Diodes have a tresholdvoltage of 0.3-0.7 V.
The amplitude must be greater. That is why this
type of receiver requires a large antenna or a
transmitter nearby.


Regards,


Website: http://members.chello.nl/g.baars13




johna@m wrote:
Hello All,

I am trying to simulate a simpe AM receiver circuit with diode
detector. I am assuming that the signal received from the antenna
(simulated with a voltage source) has a weak amplitude (around 100 uV)
and a high frequency (around 600 Khz). The issue is that the current
after the diode does not get rectified. The output current is very weak
(less than 250pA) and still contains the full sin signal (both halves
of signals).

When I try the simulation with smaller frequencies (around 5kHz) and
higher amplitude (around 0.2 v), the signal gets correctly
half-rectified, but not anymore when I work with higher frequencies and
smaller signals.

In real shematic for AM simple receiver, there is no ampification
bewteen the antenna (and the tuning LC circuit) and the diode. So how
the diode manage to half-rectifies correctly in real operating mode
when the signal is weak and high frequencies, which is the case of real
radio signals.

I use Ansoft Simplorer mainly. Any other simulators recommended ?
Thanks in advance and best regards,

John.

"johna@m" wrote in message
oups.com...
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.


There is nothing wrong with the simulator... the problem is with your idea
of a diode. The general definition of a diode is a component that conducts
normally in one direction, but does not conduct in the other. That
definition only applies to a "perfect" diode. The reality of semiconductor
diodes is that a 'barrier potential" exists across the junction. In
germanium diodes, this is around 0.3 volts; in silicon diodes, it's around
0.6 volts.
In order for the diode to conduct, this barrier potential must be exceeded
by an externally applied voltage. Until that potential is reached, the
diode is said to be reverse biased, and only a very small leakage current
flows. When the barrier potential is reached, the junction becomes forward
biased and conducts heavily.

The small signal voltage that you are trying to simulate may not be enough
to reach the barrier potential of the diode junction, thus, no conduction
(rectification) in either direction. The simulator is aware of the barrier
potential of the diode. If the peak value of your signal voltage is less
than the barrier potential, no rectification occurs. If you increase the
amplitude of the signal applied to the defined barrier potential of the
particular diode in your model, you will see rectification begin. The
higher the signal amplitude, the more rectified signal appears on the
output.


You can make a diode rectify a signal amplitude lower than the barrier
potential by applying a forward voltage that is just under the barrier
potential, so that the signal doesn't have to overcome the full barrier
potential. Fer instance, if you apply a 0.5 volt DC voltage to a silicon
diode, it will start to rectify signal levels as low as 0.1 volts.

--
Dave M
MasonDG44 at comcast dot net (Just subsitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!

DaveM wrote...

"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 ?

There is nothing wrong with the simulator... the problem is with your
idea of a diode. The general definition of a diode is a component that
conducts normally in one direction, but does not conduct in the other.
That definition only applies to a "perfect" diode. The reality of
semiconductor diodes is that a 'barrier potential" exists across the
junction. In germanium diodes, this is around 0.3 volts; in silicon
diodes, it's around 0.6 volts. In order for the diode to conduct, this
barrier potential must be exceeded by an externally applied voltage.
Until that potential is reached, the diode is said to be reverse biased,
and only a very small leakage current flows. When the barrier potential
is reached, the junction becomes forward biased and conducts heavily.

The small signal voltage that you are trying to simulate may not be enough
to reach the barrier potential of the diode junction, thus, no conduction
(rectification) in either direction. The simulator is aware of the barrier
potential of the diode. If the peak value of your signal voltage is less
than the barrier potential, no rectification occurs. If you increase the
amplitude of the signal applied to the defined barrier potential of the
particular diode in your model, you will see rectification begin. The
higher the signal amplitude, the more rectified signal appears on the
output.

You can make a diode rectify a signal amplitude lower than the barrier
potential by applying a forward voltage that is just under the barrier
potential, so that the signal doesn't have to overcome the full barrier
potential. Fer instance, if you apply a 0.5 volt DC voltage to a silicon
diode, it will start to rectify signal levels as low as 0.1 volts.

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

For example, these plots show that an ordinary 1n4148 class of silicon
signal diode, which conducts about 0.5mA at 600mV, is still working at
250mV, conducting 1uA in my measurements. In fact, this diode was still
conducted at 100mV. See http://www.fairchildsemi.com/ds/1N/1N4148.pdf
where Fairchild's datasheet also shows this exponential relationship,
albeit drawn with a draftsman's straight line.

So, as others have pointed out, diodes can rectify very small signals.
They may not be very efficient, but they will work. These plots also
show how Schottky diodes (e.g., 1n6263 and 1n5819) are better than
ordinary silicon diodes at low voltages, even below 100mV. The 1n6263
may be hard to get, but other parts, like the sd101 or bat17 may not.
http://www.vishay.com/docs/85629/85629.pdf

There are other diodes that work well at very low voltages, notably
some made by Agilent (see an1090), but we won't go into them here.


--
Thanks,
- Win

The difficulty is "what to do with that 1uA current". To put it to practical
use, a signal processor is needed that has a useful output. For example, a
MOSFET amplifier with a 10 megohm input resistance and negligible input
capacitance (for low frequency sigs) could be used.

In this case it would be better to rethink the project.
In addition to the diode, some system design is indicated.

Bill W0IYH

"Winfield Hill" -edu wrote in
message ...
DaveM wrote...

"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 ?

There is nothing wrong with the simulator... the problem is with your
idea of a diode. The general definition of a diode is a component that
conducts normally in one direction, but does not conduct in the other.
That definition only applies to a "perfect" diode. The reality of
semiconductor diodes is that a 'barrier potential" exists across the
junction. In germanium diodes, this is around 0.3 volts; in silicon
diodes, it's around 0.6 volts. In order for the diode to conduct, this
barrier potential must be exceeded by an externally applied voltage.
Until that potential is reached, the diode is said to be reverse biased,
and only a very small leakage current flows. When the barrier potential
is reached, the junction becomes forward biased and conducts heavily.

The small signal voltage that you are trying to simulate may not be
enough
to reach the barrier potential of the diode junction, thus, no conduction
(rectification) in either direction. The simulator is aware of the
barrier
potential of the diode. If the peak value of your signal voltage is less
than the barrier potential, no rectification occurs. If you increase the
amplitude of the signal applied to the defined barrier potential of the
particular diode in your model, you will see rectification begin. The
higher the signal amplitude, the more rectified signal appears on the
output.

You can make a diode rectify a signal amplitude lower than the barrier
potential by applying a forward voltage that is just under the barrier
potential, so that the signal doesn't have to overcome the full barrier
potential. Fer instance, if you apply a 0.5 volt DC voltage to a silicon
diode, it will start to rectify signal levels as low as 0.1 volts.

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

For example, these plots show that an ordinary 1n4148 class of silicon
signal diode, which conducts about 0.5mA at 600mV, is still working at
250mV, conducting 1uA in my measurements. In fact, this diode was still
conducted at 100mV. See http://www.fairchildsemi.com/ds/1N/1N4148.pdf
where Fairchild's datasheet also shows this exponential relationship,
albeit drawn with a draftsman's straight line.

So, as others have pointed out, diodes can rectify very small signals.
They may not be very efficient, but they will work. These plots also
show how Schottky diodes (e.g., 1n6263 and 1n5819) are better than
ordinary silicon diodes at low voltages, even below 100mV. The 1n6263
may be hard to get, but other parts, like the sd101 or bat17 may not.
http://www.vishay.com/docs/85629/85629.pdf

There are other diodes that work well at very low voltages, notably
some made by Agilent (see an1090), but we won't go into them here.


--
Thanks,
- Win

On 2 Feb 2005 05:07:50 -0800, "johna@m" wrote:

Hello All,

I am trying to simulate a simpe AM receiver circuit with diode
detector. I am assuming that the signal received from the antenna
(simulated with a voltage source) has a weak amplitude (around 100 uV)
and a high frequency (around 600 Khz). The issue is that the current
after the diode does not get rectified. The output current is very
weak
(less than 250pA) and still contains the full sin signal (both halves
of signals).

When I try the simulation with smaller frequencies (around 5kHz) and
higher amplitude (around 0.2 v), the signal gets correctly
half-rectified, but not anymore when I work with higher frequencies
and
smaller signals.


Yes its what you'd expect. As I explained on the thread "Junction
capacitance of diodes and zeners" the Capacitance of a forward biased
diode increases with the applied bias or if you do the math you'll
see it increases exponentially with it. They are actually 2
capacitances in consideration but thats another question. Its really
hard modelling a non linear component with linear components but the
following can be said to be true for small variations of voltage. You
get your desired precision by adding up together pieces of this model.
If you look at a piece wise linear approximation model of the diode
you'll see it has a conductance or a linear dependent current source in
parallel with 2 condensers and the current through the conductance is
Is*exp(Vapp/Vt). Vapp is the voltage across the condensers. At high
Vapp and low frequencies from above C is high and has a lower
impedance so Idiode is high. At higher frquencies Vapp on the
condensers is pretty low so Idiode is also low. If Vapp is lower you
get an even lower value.

In real shematic for AM simple receiver, there is no ampification
bewteen the antenna (and the tuning LC circuit) and the diode. So how
the diode manage to half-rectifies correctly in real operating mode
when the signal is weak and high frequencies, which is the case of
real
radio signals.

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.
Good luck.

Hi,

I have re-done the simulation with the diode 1N41481, and it rectified.
Thanks.

For the 600mv, it did give indeed around 0.5mA. I even tried at 100mv
and at 100uv, and it still conducts (respectively 60uA and 25nA).

Regards,

John.

"Winfield Hill" -edu wrote in
message ...
For example, these plots show that an ordinary 1n4148 class of silicon
signal diode, which conducts about 0.5mA at 600mV, is still working at
250mV, conducting 1uA in my measurements. In fact, this diode was still
conducted at 100mV. See http://www.fairchildsemi.com/ds/1N/1N4148.pdf
where Fairchild's datasheet also shows this exponential relationship,
albeit drawn with a draftsman's straight line.

So, as others have pointed out, diodes can rectify very small signals.
They may not be very efficient, but they will work. These plots also
show how Schottky diodes (e.g., 1n6263 and 1n5819) are better than
ordinary silicon diodes at low voltages, even below 100mV. The 1n6263
may be hard to get, but other parts, like the sd101 or bat17 may not.
http://www.vishay.com/docs/85629/85629.pdf

There are other diodes that work well at very low voltages, notably
some made by Agilent (see an1090), but we won't go into them here.


--
Thanks,
- Win


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

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

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

In real shematic for AM simple receiver, there is no ampification
bewteen the antenna (and the tuning LC circuit) and the diode. So how
the diode manage to half-rectifies correctly in real operating mode
when the signal is weak and high frequencies, which is the case of
real
radio signals.

You'll find that in reality something your simulator won't take into
account that its preferable to transmit at higher frequencies because
reception is better as the 1/f flicker noise is reduced to lower
values. Thats just 1 of the several reasons why frequency multipliers
are used in transmitters and downconverters in receivers.
I use Ansoft Simplorer mainly. Any other simulators recommended ?
Thanks in advance and best regards,