Diode and very small amplitude high frequencies signals

You're going to get some flak from the crystal radio crowd on that one.

Crystal radios generally step up the voltage from the antenna, and use a sensitive, high-impedance earphone -- and still require relatively strong signals to receive well. For packaged diodes consider a point-contact germanium or a zero-bias schottkey (although the real crystal fans will want you to stick with germanium or other material, per the other poster). Clever tricks to play include audio amplification after the diode and a very slight forward bias applied to the diode.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.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.

Information on diodes for small signal detector use

Many crystal set devotees prefer iron pyrite with a cats whisker over today's diodes for sensitivity to small signals.

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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

One of the Profs in college built a "crystal" radio that was powerful enough to drive a loudspeaker. It played day and night in the lab. He had it set up so he could change the detector, but left the LED in there because it worked as an "on" indicator too. ;-)

--
  Keith

"johna@m" schreef in bericht news: snipped-for-privacy@c13g2000cwb.googlegroups.com...

The answer to your question is that real, simple AM receivers - crystal sets - only work where the received signal is quite high. They tend to use fast, low capacitance diodes. The term "rectifier" tends to be used for bigger, slower diodes basically intended to handle around an ampere of current from a 50/60Hz source, that look like capacitors in RF circuits.

Practical AM receivers always amplify the signal before they detect it, and usually "mix" the amplified signal from the antenna with the output from a local oscillator at different, if similar, frequency chosen to be 455kHz away for the transmitted signal. The nominally 455kHz component coming out of the mixer is then filtered by an elaborate bandpass filter to reject all the other components, further amplified, and only then detected.

Search on "superheterodyne".

------------- Bill Sloman, Nijmegen

What makes you sure the simulation and the reality agree ? There is capacive coupling over the PN structure of the diode to start with. And then at one point a diode has an exponential characterization instead of binary on/off.

Rene

--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net

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!!

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:

johna@m wrote:

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,

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

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.

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

weak

and

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.

real

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.

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

Thanks!.

from the

these

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

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

"W>>

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

real

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.

John

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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.

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