A Crystal Radio for Casual AM Listening

David Wagner 2008/01/11 17:40

Antenna

  • Standard Dummy Antenna for AM Broadcast Band: 200 pF, 25 ohm resistor and 20 uH in series + ground resistance 0f 15-50 Ohm.1)

Tank Inductor

Micrometals Mixes: 1, 15, 22)

Design

American AM radio broadcasts 520-1700 kHz, and each station can cover 10 kHz. The resonant circuit Q needed is 52-170.

Frequency (kHz)Bandwidth (kHz) Q L (µH)C (pF)Rp (Ω)
520 10 52 130 720 22,000
1000 10 100 130 195 80,000
1720 10 172 130 66 120,000

If a strong local station provides 50 µW at 1000 kHz to the tank through an antenna, a 10 kHz bandwidth is possible when the load impedance is 80 kΩ 3).

The tank is loaded by drawing off most of the radio power coming from the antenna. The detector rectifies the RF signal and the remaining circuitry converts it to DC current and audio. Since the tank is resonating in the broadcast band, the audio frequency loading appears as if the DC current is just varying slowly. Since the power factor of the tank is near unity, the current needed to load the tank at 80 kΩ is easy to calculate.

  • overline{P}=R overline{I}^2 right overline{I}^2 = 0.000050/80000 right overline{I}=0.000025=25 µA (@ 2 V).

When a detector draws 25 µA off the tank it will load the tank at 80 kΩ. At this current, a modern 1N34A germanium diode has a voltage drop of about 0.1 V with a power loss of 2.5 µW. While 25 µA loads the tank at 80 kΩ and keeps the bandwidth narrow, the 47.5 µW of power transferred appears on the other side as a combined DC and audio (AC) current source of 25 µA.

Also, at this current level this detector has a series source impedance of about 2 kΩ, so the DC and audio load impedance must also be at least 2 kΩ. Since this is rectified current, at least half is 12.5 µA DC and less than half is audio signal.

A simple LCR circuit sends the current through an inductor, then branches so audio frequencies go through a capacitor to a piezoelectric speaker and DC goes through a resistor before it is grounded. The resistor passes the DC portion of the the signal.

Impedance Matching

FIXME The following needs work.

BN-61-202
TurnsInductance
Tank 120 kΩ 6 18 120
Antenna 40 kΩ 2 10
Audio 20 kΩ 1 7
BN-61-202
TurnsInductance
Tank 120 kΩ 6 21 170
Antenna 40 kΩ 2 12
Audio 20 kΩ 1 8.5
BN-61-202
TurnsInductance
Tank 120 kΩ 6 24 222
Antenna 40 kΩ 2 14
Audio 20 kΩ 1 10

Tuggle-tuned Triple-contra-wound Ferrite Toroid

             \|/
              |  +-------->|----o
              |  |
 +--------UUU-+  +-UUUU---------o
 |        :::::::::::::
 |      +-UUUUUUUUUUUUU-+
 |      |               |
 |    / |    /          |
 +--||--+--||-----------+
 | /....../
 |
---
 -

I am not sure of the polarity of each winding.

Testing

The following measurements show how this works in a crystal radio set operating at a lower power level.4)

  • The average DC current through detector, inductor, and resistor is 19 µA.
  • The detector (1N34A-red) cathode average DC is 19 µA at 0.250 V for 4.75 µW (0.07V, 2200Ω).
    • At this current, this detector has a voltage drop of about 0.07 V, a power loss of 1.3 µW.
    • The tank must be providing about 6 µW and is loaded at 16.6 kΩ5) for a bandwidth of at best 36 kHz.6).
  • Average DC through a 9560 Ω resistor: 19 µA at 0.180 V for 3.4 µW.
  • The power through the capacitor and speaker is therefore around 1.35 µW, 22.5% efficiency.
  • FIXME If the RMS voltage is nearly equal to the average DC voltage, the equivalent series resistance of the capacitor and speaker is 24 kΩ7).
  • FIXME The impedance of the audio circuitry is 10 to 13 kΩ.8)

And at a higher level 9).

  • The average DC current through detector, inductor, and resistor is 63 µA.
  • The detector (1N34A-red) cathode average DC is 63 µA at 0.80 V for 50.4 µW (0.13V, 800Ω).
    • At this current, this detector has a voltage drop of about 0.13 V, a power loss of 11.3 µW.
    • The tank must be providing about 58.6 µW and is loaded at 15 kΩ10) for a bandwidth of at best 41 kHz.11).
  • Average DC through a 9390 Ω resistor: 63 µA at 0.60 V for 37.8 µW.
  • The power through the capacitor and speaker is therefore around 12.6 µW, 21.5% efficiency.
  • FIXME If the RMS voltage is nearly equal to the average DC voltage, the equivalent series resistance of the capacitor and speaker is 28.6 kΩ12).
  • FIXME The impedance of the audio circuitry is 9.5 to 10 kΩ.13)

With antenna improvements and a 33 pF RF bypass capacitor14).

  • 118 µA at 0.83 V is 97.94 µW.
  • 0.76V at 1N60 cathode, 89.68 µW net; 8.26 µW goes through the RF bypass capacitor.
  • VF≈0.15 V, 17.7 µW power lost means the tank provides 115.64 µW and is loaded at 7712 Ω15). The DC source impedance is 6441 Ω.
  • The pot dissipates 90.23 µW at 6480 Ω. Oops! Didn't measure across just the pot (at 6480 Ω).

Notes

“The David Clark voice powered telephone sets produce 20 microamps at 60 millivolts on average on a 600 ohm loop.”–Forum Thread

3) “If power dissipated by the coupled load is greater than 10 times the power lost in the tank circuit reactive components, the tank impedance is high compared to the load impedance. Thus, for all practical purposes, the impedance of the parallel combination of the tank and load virtually equals the load resistance value.”–Foundations of Electronics: Circuits, and Devices By Russell L. Meade, Robert Diffenderfer
4) Data collected 2008-01-13 at night, probably tuned to AM 860, KONO.
6)
  1. 2 pi f = 1/sqrt{LC} right C=1/{4*pi^2*860000^2*0.000130}=263.45 pF
  2. BW=1/{2 pi R_p C} > 1/{2*pi*16620*263x10^{-12}}=36350=36 kHz
7) R=V^2 / P = 0.18^2 / 0.00000135 =24000
9) 2008-01-14T17:30+06:00, AM 860, KONO
11)
  1. 2 pi f = 1/sqrt{LC} right C=1/{4*pi^2*860000^2*0.000130}=263.45 pF
  2. BW=1/{2 pi R_p C} > 1/{2*pi*14762*263x10^{-12}}=40993=41 kHz
12) R=V^2 / P = 0.60^2 / 0.0000126 =28571
14) 2008-01-19, AM 860, KONO

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