A Crystal Radio Constant Current Hypothesis

David Wagner 2008/01/24 22:53

This is a casual examination of how crystal radios work, a bit different from most explanations available1).

AM Broadcast Signal

Antenna

Detector

Matching Network

First, assume the electronics have an inductor large enough to assume a nearly constant current (I) at radio frequencies (RF). This is almost certainly a good assumption, since most crystal radio circuits have at least one large coil, either a transformer or a transducer voice coil.2) The current may vary with the amplitude of the audio signal as the detector responds to this gradually increasing and decreasing source RF power, but to understand detector behavior, think of the detector current as constant through each RF cycle.

Now, when the current is constant, the voltage drop across the detector must be constant, so the voltage waveform on the cathode will look just like it does on the anode, but shifted down by the forward voltage (Vf). This waveform can be described as the superposition of a radio frequency sine wave and a constant voltage. The constant voltage requires an applied DC resistance greater than Vf/I to maintain.

For the audio frequencies to be filtered from the RF and passed to the transducer efficiently, the audio impedance should be much less than both the DC resistance and the RF impedance. Even a fairly small inductor (for audio components) has high impedance at radio frequencies and this level of source equivalent resistance,3) so the inductor will be a low-pass filter and smooth the RF from the detector cathode into AF.

Now, the power of the tuned station available from the tank determines the AF impedance needed to limit the tank loading to an acceptable level.

FIXME

Suppose a crystal radio set with an inductive transducer is operating as follows.4)

        \  /
o-->|----UU--+
   D     HP  |
             |
o------------+

Basic Magnetic Headphones Circuit


                \  /
       +--||--+--UU--+
       |  C   |  HP  |
       |      |      |
o-->|--+--VV--+      |
   D      R          |
                     |
o--------------------+

Magnetic Headphones Circuit with a Benny

  • fTank = 1000 kHz.
  • I = ITank = ID = IHP = 10 µA.
  • RHP,DC ≈ 2 kΩ; XHP,AF ≈ 18 kΩ; XHP,RF→∞ ⇒ IHP,RF ≈ 0. 5)
  • VD = 100 mV → PD = 1 µW; RD = 10 kΩ.

The current flow is fairly easy to calculate.

  1. I = IHP = IHP,DC + IHP,AF + IHP,RF= IHP,DC + IHP,AF = 10 µA.
  2. IHP,DC = 10 µA x 18/20 = 9 µA.
    • PHP,DC = 0.000009² x 2000 = 162 nW.
  3. IHP,AF = 10 µA x 2/20 = 1 µA.
    • PHP,AF = 0.000001² x 20000 = 20 nW.

Add a benny with R=10k and a capacitor large enough for its AF impedance to be near zero so IR,AF=0.

  1. I = IHP = IHP,DC + IHP,AF = 10 µA.
  2. I = IC,AF + IR,DC + IR,AF = 10 µA.
  3. IC,AF + IR,AF = IHP,AF.
  4. IR,DC = IHP,DC.

The current rejoins through the headphones.

  1. XAF = 18 kΩ; RDC = 10+2 = 12 kΩ.
  2. IHP,DC = IR,DC = 10 µA x 18/30 = 6 µA.
    • PHP,DC = 0.000006² x 2000 = 72 nW.
    • PR,DC = 0.000006² x 10000 = 360 nW.
  3. IHP,AF = 10 µA x 12/30 = 4 µA.
    • PHP,AF = 0.000004² x 18000 = 288 nW.

The total power consumed is the power coming from the tank.

  • PTank = 288+72+360+1000 nW = 1.72 µW.
  • RTank,load = 0.0000017/0.000001² = 1.7 MΩ.

Add a benny with R=80k and a capacitor large enough for its AF impedance to be near zero so IR,AF=0.

The current rejoins through the headphones.

  1. XAF = 18 kΩ; RDC = 80+2 = 82 kΩ.
  2. IHP,DC = IR,DC = 10 µA x 18/100 = 1.8 µA.
    • PHP,DC = 0.0000018² x 2000 = 6.48 nW.
    • PR,DC = 0.0000018² x 80000 = 259.2 nW.
  3. IHP,AF = 10 µA x 82/100 = 8.2 µA.
    • PHP,AF = 0.0000082² x 18000 = 1.21 µW.

The total power consumed is the power coming from the tank.

  • PTank = 1.21+0.259+0.006+1 = 2.5 µW.
  • RTank,load = 0.0000025/0.000001² = 2.5 MΩ.

Consider this example.

                \  /  
       +-----||--[]--+
       |     C   PZ  |
   D   |  L          |
o-->|--+--UUUUU--VV--+
          = = =  R1  |
                     |
o--------------------+

This configuration requires an enormous inductor.


                \  /  
              +--[]--+
              |  PZ  |
   D   L      |      |
o-->|--UUUUU--+--VV--+
       = = =     R1  |
                     |
o--------------------+

The most efficient configuration requires an enormous inductor to prevent audio frequencies from going through the resistor.

  1. R1 = 10 kΩ.
  2. X_{L,AF} = 2 pi f_{AF} L >> 40000 right L approx 400000/{2*pi*1000}=64 H.
  3. X_{C,AF} approx 1/{2 pi f_{AF} C} right C approx 1/{40000*2*pi*1000} approx4 pF.

Try this. FIXME

  1. RPZ,DC ≈ 20 MΩ, XPZ,AF ≈ 100 Ω.
  2. R1=20 kΩ → R1||PZ,DC = 1/{{1/R_1}+{1/R_{PZ}}} approx 20 kΩ.
  3. I_{PZ}+I_{AF}=

Audio Transducer

Ground

1) This is somewhat the reverse of Ben Tongue's analyses in A New Way to look at Crystal Radio Set Design. Get Greater Sensitivity to very Weak Signals, and Greater Volume, less Audio Distortion and Improved Selectivity on Strong Signals and elsewhere, though the conclusions are similar. FIXME: More references needed.
2) Even if the circuit does not have an inductive element, the transducer will likely act as one to the driving circuit due to the mass of the driver.
3) For a 10 mH inductance at the low end of the broadcast band, Z_I = omega L = 2 pi f L = 2*pi*520000*0.010 = 32,673 Ω.
4) The nice round numbers for the detector are typical for old style 'black band' 1N34A point-contact germanium diodes.
5) These are typical for good high-impedance headphones.http://www.klimaco.com/HAMRADIOPAGES/xtal_how_to.htm

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