Dave, your considerable effort to explain the nuances of diodes to
John
is commendable, but your explanation is rather misleading. It's not
true
that for a diode to conduct, the "barrier potential must be
exceeded,"
and "the junction becomes forward biased and conducts heavily."
Instead
the diode current has an exponential relationship to the voltage
across
it, and gradually turns on over many hundreds of millivolts, not
abruptly
at say 600mV. Here, examine some diode measurements I made a long
time
ago, http://www.picovolt.com/win/elec/com...de-curves.html

The theory confirms your results.

I diode ~K*exp(Vsignal/Vt) =K+ a*Vsignal + b*Vsignal^2+ .... by the
binomial theorem

So at low signal amplitudes the diode current follows a linear
relationship as predicted by the diode equation and confirmed by the
graphs there. At higher values the higher terms start dominating and
the exponential term takes over.

"Mark" wrote in message
oups.com...



Must be my lucky week!. I needed to make some similar readings.
Thanks!.
I still puzzle over the oft quoted "up to about 30mVrms the output
from the
diode offers a square law response and will approximate a true RMS
measurement.
As the OP found, there's very little happening down there. Who makes
these
sweeping statements?.
regards
john

No it's true. Look at the curves and notice that the current scale on
the x axis is a log scale. When the RF input is very small, the DC out
is proportional to the log of the RF level i.e. the RF in dB. This is
how the normal power meter works. It also provides a true RMS value
for modulated RF signal. Once the signal gets too big and the diode
begins to work as a converntoin rectifier, this relationship no longer
holds true. Notice the curves break upeards. When the RF volatge to
log I curves are straight line, this is rhe square law region where the
diode current gives you true RMS readings of the RF voltage.
Think of it as a voltage in dB to current converter.

Mark

I know where you're coming in from, but whichever way I look at Win's graphs
I'm seeing a straight line for that 1N4148 at 0-30mv levels. Yes, the scales
are log-log but the constant of proportionality is dead straight linear. I.e
1mv RF in gives 1e-10 amps and 10mV in gives 1e-9 amps and pro-rata for all
points in between (you did see the double decade increments?). This agrees
with the 10Mohm value that's marked on the graph.
It shouldn't matter if the RF signal is swept over a 1 to 30mV range or just
a 5 to 5.01mV range, the DC out will be directly proportional to the 'DC' in
and no distortion of the waveform can occur, hence no dc offsets or
harmonics.
The graph next door though , the 1n5819, looks like it could offer up a tad
of rf dB-I rectification. Though to my eyes it still looks way more like a
resistor than anything with a square or log law response.
There'll be better devices out there that offer (say) quadratic like
classical responses at these low levels but it's moving out of the 'common
or garden' playground.
regards
john

"K7ITM" wrote in message
oups.com...
Indeed, as Win says, you can get some signal out of a diode detector
even for very low input levels. With fairly simple home-brew
techniques but a lot of attention to the details of leakage currents
and op amp offset voltages, I'm able to detect RF signals down to a
very few tens of microvolts. That's using either a zero-bias Schottky
detector diode such as the Agilent HSMS-2860, or an old germanium point
contact diode. At very low signal levels, the optimum load resistance
is quite high. (See Agilent detector diode ap notes for details.)
Things are actually easier if you're only interested in the modulation
component of an AM signal, and not in trying to detect the carrier
level, since the offsets aren't particularly important for AC signals.
A JFET audio amplifier, or even a carefully-designed bipolar amplifier,
can give you a very low noise figure for the high source resistance
that the diode detector running at low input levels gives you.

There are tricks you can play to make a receiver that works from the
power received by the antenna. If you live near a transmitter that's
putting out significant power in your direction, you may be able to set
up a rectifier for that received power and use it to run a micro-power
amplifier following the detector for the station you wish to receive.
If you want to hunt for weak stations, you'll need a carefully designed
and built RF input tank/filter circuit. At night, especially, it's
possible to listen to stations quite a ways away using no active
components in the RF path before the detector.

Cheers,
Tom

10's of uV. Egad!. Wish I had your patience!. Built a feedback linearised RF
probe head last year, using a couple of dual BAT85 SM packages in its tip.
They were used as voltage doublers working into 10Mohms. I tried really,
really hard, (well, about an hour) to see a mV of RF i/p but random DC
shifts, thermocouple effects and second order temperature drifting called a
halt to the project. Wish I'd thought about these things before starting :-(
regards
john

john jardine wrote...

I know where you're coming in from, but whichever way I look at Win's graphs
I'm seeing a straight line for that 1N4148 at 0-30mv levels. Yes, the scales
are log-log but the constant of proportionality is dead straight linear. I.e
1mv RF in gives 1e-10 amps and 10mV in gives 1e-9 amps and pro-rata for all
points in between (you did see the double decade increments?). This agrees
with the 10Mohm value that's marked on the graph.

Right, but that part of my measurements cries out for further bench
exploration. It represents only one part, and is unconfirmed. Also,
what happens if the voltage is reversed? Are we to believe the diode
is a 10M resistor, shunted by a diode? I'm not comfortable with that.


--
Thanks,
- Win

"Winfield Hill" -edu wrote in
message ...
john jardine wrote...

I know where you're coming in from, but whichever way I look at Win's
graphs
I'm seeing a straight line for that 1N4148 at 0-30mv levels. Yes, the
scales
are log-log but the constant of proportionality is dead straight linear.
I.e
1mv RF in gives 1e-10 amps and 10mV in gives 1e-9 amps and pro-rata for
all
points in between (you did see the double decade increments?). This
agrees
with the 10Mohm value that's marked on the graph.

Right, but that part of my measurements cries out for further bench
exploration. It represents only one part, and is unconfirmed. Also,
what happens if the voltage is reversed? Are we to believe the diode
is a 10M resistor, shunted by a diode? I'm not comfortable with that.


--
Thanks,
- Win

Yes. Ive been mulling that over. If there is no static bias current flowing
through the diode and it magically switches to 'open circuit' when the
incoming voltage reverses sign, then it would still make a perfect 'average
respsonding' rectifier even at these mV or even uV levels. Just how does the
reverse current start to act in the reverse direction?. On the face of it, a
IN4148 seems easy enough to check out.
regards
john

On 3 Feb 2005 07:49:33 -0800, "lemonjuice"
wrote:

Well if you want to cheat you can have more turns on the primary then
the secondary of the input transformer and you get a higher voltage
(grin). I'd have to see the exact circuit you are talking about to be
of more help.

Input transformers are all very well, but some good voltage step-up
can be obtained by carefully chosen values of hi-Q capacitor and
inductor in series between the aerial and the diode. Of course this
makes the impedance even higher, just as the transformer would, but
how strong's your signal? It might be the cheapest alternative.
--

"What is now proved was once only imagin'd." - William Blake, 1793.

In article , hill_a@t_rowland-dotties-
harvard-dot.s-edu says...
of my measurements cries out for further bench
exploration. It represents only one part, and is unconfirmed. Also,
what happens if the voltage is reversed? Are we to believe the diode
is a 10M resistor, shunted by a diode? I'm not comfortable with that.

I'm confused. Is there some reason to expect the semiconductor material
to be a perfect insulator with no resistivity at all? Nothing's
perfect, and those diodes probably aren't made in the most exacting
processes.

I would be blown away if you *couldn't* measure some ohmic current flow
in a diode at any particular voltage level.

-- jm

------------------------------------------------------
http://www.qsl.net/ke5fx
Note: My E-mail address has been altered to avoid spam
------------------------------------------------------

"Winfield Hill" -edu wrote in
message ...
john jardine wrote...

I know where you're coming in from, but whichever way I look at Win's
graphs
I'm seeing a straight line for that 1N4148 at 0-30mv levels. Yes, the
scales
are log-log but the constant of proportionality is dead straight linear.
I.e
1mv RF in gives 1e-10 amps and 10mV in gives 1e-9 amps and pro-rata for
all
points in between (you did see the double decade increments?). This
agrees
with the 10Mohm value that's marked on the graph.

Right, but that part of my measurements cries out for further bench
exploration. It represents only one part, and is unconfirmed. Also,
what happens if the voltage is reversed? Are we to believe the diode
is a 10M resistor, shunted by a diode? I'm not comfortable with that.


--
Thanks,
- Win

Test on a 1N4148.
ForwardV DiodeR
+50mV 8megs.
+30mV 9megs.
+20mV 10megs.
+10mv 12megs.
+5mV 21megs.
ReverseV
-5mV 21megs.
-10mV 30megs.
-30mV 270megs.

All the figures are suspect as the noise filter had a long settling time
and I was too lazy to wait but they do show a very sharp reverse cutoff, at
about 10-15mV.

And a bonus, a rectification test feeding the diode from a 60Hz source and
10k series R ...
ACi/p DCo/p
430mV 59mV
300mV 12mV
200mV 2.4mV
100mV 140uV
60mV 66uV
30mV 18uV
20mV 9uV
10mV 1uV
regards
john

John Miles wrote...

Winfield Hill wrote...
http://www.picovolt.com/win/elec/com...de-curves.html ...
that part of my measurements cries out for further bench exploration.
It represents only one part, and is unconfirmed. Also, what happens
if the voltage is reversed? Are we to believe the diode is a 10M
resistor, shunted by a diode? I'm not comfortable with that.

I'm confused. Is there some reason to expect the semiconductor material
to be a perfect insulator with no resistivity at all? Nothing's
perfect, and those diodes probably aren't made in the most exacting
processes.

I would be blown away if you *couldn't* measure some ohmic current flow
in a diode at any particular voltage level.

Agreed. It's the rather low 10M value that raises my eyebrows.
Hence my suggestion that the measurements be revisited. Picked up
by John Jardine, who obtained similar values, copied below:

Test on a 1N4148.
ForwardV DiodeR
+50mV 8megs.
+30mV 9megs.
+20mV 10megs.
+10mv 12megs.
+5mV 21megs.
ReverseV
-5mV 21megs.
-10mV 30megs.
-30mV 270megs.

John also suggests the measurements may need further refinement.

Oops! I can think of several circuits I've designed over the years
using diodes for discharge protection that might not work exactly as
I intended, given this observation. And I recall several circuits
where I intentionally back biased the diode a few hundred millivolts
to insure an open circuit.


--
Thanks,
- Win

I read in sci.electronics.design that john jardine
wrote (in
k) about 'Diode and very small amplitude high frequencies signals', on
Sat, 5 Feb 2005:
Test on a 1N4148.
ForwardV DiodeR
+50mV 8megs.
+30mV 9megs.
+20mV 10megs.
+10mv 12megs.
+5mV 21megs.
ReverseV
-5mV 21megs.
-10mV 30megs.
-30mV 270megs.

How did you measure the resistance? Is it an incremental resistance
(slope of the V/I curve at the data point) or the slope of a line
joining the origin to the data point on the curve.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk