"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

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

"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

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.

"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

Paul Hovnanian P.E. wrote:

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.


Exactly. At very small currents, the diode is just a resistor (shunted
by a capacitance). At slightly small currents, it's a very poor diode,
with reverse current almost equal to the forward current, so on each
negative half cycle you suck out nearly all the charge you delivered
during the positive half cycle.

Roy Lewallen, W7EL