Quick summary:
Greater sensitivity to very weak signals can be attained by lowering the RF signal power at which the detector changes from the linear to the square-law mode of operation (See Article #10, Figs. 3 &4 and part #3).  This is accomplished by connecting the highest impedance point of the RF tuned circuit to a diode having the proper Saturation Current (See Article #15).  The output resistance of the detector must be impedance matched to the headphones, or selectivity and sensitivity will be impaired.

Greater volume, less audio distortion and improved selectivity can be attained on strong signals by properly impedance matching the diode to the RF source and headphone impedance as stated above. The DC and audio AC loads on the detector must also be equalized. 
 

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

• Antenna:  Signal source 
• RF tuned circuit:  To provide selectivity and impedance matching between the resistance of the antenna-ground circuit and the RF input resistance of the diode detector.  The tuned circuit has some power loss.  The greater the selectivity, the greater this loss. 
• Diode detector:  Characterized as a black box that accepts RF input power and converts it to DC output power.  It has an input resistance, an output resistance and an insertion loss (dB).  These three characteristics are interrelated with Input RF power, RF source resistance driving the detector, Audio Load resistance and the parameters of the diode used. 
• Output transformer:  To impedance transform the average headphone impedance to that required by the diode. 
• Audio load:  Headphones, what else? 

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 to the values seen by the diode detector.  X1 represents the reactance of the tuned circuit(s) seen at the Output terminals.  Its impedance is considered to be zero at harmonics of the frequency to which it is tuned.  Its impedance is also zero at DC and at Audio frequencies.  R2 represents all the losses in the tuned circuits at resonance, as seen by the diode.

The transformed V1 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".  To get this power, the load itself must equal R1, and it is called a 'Matched Load'.  Changing the impedance transformation in the tuned circuit(s) can change the values of V1 and R1.  This does not change the incident power.  For instance, if V1 is doubled, R1 must quadruple to keep 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 as well as between the Detector Output Resistance and the headphones.  We will show that the Diode Detector Power Loss (DDPL), for 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 impedance.

The most important diode parameters to consider for Xtal set operation are saturation current 'Is' and 'n'.  They show up in the Schottky diode equation: Id = Is*(exp((Vd-Id*Rs)/(0.026*n)) -1), at room temperature.  This equation is used in the circuit simulation program called 'SPICE".  In crystal set applications, the Id*Rs term can be neglected because it is usually much smaller than V.  The equation then becomes: Id=Is*(exp(Vd/(0.026*n))-1).  The equation provides a good approximation of the V/I relationship for most diodes if the parameters Is, n, and Rs are really constant. Some diodes, especially germaniums seem to have Is and n values which increase at very high currents.

     n = ideality factor, sometimes called emission coefficient (about 1.05 for silicon Schottky diodes, and about 1.15 for germaniums 
           depending upon the DC current through the diode).  The lower the value of n, the higher will be the sensitivity to weak signals.
     Vd = Diode voltage in Volts
     Id = Diode current in Amps
     Is = Diode Saturation current in Amps
     Rs = Diode parasitic series resistance in ohms (usually small enough to have no effect in Xtal sets)

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 etc.), I have used 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’s program runs on Mathcad 6.0.  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 currrents (0.1 and 1.0 mA) to cover a smaller range, say, two or three-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 out 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.  This satisfies the condition of very low mismatch Power Loss but is only true for diode rectified currents of up to about 5 *Is.  The 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 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.

In general, headphones should be impedance matched to the output resistance of the diode detector by a transformer.  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 small (maximum dimension of less than one inch) , high transformation ratio transformers because they may have 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.  The 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.  One of the best transformers I have found for Xtal set purposes is supplied by Antique Radio Supply.  The model number is P-T157.  It has enough shunt inductance on the low impedance winding to drive a 12k Ohm load well.

Headphones such as the 2000 DC ohm Brandes Superior  have effective AC impedance of 12,000 ohms (PHI), but a DC resistance of 2000 ohms.  If the Brandes' impedance is considered to be 12,000 ohms and is used in a 12,000 ohm circuit (without a transformer), too high a diode DC current will be drawn and the diode detector will exhibit a lower RF and Audio resistance than is calculated.  This will load down the output RF tuned circuit thus reducing selectivity and also give increased insertion power loss.  For 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 Brandes a parallel combination of a 10,000 Ohm resistor shunted by a cap large enough to bypass the lowest audio frequency desired.  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 headphone impedance (PHI).  Another advantage that accrues from adjusting the diode DC load to equal the AC load has to do with 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.  Make the DC and AC diode load impedances equal and have the diode 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.  Impedance matched conditions also result in less power loss with consequently higher sound volume.  If the headphone average impedance over the frequency range 0.3-3.3 kHz is transformed to a value lower than the output resistance of the diode, these effects are less.  If no transformer is used, these effects may be hard to observe because the headphone average 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 average impedance. 

Note that we have stayed away from the concepts of "diode breakpoint, or knee" and "square law vs. linear operation" of the detector diode.  We have instead used the concept of "diode detector insertion power loss" with the diode used between matched Input and Output resistances.  These resistances are: n*0.026/Is Ohms for the Input (RF) and the same value, n*0.026/Is Ohms for the Output (Audio), for signal levels that cause  the Diode Detector insertion power loss to be above about 2 dB. For insertion power losses less than 2 dB, the output load should rise toward n*0.052/Is ohms.

What is the advantage of using a diode with a low Is?  We will see that if input and output impedance matched conditions are maintained, diodes with lower Is give higher Xtal 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 Diode Detector Power Losses.  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 a HP 5082-2835.  Note that the graph data is valid only under the condition that the Input and Output are power matched.

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 average 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 average 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 sets.  Germanium diodes are somewhat worse and cats whisker diodes vary a lot. This resistance increases detector loss and reduces selectivity. "n" in the Diode equation is usually close to 1.05 for Schottky hot carrier diodes.  It is about 1.15 in Germanium diodes.  All diodes have a fixed parasitic series resistance Rs.

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.  A Q of above 350 is hard to get at fo=1.0 mHz, and with a circuit capacitance of, say, 100 pF, the unloaded (by the antenna) resonant resistance will be 557k Ohms.  Single tuned RF bandwidth will be fo/Q = 2.86 kHz. If the antenna is coupled in to drop the resonant resistance by half to 278k, one half of the received RF power will be dissipated in the resonator losses (3 dB) and the bandwidth will open up to 5.72 kHz (loaded Q of 175). Now, if a diode is selected to match the 278k ohm source, it will present a 278k load and result with an output tuned circuit loaded Q of 87 giving a RF bandwidth of 11.5 kHz.  This diode should have an Is of n*0.026/278k = 100 nA. (assuming a Schottky barrier diode is used)

One way to get an impedance transformation to the 12k Ohm AC load of a 2k ohm DC resistance Brandes Superior headset is to use an Antique Electronic Supply # P-T157 transformer.   I measure an insertion power loss of only 0.5 dB with this transformer connected as follows:

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.

What is the effect on the sound in the headphones of a loss 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 had 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 time delay, only half could tell if the level of the sound was changed.  Incidentally, the listeners were not golden eared high listeners.

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 – 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 pot, 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 AC audio impedance of the headset.  When receiving very strong signals, R3 should be set for minimum distortion and maximum volume.

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.

3. SUMMARY

This design approach for crystal sets provides the following benefits:

The volume from very low strength (DX) signals is increased (less detector power loss). Louder sound volume with less audio distortion when very strong signals are received. Improved high signal level selectivity without changing coupling or coil taps.  Less variation of selectivity with signal strength.  No need to tap the diode down on the output tuned circuit. Highest weak signal sensitivity is always achieved by connecting a good diode of the proper Is to the highest impedance point  (assuming that the correct audio impedance transformation to the headphones is used and that the transformer has low loss). Enables diodes with too high an Is to be used without a large reduction in selectivity, by increeasing R3.

 
Achieve the benefits by doing the following:

Use a diode with an appropriate Is to impedance match the resonant resistance of the "antenna loaded RF tuned circuit" that drives the diode.  See Article # 15 for new information on this. Match the audio output resistance of the diode to the impedance of the headphones by using a low loss audio transformer. Use a bypassed variable resistance in series with the cold end of the primary of audio transformer to enable the diode DC load resistance to be made equal to the AC load impedance.

Published: 07/03/99;  Last revision: 08/16/01