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Diode and very small amplitude high frequencies signals
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" schreef in bericht oups.com... Hello All, I am trying to simulate a simple 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 sine 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 schematic for AM simple receiver, there is no amplification 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, 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 |
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, 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 |
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. 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. http://uweb.superlink.net/~bhtongue/.../10npddec.html http://uweb.superlink.net/~bhtongue/.../16MeaDio.html 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. |
Bill Sloman wrote:
"johna@m" schreef in bericht oups.com... Hello All, I am trying to simulate a simple 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 sine 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 schematic for AM simple receiver, there is no amplification 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, 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, You're going to get some flak from the crystal radio crowd on that one. 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. 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 |
Hi,
In days of yore a crystal set would have had a longish wire aerial (several millivolts output), a high-Q tuned circuit (further magnification) and a high-Z load on the detector all, of which lessened the need for a perfect characteristic. And that's before you placed the headphones in a pudding bowl. Cheers - Joe |
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. |
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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. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com |
"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 |
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, |
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 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. |
"Mark" wrote in message oups.com... 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 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 |
"K7ITM" wrote in message oups.com... 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 |
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 |
"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 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 |
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. -- "What is now proved was once only imagin'd." - William Blake, 1793. |
In article , hill_a@t_rowland-dotties-
harvard-dot.s-edu says... 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 ------------------------------------------------------ http://www.qsl.net/ke5fx Note: My E-mail address has been altered to avoid spam ------------------------------------------------------ |
"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 |
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. -- Thanks, - Win |
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 |
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 ... 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 ... 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 |
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