Diode and very small amplitude high frequencies signals

[lemonjuice]

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/comp/diode/diode-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.

 

[john jardine]

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

 

[john jardine]

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

 

[Winfield Hill]

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.

 

[john jardine]

john jardine wrote...

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.

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

 

[Paul Burridge]

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.

 

[John Miles]

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

 

[john jardine]

john jardine wrote...

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.

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

 

[Winfield Hill]

John Miles wrote...

Winfield Hill wrote...

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.

 

[John Woodgate]

I read in sci.electronics.design that john jardine
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.

 

[Winfield Hill]

Winfield Hill wrote...

John Miles wrote...

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..."

 

[lemonjuice]

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 jardine]

I read in sci.electronics.design that john jardine
k>) about 'Diode and very small amplitude high frequencies signals', on
Sat, 5 Feb 2005:

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 shite) 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

 

[Roy Lewallen]

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

 

[john jardine]

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

 

[Mike Monett]

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/ArticleID/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

 

[Paul Hovnanian P.E.]

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 P.E.]

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 shite) 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.

 

[Winfield Hill]

Mike Monett wrote...

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/ArticleID/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/comp/diode/diode-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!

 

[john jardine]

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.

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