Winfield Hill wrote...

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.

And others where I used a transistor collector or JFET gate instead.

Pease Porridge in the Feb 3rd issue of Electronic Design mentions
this problem, and Bob suggests using a transistor. "Using 2n3904s
as diodes is very important because most ordinary diodes are much
too leaky around +/-60mV to work well. Ordinary gold-doped 1n914s
and 1n4148s are quite unsuitable..."


--
Thanks,
- Win

On Fri, 04 Feb 2005 17:15:53 +0000, Paul Burridge
wrote:

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.

Yes if you use a serial resonant with a capacitor, inductor and a
resistor as you're suggesting you get voltage amplification factor
exactly equal to the Q of the circuit plus you get frequency and
bandwidth selectivity

With a transformer you get all 3 of the above without having to add an
inductor (as you use the inductors in the windings of the transformer)
plus you get impedance level shifting of the capacitance and
resistance in the secondary to the primary multiplied by the square of
the turns ratio multiplied by the capacitance and resistance in the
secondary of the transformer.
Is it worth it. I can't tell but I see more parallel resonant circuits
then serial ones .

I've actually seen implementations of the above using positive and
negative feedback circuits with an opamp to get some really interesting
results.

"John Woodgate" wrote in message
news
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

I didn't measure the resistance. The values just come from the static V and
I plot points on the graph.
Having had my remaining bench DVM, (good ol'e UK, Datron ****e) pack in on
me and 2 battery DVMs keel over with flat batteries and the CMOS buffers
floating off to la la land and 2 crocodile clips secretly fail and finally
my electric pencil sharpener going tits up, I was not of a mind to press
on :-)
regards
john

I haven't followed this thread very thoroughly, so this might not be
directly relevant. But it should be of interest to anyone trying to
detect small signals with a diode.

There are several reasons why diodes do poorly with small AC signals.

The first is, of course, the forward drop. However, this can in theory
be reduced to an arbitrarily low value by reducing the current to a low
enough value (by, for example, making the load impedance high enough).

The second is that the ratio of reverse to forward current increases as
the signal gets smaller and smaller, reaching one at the limit. This can
be observed by looking at the I-V curve of a diode. At the origin, the
curve is a straight line -- the diode behaves just like a resistor.

The third reason is the diode capacitance. This shunts the diode,
effectively lowering the reverse impedance. It also lowers the forward
impedance, but when the forward Z is lower than the reverse Z, the net
effect is to further degrade the forward/reverse impedance ratio.

You can make all the DC measurements you want, but they only tell half
the story. When you apply AC, you charge the load capacitor during half
the cycle according to the diode's forward impedance, and charge is
removed from it during the other half according to the diode's reverse
impedance. As the forward/reverse impedance ratio degrades due to the
two effects mentioned above, the net charge you get in the load
capacitance decreases, hence the voltage it's charged to decreases. This
ends up looking like a larger diode forward drop.

I spent a lot of time thinking about this some years ago when designing
a QRP wattmeter, and some of the conclusions I came to appear in the
resulting article, "A Simple and Accurate QRP Directional Wattmeter",
published in QST, February 1990. See the analysis on p. 20, "Ac v Dc:
Why the Difference?"

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
I haven't followed this thread very thoroughly, so this might not be
directly relevant. But it should be of interest to anyone trying to
detect small signals with a diode.

I spent a lot of time thinking about this some years ago when designing
a QRP wattmeter, and some of the conclusions I came to appear in the
resulting article, "A Simple and Accurate QRP Directional Wattmeter",
published in QST, February 1990. See the analysis on p. 20, "Ac v Dc:
Why the Difference?"

Roy Lewallen, W7EL

Your article sounds interesting. Is there a link available to see it?.
The simplest approach I've seen, was is in the 'Levell TM6A broadband
voltmeter'(UK). Designer chopped the low level diode output at 20Hz,
allowing a 1mVac FSD.
regards
john

Roy Lewallen wrote:

[...]

The second is that the ratio of reverse to forward current increases as
the signal gets smaller and smaller, reaching one at the limit. This can
be observed by looking at the I-V curve of a diode. At the origin, the
curve is a straight line - the diode behaves just like a resistor.

[...]

Roy Lewallen, W7EL

Excellent description - thanks.

Only one small problem - as Win pointed out, Bob Pease feels a
diode-connected 2N3904 has lower leakage at low voltage than a 1N4148:

"What's All This Comparator Stuff, Anyhow?"

http://www.elecdesign.com/Articles/A...9517/9517.html

Does this mean a 2N3904 has a shallower slope than a 1N4148 through zero, or
perhaps one or the other has an offset, such as the Agilent Zero Bias
Schottky Detector Diodes shown in AN969?

http://www.spelektroniikka.fi/kuvat/schot8.pdf

Regards,

Mike Monett

Roger Lascelles wrote:

"Tim Wescott" wrote in message
...
johna@m wrote:
Should not we expect that the current, even at very small level, to be
half rectified by a diode, since the reverse resistance of the diode is
supposed te be far greater than the forward resistance?

Why can't we found this result in smulation. Is it a flaw in the
simulator (Simplorer) or is the theoric behavior of a diode that
changes in case of very small input ?

Regards,

John.

The diode behavior is a continuous curve, so for a small AC voltage you
won't see much change in the diode's resistance even at zero bias.
Unless you're modeling a really leaky diode, however, you are probably
seeing a situation where the diode's resistance is effectively shunted
by it's capacitance and you are seeing capacitive coupling rather than
conduction.

The point about continuous curve is well made.

The diode doesn't have to hard rectify. As long as it has a non-linear V-I
graph it will produce some audio. The more sharply curved the
characteristic, the more audio is produced.

In the valve days, the anode bend detector worked that way, using a valve
biased to operate on the curved part of the characteristic.

Roger

Right. Take a look at a diode curve across a 100 uV region. Most diodes
will look pretty flat, even without considering the effects of
capacitance and other parasitics.

--
Paul Hovnanian
------------------------------------------------------------------
My inner child can beat up your inner child.
-- Alex Greenbank

john jardine wrote:

"John Woodgate" wrote in message
news
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

I didn't measure the resistance. The values just come from the static V and
I plot points on the graph.
Having had my remaining bench DVM, (good ol'e UK, Datron ****e) pack in on
me and 2 battery DVMs keel over with flat batteries and the CMOS buffers
floating off to la la land and 2 crocodile clips secretly fail and finally
my electric pencil sharpener going tits up, I was not of a mind to press
on :-)
regards
john

In other words, this data is just a plot of a diode's DC I vs V
characteristic, right?

What is of more interest is the slope at a given DC operating point. If
we pick 0V, for example, the above data (within the limits of its
precision) gives a flat line around that point (+5mV 21 Mohms, -5mV 21
Mohms). With a 100 uV signal, you might as well throw a 21 M ohm
resistor in there instead.

--
Paul Hovnanian
------------------------------------------------------------------
Think honk if you're a telepath.

Mike Monett wrote...

Roy Lewallen wrote:

The second is that the ratio of reverse to forward current increases as
the signal gets smaller and smaller, reaching one at the limit. This can
be observed by looking at the I-V curve of a diode. At the origin, the
curve is a straight line - the diode behaves just like a resistor. ...

Excellent description - thanks.

Only one small problem - as Win pointed out, Bob Pease feels a
diode-connected 2N3904 has lower leakage at low voltage than a 1N4148:
"What's All This Comparator Stuff, Anyhow?"
http://www.elecdesign.com/Articles/A...9517/9517.html

Does this mean a 2N3904 has a shallower slope than a 1N4148 through zero,
or perhaps one or the other has an offset, such as the Agilent Zero Bias
Schottky Detector Diodes shown in AN969?

No, it means its a better diode at low currents. See my curves again,
http://www.picovolt.com/win/elec/com...de-curves.html Note the
1n458 and the JFET diodes, which follow the theoretical 60mV/decade rule
down to very low currents. As for Roy Lewallen's "ratio of reverse to
forward current" argument, there is no reverse current for these fine
fellows, at least for DC and reasonably low frequencies. It's the very
crummy gold-doped 1n4148 that falls over. Awwkk!


--
Thanks,
- Win

"Paul Hovnanian P.E." wrote in message
...

In other words, this data is just a plot of a diode's DC I vs V
characteristic, right?

What is of more interest is the slope at a given DC operating point. If
we pick 0V, for example, the above data (within the limits of its
precision) gives a flat line around that point (+5mV 21 Mohms, -5mV 21
Mohms). With a 100 uV signal, you might as well throw a 21 M ohm
resistor in there instead.

--
Paul Hovnanian
------------------------------------------------------------------
Think honk if you're a telepath.

Yep. Straight as a die between +/- 4mV. Incremental resistance dV/di =10Mohm
regards
john