower voltages than at the quiescent
operating point."

Given my perfect diode and (let's say) a 0.25V peak signal, I contend
that my output envelope will be an exact reproduction of the input
envelope; but (see his graph #2) his required nonlinearity will seriously
distort the output envelope.

He seems to be saying that without that, you can't demodulate the signal.
I assert that without that, I demodulated it better than he did.

I think you and the author are saying the same thing. Consider that
in your ideal diode case, the quiescent operating point is zero volts.
Above that, the slope of your diode is infinte (zero ohms) and below
that it's zero (infinite ohms). The author is suggesting that you do
NOT need it to be so sharp, and in fact, I find that I can detect RF
down to below 100 microvolts or so with a Schottky diode designed for
zero-bias detector service, though the output voltage is very tiny
down there. If I could buy one of your perfect diodes, 100uV (RMS) of
RF would give me around 141uV of output, instead of the sub-microvolt
I do get. The low output is the result, of course, of the fact that
the diode dynamic resistance at -140uV is only very slightly different
from its dynamic resistance at +140uV.

In a real diode, the change in slope is gradual with no sharp corners,
but if you could find a diode which looked like, say, 1000 ohms in the
reverse direction for all voltages and 1000.1 ohms in the forward
direction, it would still work as a detector, albeit a poor one.

Cheers,
Tom