Practical considerations, helpful definitions of terms and useful explanations of some concepts used in this Site

1.  An explanation as to why some diodes that work well in the Broadcast Band cause low sensitivity and selectivity when used at Short Waves:  The parasitic (approximately fixed) series resistance Rs of a diode is in series with the parallel active elements. The nonlinear active elements are the junction resistance Rj, which is a function of current through the diode, and the junction capacitance Cj, which is a function of the voltage across it.

The parasitic series resistance of some 1N34 diodes can be pretty high, and in series with the junction capacitance, make that capacitance have a rather low Q at high frequencies.  This capacitance is, in a crystal radio set, effectively in parallel with the RF tank. The tank usually has a small value tuning capacitor itself, so the overall tank circuit Q is reduced at high frequencies.  This is the main reason why diodes having large values for Rs and CJ perform poorly at high frequencies. 

2.  An explanation of the meaning and use of dB and dBm:  In the acronym dBm, "d" means one-tenth.  "B" refers to the Bel and is named after Alexander Graham Bell.  The Bel is used to express the ratio of two powers, say (Output Power)/(Input Power).  Let's call this power ratio "(pr)".  Mathematically, a power ratio, expressed in Bels, is equal to the logarithm of the ratio of the two powers.  B=log (pr).  If the two powers are equal, the power ratio expressed in Bels is 0 B.  This is because the log of one is zero.  Another illustration:  Assume that the power ratio is twenty. (Pr)=20.  The logarithm of 20 is about 1.3. This power ratio in Bels is 1.3 B.  One decibel is equal to 0.1 Bel.  That is, 10 dB=1 B.  If we express the two power ratios mentioned above (1 and 20) in dB, we get 0 dB and about 13 dB.

So far, we have seen that the decibel is used to express the ratio of two powers, it is not a measure of a power level itself.  A convenient way to express an actual power level using dB is to use a standard implied reference power for one of the powers.  dBW does this.  It expresses the ratio of a power to the reference power (One Watt in this case).  dBm uses a reference power of one milliwatt.  A power level of, say 100 milliwatts, can be said to be a power level of +20 dBm (twenty dB above one milliwatt).  Why?  (100 milliwatts)/(1 milliwatt)=100.  The logarithm of 100 is 2.  10 times 2 equals 20.

The convenient thing about using dB comes from a property of logarithms:  The logarithm of the product of two numbers is equal to the sum of the logarithm of each number, taken separately.  An illustration:  If one has a power source of, say 2.5 mW and amplifies it through an amplifier having a power gain of, say 80 times, the output power is 2.5 X 80=200 mW.  2.5 mW expressed in dBm is +4 about dBm.  A power gain of 80 times is about +19 dB.  The output power is 4+19=+23 dBm.

3.  Maximum Available Power:  If one has a voltage source Vs with an inaccessible internal resistance Rs, the load resistance to which the most power (Pa) can be delivered is equal to Rs.  Pa is called the 'maximum available power' from the  source Vs, Rs.  Any load resistance other than one equal to the source resistance, Rs, will absorb less power.  This applies whether the voltage is DC or AC (RMS).  The formula for power absorbed in a resistance is "voltage-squared divided by resistance".  In the impedance matched condition, because of the 2 to 1 voltage division between the source resistance and load resistance, one-half of the internal voltage Vs will be lost across the internal source resistance.  The other half will appear across the load resistance.  The actual power available to the load will be, as indicated in the preceding relation: Pa = [(Vs/2)^2]/Rs = (Vs^2)/(4*Rs).  Again, in the impedance matched condition, the total power delivered to the series combination of source and load resistance is divided up into two halves.  One half is unavoidably lost in the internal source resistance.  The other half is delivered as "useful output power" to the load resistance.

The 'maximum available power' approach is useful when measuring the insertion power-loss of two-port devices such as transformers, amplifiers and crystal radio sets, which may not exhibit an input or output impedance that is matched to the power source.  The input impedance may be, in fact a combination of resistive and reactive components.  If the Vs,Rs source is connected to a resistive load (Ro) of value equal to Rs ohms, it will receive and dissipate a power of Pa Watts.  This is the maximum available power from the Vs, Rs source, so we can say we have a 'no loss' situation.  Now, assume that a transformer or other two-port device is inserted between the Vs,Rs source and Ro, and that an output voltage Vo is developed across Ro.  The output power is (Vo^2)/Ro.  The 'insertion power loss' can now be calculated.  It is: 10*log (output power)/(maximum available input power) dB.  After substituting terms, the equation becomes:  Insertion power loss =10*log [(Vo/Vs)^2)*(4*Rs/Ro)] dB.

If the input voltage is referred to by its peak value (Vsp) as it is in a SPICE simulation, instead of by its RMS value, the equation changes.  The RMS voltage of a sine wave is equal to the peak value of that wave divided by the "square root of 2".  Since the power equation squares the voltage, the equation for the 'available input power' changes to Pa = (Vsp^2)/(8Rs).

4.  Diode Saturation Current and Ideality Factor:  Saturation current is abbreviated as Is in all of these articles.  Assume that one connects a DC voltage source to a diode with the polarity of the voltage source such as to bias the diode in the back direction.  Increase the voltage from zero.  If the diode obeys the classic Shockley ideal equation exactly, the current will start increasing, but the increase will flatten out to a value called the saturation current as the voltage is further increased.  That is, as the voltage is increased, the current will asymptotically approach the saturation current for that diode.  A real world diode has several mechanisms that cause the current to actually keep increasing somewhat and not flatten out as the back direction voltage is further increased.  Diode manufacturers characterize this as reverse breakdown and specify that the back current will be less than a specified value, say 10 uA at a specified voltage, say 30 V, called the reverse breakdown voltage.  BTW there are other causes of excessive reverse current that are collectively referred to as reverse bias excess leakage current.  Some diodes have a sharp, controlled increase in reverse current at a specified voltage.  These diodes are called Zener diodes.

Diode Saturation Current is a very important SPICE parameter that, along with the diode Ideality Factor, n determines the actual diode current when it is forward biased by at particular DC Voltage.  Id=Is*(e^(Vd/(0.026*n)-1) at room temperature.  This expression ignores the effect of the parasitic series resistance of the diode because it has little effect on the operation of crystal radio sets at the low currents usually encountered.  Here Id is the diode current, e is the base of the natural logarithms (2.7183...),  ^ means raise the preceding symbol to the power of the expression that follows (Sometimes e^ is written 'exp'), * means multiply the preceding and following symbols, VD is the voltage across the diode and n equals the "Ideality factor" of the diode.  At low signal levels, most detector diodes have an n of between 1.05 and 1.2).  The lower the value of n, the higher will be the weak signal sensitivity.  One can see that Is is a scaling factor for the actual curve generated by the factor (e^(VD/(0.026*n)-1).

Diode ideality factor (n):  The value of n affects the low signal level sensitivity of a diode detector and its RF and audio resistance values.  n can vary between 1.0 and 2.0.  The higher the value of n, the worse the low signal level detector sensitivity.  The low signal level RF and audio resistances of a diode detector vary directly with the value of n.  Schottky diodes usually have a value of n between 1.03 and 1.10.  Good germanium diodes have an n of about 1.07 to 1.14 when detecting weak signals.  Silicon p-n junction diodes such as the 1N914 have values of n of about 1.8 at low currents and therefore have a lower potential sensitivity as diode detectors than Schottky and germanium point contact diodes.  The value of n in Schottky diodes seems to be approximately constant over the full range of currents and voltages encountered in crystal radio set operation, but varies with diode current in silicon pn junction and germanium point contact diodes.  A way of thinking about n is to consider it as a factor that effectively reduces the applied signal voltage to a diode detector compared to the case of using an ideal diode having an n of 1.0.  Less applied signal, of course, results in less detected output.

Here are a few bits of information relative to diodes:

Typically, if a diode is biased at 0.0282*n volts in the forward direction, it will pass a current of 2 times its Is.  If it is biased at 0.0182*n volts in the reverse direction, it will pass a current of 0.5 times its Is.  If a diode is biased at 0.0616*n volts in the forward direction, it will pass a current of 10 times its Is.  If it is biased at -0.0592*n volts, it will pass a current of -0.9 times its Is. These values are predicted from the classic Shockley equation.  In the real world, reverse current can depart substantially from values predicted by the equation because of effects not modeled (the reverse current becomes higher).  Gold bonded germanium diodes usually depart somewhat from the predicted values when operated in the forward direction.  The effect appears as an increase of Is when measurements are made at currents above about 6 times the low-current Is.

Values of Is and n determine the location of the apparent 'knee" on a linear graph of the diode forward current vs. forward voltage. See Article #7.  An easy way to estimate the approximate value of Is can be found in Article #4, section 2.  A method of measuring Is and n is given in Article #16.

If one connects two identical diodes in parallel, the combo will behave as a single diode having twice the Is, and the same n as one of them.  If one connects two identical diodes in series, the combo will behave as a single diode having twice the n and the same Is as one of them.  This connection results in a diode having 3 dB less potential weak signal output than one of the diodes by itself.

5.  Explanation of why, in a diode detector, and by how much, the RF input resistance and audio output resistances change as a function of input signal power.  Consider first, a diode detector that is well impedance-matched both at its input and its output when driven by a very low power RF input signal.   There will then exist an appreciable power loss in the detector.  (The audio output power will be appreciably less than the input power.).  The input and output impedances of the detector will approximately equal each other and approach:  Rd = 0.026*n/Is.  See part 3 above for a definition of terms.  For this illustration, let the diode have an Is of 38 nA and an n of 1.02.  Rd will be 700k Ohms.  The well impedance-matched condition will hold if the input power is raised from a low value, but only up to a point.  After that , the match will start to deteriorate.  At an input power about 15 dB above that of the square-law-linear crossover point, the match will have deteriorated to a VSWR of about 1.5:1  (VSWR = Voltage Standing Wave Ratio.).  A further increase of input signal power will result in a further increase of VSWR.  This means that the input and output resistances of the detector have changed from their previously matched values.  The input resistance of the diode detector decreased from the value obtained in the well impedance matched low power level situation.  The output resistance increased.  The reason for this change is that a new law now governs input and output resistance when a diode detector is operated at a high enough power level to result in a low detector insertion power loss.  It now operates as a peak detector.  The rule here is that the CW RF input resistance of a diode peak detector approaches ½ the value of its output load resistance.  Also, the audio output resistance approaches 2 times the value of the input AC source resistance.  Further, since the detector is now a peak detector, the DC output voltage is the "square root of 2" times larger than of the applied input RF RMS voltage. (It's equal to the peak value of that voltage).  These existence of these relationships is necessary so that in an ideal peak detector, the output power will equal the input power (No free lunch).  Summary:  Output DC voltage equals sqrt2 times input RMS voltage.  Since the output power must equal the input power, and power equals voltage squared divided by resistance, the output load resistance must equal two times the source resistance, assuming impedance matched conditions prevail.  If we were to adjust the input source resistance to, say 495k ohms (reduce it by sqrt2) and the output load resistance to 990k ohms (increase it by sqrt2) by changing the input and output impedance transformation ratios, the insertion loss would become even lower than before the change and the input and output impedance matches would be very much improved (remember we are now dealing with high signal levels).

A good compromise impedance match, from one point of view, occurs if one sets the RF source resistance to 0.794*Rd and the audio load resistance to 1.26*Rd.  With this setup, theoretically, the impedance match at both input and output remains very good over the range of signals from barely readable to strong enough to produce close to peak detection.  A measure of impedance match is "Voltage Reflection Coefficient", and in this case it is always better than 18 dB (VSWR better than 1.3).  Excess insertion loss is less than 1/3 dB and selectivity is largely independent of signal level.  Information presented in Article #28 shows that, if the diode load resistance is made equal to Rd and the RF source resistance is made equal to Rd/2, the weak signal output of the detector will be about 2 dB greater than if both ports are impedance-matched!  There is little benefit when strong signals are received, since both input and output ports become impedance matched.

Here is an interesting conceptual view of a high signal level diode detector circuit:  Assume that it is driven with a sufficiently high level sine wave voltage so it operates in its peak detection mode, and is loaded with a parallel RC of a sufficiently long time const ant.  This detector may be thought of as a low loss impedance transformer with a two-to-one impedance step up from input to output, BUT having an AC input and a DC output, instead of the usual AC input and output.   The DC output power will approximately equal the AC input power and the DC output voltage will be about sqrt 2 times the RMS AC input voltage.

5A.  A comparison of conventional half-wave and half-wave voltage-doubling detectors:  Here is some info that may be of interest re conventional half-wave detectors vs. voltage doubling half-wave detectors when each is terminated with an output load of Ro. For illustration purposes we will assume the input voltage to the detector to be 1.0 volt RMS. The RF input resistance of the detector will be designated as Ri. All diodes have the same Is and n. It is assumed that good diodes such as a 5082-2835 Schottky, ITT FO-215 germanium or other are used. The info relates to the RF input resistance of detectors (it has a large effect upon selectivity) and their output audio resistance. See Point 4 in this Article for info on diode Is and n.

A high input power level is defined as one that is high compared to that at the LSLCP of the detector. A low input power level is defined as on that is low compared to that at the LSLCP of the detector. See "Quick Summary" in Article #15 for info on LSLCP.

1) Conventional half-wave detector operating at a high input signal power level: The detector, in this case, operates as a peak detector. Since it is a passive device, its output power will approximately equal its input power, under impedance-matched conditions. The output DC voltage will approach sqrt2 times the input RMS voltage, since the peak value of a sine wave is sqrt2 times its RMS value. For the input power, (1.0^2)/Ri, to equal the output power, [(1.0*sqrt2)^2]/Ro, the input RF resistance (Ri) must equal 1/2 Ro. That is, Ri=Ro/2. This illustrates the direct interaction between the RF input resistance and output audio resistive load. At high input power levels selectivity drops when the resistive audio output load value is lowered.  The audio output resistance of the detector approaches 2 times the RF source resistance driving it. If the diode were an ideal diode, the word "approximately" should be eliminated, and "approaches" should be changed to "becomes" .

2) Conventional half-wave detector operating at a low input signal power level: The detector, in this case, does not operate as a peak detector, and exhibits significant power loss. At low input signal power levels Ri approaches 0.026*n/Is ohms (diode axis-crossing resistance) and becomes independent of the value of Ro.

The audio output resistance of the detector approaches the same value as the axis-crossing resistance (see above).

3) Half-wave voltage doubling detector operating at a high input signal power level: The detector, in this case, operates as a peak detector. Since it is a passive device, its output power will approximately equal its input power, under impedance-matched conditions. The output DC voltage will approach 2.0*sqrt2 times the input RMS voltage, since the peak of a 1.0 volt RMS sine wave is sqrt2 times its RMS value. For the input power (1.0^2)/Ri to equal the output power [(1.0*2*sqrt2)^2]/Ro, the input RF resistance (Ri) must equal 1/8 Ro. That is, Ri=Ro/8. This illustrates the direct interaction between the RF input resistance and output audio resistive load. At high input power levels selectivity drops substantially if the output resistive audio load value is lowered.

The audio output resistance of the detector approaches 8 times the RF source resistance driving it. This fact is seldom recognized and it may be the cause of some of the problems encountered by those experimenting with doublers.

4) Half-wave voltage doubler operating at a low input signal power level: The detector, in this case, does not operate as a peak detector, and it has significant power loss. At low input signal power levels Ri approaches (0.026*n/Is)/2 ohms and becomes independent of the value of Ro.

The audio output resistance of the detector approaches twice the axis-crossing resistance of the diode.

5) Summary:  At high input power levels, and with both input and output matched, power loss in both half wave and half wave voltage doubling detectors approaches zero dB.  Sound volume should be the same with either detector.  At low input power levels both detectors exhibit substantial power loss.  I believe, but have not proven, that at low input power levels the doubler has a higher power loss than the straight half wave detector, and should deliver less volume.

6.  Some misconceptions regarding Impedance matching and Crystal Radio Sets:  To understand the importance of impedance matching, one must first accept the concept of power.  A radio station accepts power from the mains and converts some of it to RF power which is radiated into space.  This power leaves the transmitting antenna at the speed of light and spreads out as it goes away from the antenna.  One can prove that power is radiated by substituting a LED diode for the regular diode, getting physically close enough to the station and then tuning it in.  The LED will light up (give off light power), showing that some power is being broadcast and that it can be picked up. Now back at home, if one tunes in the station one gets sound in the headphones.  What activates one's hearing system is the power of the perceived sound.  BTW, if one gets too much sound power in the ear for a long enough time, the power can be strong enough to break off some of the hair cells in the inner ear and reduce one's hearing sensitivity forever.  The theoretical best one can do with a crystal radio set setup is the following:  (1) Use an antenna-ground system to pick up as much as possible of the RF power passing through the air in its vicinity .  In general, a higher antenna will pick up more power from the passing RF waves than will a lower one.  (2) Convert the intelligence carrying AM sideband RF power into audio electrical power.  (3) Convert the electrical audio power into sound power and get that power  into the ear.

There are power losses at each of the three steps and our job is to minimize them in order to get as much of the sideband RF power passing through the vicinity of the antenna (capture area)  changed into audio power for our ears.  We want all of the "available power" at the antenna-ground system to be absorbed into the crystal radio set then  passed on through it to our headphones as sound.  However, some of it will be unavoidably lost in the RF tuned circuit.  If the input impedance of the crystal radio set is not correctly matched to the impedance of the antenna, some of the RF power hitting the input to the crystal radio set will be reflected back to the antenna-ground system and be lost.

An impedance-matched condition occurs when the resistance component of the input impedance of the crystal radio set equals the source resistance component of the impedance of the antenna-ground system.  Also, the reactive (inductive or capacitive) component of the impedance of the antenna-ground system must see an opposite reactive (capacitive or inductive) impedance in order to be canceled out.  In the impedance-matched condition, all of the maximum available power (See section on "Maximum Available Power" above) intercepted by the antenna-ground system is made available for use in the crystal radio set and none is reflected back towards the antenna to be lost.

Now we are at the point where confusion often exists:  The voltage concept vs. the power concept.  Let's assume that the diode detector has a RF input resistance of 90,000 Ohms.  Assume that the antenna-loaded resonant resistance of the tuned circuit driving it is 10,000 Ohms.  If one uses voltage concepts only, one might think that this represents a low loss condition.  NOT SO!  After all,  9/10 of the actual source voltage is actually applied to the detector.  If one impedance matches the 10k ohm source RF resistance to the diode 90k ohm RF resistance via RF impedance step-up transformation (maybe connecting the antenna to a tap on the tuned circuit, and leaving the diode on the top), good things happen.  (We will assume here that, in the impedance transformation to follow, the ratio of loaded-to-unloaded Q of the tuned circuits is not changed.)   For an impedance match, the tuned circuit resonant resistance should be transformed up by 9 times.  If this was done by a separate transformer (for ease of understanding) it would have a turns ratio of 1:3, stepping up the equivalent source voltage by 3 times and changing the equivalent source resistance to 90,000 Ohms.  What now?  Before matching, the diode got 9/10 of the source voltage applied to it.  Now it gets 1/2 the new equivalent source voltage (remember the equivalent voltage is 3 times the original source internal voltage).  The 1/2 comes from the 2:1 voltage division between the resistance of the equivalent source of 90,000 Ohms and the detector input resistance of 90,000 Ohms.  The ratio of the new detector voltage to the old  is: 3 times 1/2 divided by 0.9 = 1.67 times.  This equates to a 4.44 dB increase in power applied to the detector.  If the input signal to the detector is so weak that the detector is operating in the square-law region, the audio output power  will increase by 8.88 dB!  This is about a doubling of volume.

7.  Caution to observe when cutting the leads of a glass Agilent 5082-2835 Schottky diode (or any other glass diode):  When it is necessary to cut the leads of a glass packaged diode close to the glass body, use a tool that gives a scissors type of cut.  Diagonal cutters give a sudden physical shock to the diode that can damage its electrical performance.  This physical shock is greater than one might expect because of the use of plated steel instead of  more ductile copper wire. Steel is used, in part, because of its lower heat conductivity, to reduce the possibility of heat damage during soldering.

8.  Several different ways to look at a diode detector:  A diode detector can be thought of as a mixer, if one thinks of its input signal as consisting of two identical signals of equal power, in phase with each other.  It is well known that if a common AM mixer is fed with two signals of frequencies f1 and f2 Hz, most of the output it generates will consist of the second harmonic of each signal and two more signals at other frequencies.  One is at the sum frequency (f1+f2) Hz and one at the difference frequency (f1-f2).  Additional mixer products can be generated, but they will be weaker than those mentioned and will be neglected in this discussion.  In the case of an AM diode detector, we may consider that its input signal of power P Watts is in reality the sum of two equal in-phase signals, each of power P/2 and that there will be four output components, as stated above. They are:

9.  Using surface mount components in crystal radio sets:  A convenient way to connect to the tiny leads of small surface-mount diode and IC devices is to first solder them to a "Surfboard".  Pigtail leads can them be soldered through holes drilled in the Surfboard conducting races for connection to a circuit.

A surface mount device such as the OPA-349 integrated circuit (Eight lead SOIC package) can be soldered to a surfboard such as that manufactured by Capital Advanced Technologies ( http://www.capitaladvanced.com ).  Their Surfboards #9081 or #9082 are suitable and are available from various distributors such as Alltronics, Digi-Key, etc.

Surface mount diodes manufactured using the SOT-23 package can be handled using Surfboard #6103.  Diodes using the smaller SOT-323 package can be handled using Surfboard #330003.  This includes many Agilent surface mount diodes useful in crystal radio sets.  Packages containing multiple diodes exist that use the SOT-363 six lead package.  They can be handled using Surfboard #330006.  Agilent produces many of their Schottky diodes in dual, triple and quad form in the SOT-363 package.

It is recommended that anyone considering using Surfboards visit the above mentioned Website and read "Application Notes" and the "How-to Index".

10.  How to modify the tone quality delivered by headphones:  It is interesting to note that driving magnetic headphone elements with a high source resistance tends to improve the treble and reduce the bass response, compared to the response when the AC source resistance matches the effective impedance of the elements.  Conversely, driving the headphones elements from a low resistance source tends to roll off the treble, and relatively speaking, improve the bass.  With piezo ceramic or crystal elements, a high source resistance tends to reduce the treble and improve the bass response, compared to the response where the source resistance matches the effective impedance of the elements.  A low source resistance tends to reduce the bass and emphasize the treble.  Some piezo elements sound scratchy.  This condition can be minimized by driving the elements from a lower resistance source.

Here are some practical experimental ways to vary the audio source resistance of a crystal radio set when receiving weak-to-medium-strength signals.  A medium strength signal is defined as one at the crossover point between linear to square law operation (LSLCP).  See the graphs in Article #15A.

Quick introduction:
Greater sensitivity to very weak signals can be attained by lowering the RF signal power level (linear-to-square law point, or LSC point) at which the detector changes from the linear to the square-law mode of operation (See Article #10, Figs. 3 & 4 and part #3 for an explanation of the LSC point).  This is accomplished by connecting the highest impedance point of the RF tuned circuit to a diode having the proper Saturation Current (See Article #15A).  The output resistance of the detector should be impedance matched to the headphones, usually by a low-loss audio transformer, for maximum sensitivity.  Greater volume, less audio distortion and improved selectivity can be attained on strong signals by properly impedance matching the RF source resistance to the RF input resistance of the detector and also matching the output resistance of the detector to the effective impedance of the headphones. The DC and audio AC loads on the detector should also be made equal. This analysis does not involve the analysis of diode instantaneous voltage and current wave-forms, input voltage, output voltage, diode turn-on voltage or tuned circuit peak-clipping.  This analysis does consider the detector to be a black box having a linear input RF resistance and a linear output resistance, is driven from an AC power source and delivers power to an output load.  These resistances are independent of input signal power at low power levels (somewhat below the LSC point) and depend only upon the characteristics of the diode.  At high input power levels (somewhat above the LSC point), the input resistance is still linear and depends primarily on the output load resistance.  The output resistance depends primarily on the source resistance.

 

A crystal radio set may be thought of as the cascaded connection of several basic components.

The Antenna and RF Tuned Circuit will be combined into three components.  V1 and R1 represent the antenna induced voltage and resistance, impedance transformed by the tuned circuits and antenna reactance to the series-connected values seen by the diode detector.  X1 represents the reactance of the tuned circuit(s) seen at its output terminals.  Its impedance is considered to be substantially zero at harmonics of the frequency to which it is tuned.  Its impedance is also substantially zero at DC and at Audio frequencies.  R2 represents all the losses in the tuned circuits at resonance, as seen by the diode.  This is not the conventional way of viewing the signal source for a detector. 

The Detector will be represented as follows: The LC tank assures that the input is effectively shorted to ground at DC and at audio frequencies as well as all RF frequencies except that to which it is tuned.  The output is effectively shorted to ground at RF by C1.

The Output Transformer circuit will be represented as shown below.  The purpose of R3 and C2 will be covered later.

We start out with the assumption of no losses in the tuned circuits.  This condition makes R2 equal to infinity, not a practical assumption of course, but it will simplify what follows.  The input circuit then reduces to a simple series connection of the parallel tuned circuit, impedance transformed antenna voltage, and a series resistance. This resistance includes the effects of radiation, antenna, lead-in and ground circuit resistance.  A simple transformation enables us to eliminate R2 entirely by combining its effects into a changed value for R1 and a new value for V1.  The new value for V1 is: V1new = V1old*(R2old/(R1old + R2old)).  The new value for R1 is: R1new = (R1old*R2old)/(R1old + R2old).  With this transformation the new value for R2 is infinity, so it can be eliminated from the circuit.  Of course, the maximum available power from the new source 'V1new, R1new' is less than what was available from the original source 'V1old, R1old' by the amount that was dissipated in R2.  From now on, V1new and R1new will be referred to as V1 and R1.  The RF Source Voltage (V1) is assumed to be un modulated CW.

The transformed V1 (RMS) and R1 represent a Power Source of available power Pa = (V1^2)/(4*R1).  This is the most power it can deliver to a load.  It is also sometimes called the "Incident Power".  For the load to absorb this power, the load itself must equal R1, and then it is called an 'Impedance Matched Load'.  Changing the impedance transformation in the tuned circuit(s) changes the values of V1 and R1.  This does not change the available power.  That is still (V1^2)/(4*R1).  As an illustration, if V1 is doubled, R1 must quadruple thus keeping the power the same.

The approach we will use in this analysis is to minimize impedance mismatch power loss between the transformed antenna resistance and the diode detector input RF resistance as well as between the detector audio output resistance and the headphones.  We will show that the diode detector power Loss (DDPL), for very weak signal levels, can be minimized by using a diode with as low a Saturation Current (Is) as possible if all else is equal.  In addition, the lower the ideality factor (n) of the diode, the greater will be the sensitivity to weak signals.  The limitation here is that if a diode with a lower Is used, the required diode RF source and audio load resistances go up in value.  That limit is reached when the diode is connected to the top (the highest impedance point) of the tuned circuit.  The high frequency audio cutoff point may be reduced because of unavoidable winding capacitance in the audio output transformer acting against the required higher transformed headphone effective impedance.

The most important diode parameters to consider for Xtal set operation are saturation current 'Is' and 'n'.  They show up in the Shockley diode equation:  Id = Is*(exp((Vd-Id*Rs)/(0.026*n)) -1), at room temperature.  In crystal radio set applications, the Id*Rs term may be neglected because it is usually much smaller than V.  The equation then becomes:

Id=Is*(exp(Vd/(0.026*n))-1).        (1)

This equation provides a good approximation of the V/I relationship for most diodes, provided the parameters Is, n, and Rs are really constant.  Some diodes, especially germanium and silicon junction diodes seem to have Is and n values which increase at very high currents (higher than those usually encountered in crystal radio set operation).  In some of these diodes, the values of Is and n also increase at very low currents, harming weak signal reception.  Is and n are usually constant in Silicon Schottky diodes, over the current range encountered in crystal radio set use.
 

Agilent specifies the values of Is, Rs and n for Schottky diodes in their catalog.  They are listed in the table of SPICE parameters.  To find some SPICE parameters for other diodes (germanium types etc.), one can use used a neat Computer Program written by Ray Waugh of Agilent.  To use it one measures the diode forward voltage at five different currents (0.1 mA, 1.0 mA, 4.8 mA, 5.0 mA and 5.2 mA). Ray&39;s program runs on Mathcad 6.0 or higher.  One enters the five voltages and voila, out come Is, n, and Rs.  Remember this caveat:  The program assumes that Is, n and Rs are constant and do not vary with diode current.  If they do vary, one can change the first two currents (0.1 and 1.0 mA) to cover a smaller range, say, two-to-one, that bracket a desired diode operating current and get the Is and n values for that current.  Ray told me that if anyone wants a copy of this program, it would be OK for me to supply it. A simplified method of approximating Is (n must be estimated) that does not require having Mathcad is described in article #4.  A complete description of a test set-up and calculation method for determining both n and Is is shown in Article #16.

Here is what I have found experimentally through a SPICE simulation of a diode detector.  If a detector diode is fed by an RF source resistance of n*0.026/Is ohms and is loaded by an audio load resistance of n*0.026/Is, then both input and output ports are matched with a return loss of better than 18 dB,  assuming the signal is of weak to medium strength.  This satisfies the condition of very low mismatch but only holds true for diode rectified currents of up to about 5*Is.  An impedance matched diode detector insertion loss at a rectified current of 5*Is is about 3-4 dB.

The input and output impedance match starts deteriorating with a DC rectified current of over about 5*Is because of the change from square law operation towards linear response at the higher input levels.  At the highest RF Power  input level point shown in the following graph, the rectified DC current is 500 nA and the input RF Return Loss (impedance match) is -12 dB.  Diode detector power loss is 1.39 dB.  At these high levels of Input Power, good matching conditions are restored if the Input Source Resistance is kept at n*0.026/Is and The Output Load Resistance is increased to 2*n*0.026/Is.  If this is done input return loss goes to -26 dB and the insertion loss reduces to 0.93 dB.

Here is a graph of Diode Detector insertion power loss of an Agilent 5082-2835 or HSMS-2820 Schottky diode detector driven by a 1.182 megohm source and loaded by a 1.182 megohm load.  Note that these are very high resistance values for a usual Xtal set.  The SPICE simulation was done using an Intusoft ICAP/4 simulator.  Is of the diode=22 nA, n=1.03.  The plot shows the insertion power loss as a function of the resultant rectified DC current.

2.DISCUSSION

In general, headphones should be impedance matched by a transformer to the output resistance of the diode detector.  To use a diode of such a low Is as 22 nA with, say, a Brandes Superior 12k Ohm AC impedance 2k Ohm DC resistance headphones, an impedance transformation of 1,182,000/12,000 = 98.5:1 is needed (this high a ratio is hard to get).  See Article #2, "Personalized Headphone Impedance" (PHI).  One should be cautious of some small (maximum dimension of less than one inch) , high transformation ratio transformers because they may have a high insertion power loss.  They also may also show the effects of nonlinear inductance because the initial permeability of the core is not high enough.  Their shunt inductance is usually so low at low xtal set  DX  power levels, that the specified low frequency audio cut-off spec is not met.  At the transformer's rated power level, the shunt inductance is generally high enough so that the low frequency cut-off spec is met.  See Article #5 for info on various audio transformers.

Headphones such as the 2000 DC ohm Brandes Superior  have an effective AC impedance of 12,000 ohms (PHI), but a DC resistance of 2000 ohms.  If the Brandes' impedance is incorrectly considered to be 12,000 ohms at DC and audio frequencies, and is used in a 12,000 ohm circuit (without a transformer), too high a diode DC current will be drawn because the DC resistance is really 2000 ohms, not 12,000.  This will load down the output RF tuned circuit thus reducing selectivity and also give increased insertion power loss.  For best selectivity and minimum audio distortion at medium and high signal levels, the DC load resistance on the diode should be the same as the AC audio load.  The solution to this problem is to place in series with the headphones a parallel combination of a 10,000 ohm resistor shunted by a cap large enough to bypass the lowest audio frequency of interest.  When a transformer is used;  the parallel RC*  (See R3 and C2 on the schematic above.) should be connected in series with the low end of the high impedance transformer primary winding.  In this case the resistor should equal the transformed effective headphone impedance (PHI).  Another advantage that accrues from adjusting the diode DC load to equal the AC load has to do with the way selectivity varies as a function of signal level.  When the diode DC load is much smaller than the AC load (the case when using a transformer and no parallel RC), selectivity starts to reduce more and more as signal strength increases above a moderate level.  The reason is that the detector rectified current increases very rapidly because of the low DC diode load resistance.  A high rectified DC current always reduces the input and output resistances of a diode detector.  Audio distortion may also appear.  Now make the DC load higher, say equal the AC diode load impedance and have the detector impedance matched at both input and output (at low signal levels).  What happens then?  As the signal strength increases above a moderate level, the selectivity will change by a much smaller amount because the RF resistance of the diode detector will not drop as much as it did when the DC load resistance was small.  The resistance does not drop as much because the DC rectified current is less because the DC diode load resistance has been set to a higher value than before.  Impedance matched conditions also result in less power loss with consequently higher sound volume.  If the headphone effective impedance over the frequency range 0.3-3.3 kHz is transformed to a value lower than the output resistance of the diode, these beneficial effects are reduced.  If no transformer is used, these effects may be hard to observe because the headphone effective impedance will probably be lower than the output resistance of the diode. Also, headphones usually have a resistive impedance component about 1/6 the average value, and that goes part way towards being equal to 80% of the effective impedance.
* This may be the first time anyone has suggested placing a parallel RC in series with the diode to enable adjusting its DC load resistance equal its average AC load.  Some people call it a "benny". 

What is the advantage of using a diode with a low Is?  We will see that if matched input and output impedance conditions are maintained, diodes with lower Is give higher crystal radio set sensitivity (lower diode detector power loss) than diodes with higher Is, all else being equal.  The statement above is especially important when dealing with low power signals that themselves result in high DDPL.  The following graph shows the relationship between Diode Detector Power Loss at a relatively low DC Power Output Level  (-66 dBm) vs. diode Is for diodes having an n of 1.03.  Note that the graph data is valid only under the condition that the input and output are power matched.

NOTE: There is an error in the title of the graph. It should read: Detector Loss vs. Diode Is for a DC Power Output of -66 dBm.

I used the -66 dBm signal level for the graph because it is related to the weakest voice signal I can hear with my most sensitive headphones, and still understand about 50% of the words.  Here is the listening experiment that I used to determine that power level.  I fed my headphones directly from a transistor radio through my FILVORA and reduced the volume until I judged I could understand about 50% of the words of a voice radio program.  This enabled me to determine the average impedance of the headphones. (See article #2).  I then measured the p-p audio voltage (Vpp_audio) on the headphones with an oscilloscope. Assume  the AM station was running at about 100% modulation.  The peak instantaneous audio voltage at the detector will be equal to Vpp_audio since the modulation is 100%.  Now make the assumption that a CW carrier is driving the detector at such a level that the DC output voltage (Vdc) at the detector is equal to Vpp_audio.  That DC  voltage across a resistor of value equal to the detector load resistance will deliver an output power of Pdc=10*log((1000*(Vdc^2))/Rload) dBm.  Since I could not get into the radio to measure the actual detector voltages and the audio load resistance,  I used the p-p voltage measured across my 1200 Ohm headphones in place of Vdc to calculate the instantaneous power at the modulation peaks.  Pp=10*log(1000*((Vpp_audio^2)/1200= -66 dBm.  This power, Pp is that used in calculating the graph above.  In my case Vpp_audio = 0.00055 Volts and effective headphone impedance = 1200 Ohms. 

To calculate the actual audio power level I was using in the listening experiment, I assumed that the demodulated audio voltage was a sine wave (not a voice) with the same p-p value as the actual measured voice p-p voltage.  It was then a simple matter to use the p-p voltage of the assumed audio sine wave (Vpp_audio) and the effective impedance (PHI) of the headphones to calculate the power of the audio sine wave in dBm.  P=10*log ((1000*(Vpp_audio^2))/(8*PHI)) dBm.  This value comes out 9 dB less than the DC power of -66 dBm.  Of course there is an error here in assuming that a sine wave of a specific p-p voltage has the same RMS value as that of a broadcast voice waveform of an equal p-p value.  The "Audio Cyclopedia", in an article on VU meters, states that the actual power from a voice signal is 8-10 dB less than the power from a sine wave of the same p-p voltage.  I'll use 9 dB.  Bottom line:  The audio power from a voice voltage waveform is 18 dB less than the audio power from a sine wave voltage of p-p value equal to the p-p voltage of the voice waveform.  We can now calculate that the electrical power of weakest voice audio signal I can barely understand is -66 -9 -9 = -84 dBm.  This figure depends on the sensitivity of the headphones used and one's hearing acuity.  I used a  good sound powered headphone set in this test.  My hearing acuity is pretty poor. 
 

Keep in mind that diodes have an unavoidable back leakage resistance.  Schottky diodes generally are very good in this respect.  An exception is the so-called "zero bias" detector diodes.  They have very high Is values and low reverse breakdown voltages and are generally not suitable for crystal radio sets.  Germanium and cats whisker diodes are worse than Schottkys and vary greatly. This reverse resistance increases detector loss and reduces selectivity. "n" in the diode equation is usually close to 1.05 for Schottky barrier diodes.  It is about 1.15 in Germanium diodes.  All diodes have a fixed parasitic series resistance Rs.  It is usually low enough to be ignored in crystal radio sets.  One problem with Schottky diodes having a low reverse breakdown voltage and low Is is that they are more vulnerable to damage from static electricity than diodes with a higher leakage resistance.

Tuned circuit loss and bandwidth considerations:  A practical problem in using a diode of low Is is getting a high enough tuned circuit impedance for driving the diode.  Of course, the first thing to do is to tap the diode all the way up on the output tuned circuit.  An isolated tuned circuit having a  typical Q of 350 at a frequency of 1.0 MHz, with a circuit capacitance of say 100 pF, and not coupled to an antenna or detector diode will have a resonant resistance of about 560k ohms.  RF bandwidth will be fo/Q = 2.86 kHz.  If an antenna resistance is now coupled in sufficiently to drop the resonant resistance by half to 280k, all of the available received RF power will be dissipated in the resonator, resulting in a bandwidth of 5.72 kHz (loaded Q of 175).  If a diode is selected to match the now 280k ohm source resistance, it will present a 280k RF load resistance and result with a tuned circuit loaded Q of 87.5 giving an RF bandwidth of 11.4 kHz.  The overall power loss caused by the tuned circuit loss is 3 dB.  The diode will only receive 1/2 the maximum available-power at the antenna.  The diode should have an Is of about n*0.026/278k = 100 nA (assuming a Schottky barrier diode is used).  Note this:  Even though the the diode is driven from a perfectly matched source (parallel connected combo of 560k tuned circuit loss and 560k antenna resistances), now the antenna does not see a matched load.  It sees a parallel combo of the tuned circuit loss resistance of 560k and the 280k RF resistance of the diode.  This is a resistance of 187k ohms.  This mismatch power loss, included in the 3 dB above can be partially recovered by properly and equally mismatching the antenna and the diode.  If this is done by more loss-less impedance transformation (technically, with an S parameter return loss of -11.7 dB), the total tuned-circuit power loss reduces to 2.63 dB, a reduction of 0.37 dB (pretty small, but it's there).  If the ratio of unloaded to loaded tuned circuit Q was less than the 4:1 ratio used here, the loss reduction would be larger.

Audio impedance transformation:  One way to transform the 12k ohm effective impedance of a 2k ohm DC resistance Brandes Superior headset up to 280k ohms is to use an Antique Electronic Supply # P-T156, Stancor A53-C or similar 3:1 turns-ratio inter stage transformer.   I measure an insertion power loss of only 0.5 dB with the following connection (See Articles #4 and #5 for other options.):

Note that the impedance transformation ratio is 16:1 thus stepping up the impedance of the 12,000 ohm headphones to 192,000 ohms not 278,000 ohms.  This represents a mismatch of about 1.5:1. It will add a mismatch insertion power loss of only 0.15 dB. If the impedance mismatch had been 2:1, the insertion power loss would have been 0.5 dB.  A 4:1 mismatch gives an insertion loss of 1.9 dB.

The lead grounding the transformer lamination stack and frame is used if the transformer is mounted on an insulated material.  It prevents the buildup of static charge on the frame during dry weather.  Discharge of it might cause a crackling sound in the headphones or damage the diode (I got the crackling sound until I made the grounding connection).

The transformer windings start and stop leads should be connected as shown to minimize the effect of the primary to secondary winding capacitance.  If the f and s connections are reversed, the capacitance between the end of the secondary and the start of the primary winding will be across the primary and reduce the high frequency cut-off point.  The lower impedance (secondary) winding is usually wound on the bobbin first, then after winding on several layers of insulation film, the higher impedance (primary) is wound.

To determine how to connect the leads of the transformer, connect the primary and secondary windings as shown. (Disregarding the s and f notations).  Connect an audio generator set to 1.0 kHz through a 200k ohm resistor. Load the secondary with a 12,000 Ohm resistor.  Probe the input and output voltages with a scope.  The output voltage should be about 0.25 of the input voltage. If the output voltage is about 0.5 that of the input, reverse the secondary leads.  Repeat the test at 20,000 Hz and note the input and output voltages.  Now reverse both the primary and secondary leads and repeat the 20 kHz test.  The connection that gives the largest output voltage at 20 kHz is the correct one.

Note that R3 is shown above as a rheostat not a fixed resistor.  The nominal setting under the low signal level conditions discussed here is about 192k Ohms.  Setting it to zero has little effect on reception of these low level signals.  With this design approach, when receiving high level signals, RF selectivity is not reduced as much as when the DC resistance in the diode circuit is substantially below the effective impedance of the headset.  When receiving very strong signals, R3 should be set for minimum distortion.

One last comment: These design values are not critical.  If impedances vary by several times from the optimum values, usually only a small sensitivity reduction will occur.

What is the effect on the volume in the headphones of a change of X dB?  Many years ago I did a study which determined, in a blinded condition, that a +1.0 dB or a -1.0 dB change in sound level was barely discernible by most people.  Half couldn't tell if the sound level was changed or not after being told that a change might have occurred.  Another study had the listener listen to a sound.  The sound was then turned off for several seconds and then on again at the same level, at a level of +3.0 dB or at a level of -3.0 dB.  After the delay, only half the listeners could tell whether the level of the sound had changed or not.  Incidentally, the listeners were not golden eared hi-fi listeners.

 

4. SUMMARY

 

Impedance matching for magnetic and piezo-electric headphones, measurements on several audio transformers, and a transformer loss measurement method

 

Quick Summary:  This Article discusses the use of audio transformers with crystal radio sets and gives the results of loss measurements on several of them.  A method for measuring insertion power loss is also described.

Many crystal radio set designs provide impedance step down taps on the final RF tuned circuit.  If the diode is connected to one of these taps, its loading on the tuned circuit is reduced and selectivity is impro