Crystal Radio Modeling

— David Wagner 2007/10/05 14:27

Formulæ

¹⁾
${{d^2}i}/{dt^2} = -I omega^2 sin(omega t + phi)$

Component	Voltage Drop v(i)	dv/dt
Capacitor	$v_C = {1/C}int{}{}{i}dt$	$dv_C/dt={1/C}i = {1/C} I sin(omega t + phi)$
Resistor
Inductor	$v_L = L {di/dt}$	$dv_L/dt=L{{d^2}i}/{dt^2} =-L I omega^2 sin(omega t + phi)$
Diode	$v_D = n V_T ln({i/{I_s}}+1)$	$dv_D/dt= {{n V_T omega cos(omega t + phi)}/{sin(omega t + phi)+{I_s/I}}}$ ²⁾
Diode with Series Resistance	$v_{DR} = n V_T ln({i/{I_s}}+1) + R_s i$

“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.

n = Ideality factor, sometimes called emission coefficient. This parameter is usually between 1.05 and 1.15 for silicon Schottky and germanium diodes commonly used in crystal radio sets.
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)”

–A New Way to look at Crystal Radio Set Design.... by Ben H. Tongue

V_T=0.026
I_s: 45-1300 nA (consider 200 nA)

Frequencies of Interest

The frequency ranges of concern are widely separated.

DC	Direct Current	0-100 Hz
VF	Voice Frequency Range,	300 Hz -5 kHz³⁾ Hz (especially 512 to 2,048 Hz⁴⁾ or 300-3,400 Hz⁵⁾
BC	Broadcast Band	500-1,700 kHz
HF	High Frequency	>1,700 kHz (ignored)

“1. AM stations mostly use amplitude compression which enhances the low-amplitude audio portions. According to station engineers a modulation factor of m = 0.65 is a good mean value for averaged music and speech contributions if amplitude compression and a peak modulation of m = 1 are employed.”–Diode AM Conversion Efficiency in Crystal Sets and Some Implicationsby Berthold Bosch, DK6YY

Hobbydyne

FIXME Redo the schematic shamelessly ripped from the author's site. Include the trimmer cap.

Note the four resonant circuit loops in the radio, including the Benny, but not counting the loop with the diode, the headphones, or the antenna tuner.

       L1                          L1 -i3->
    +--UUUUU--+                 +--UUUUU--+  
    |  = = =  |                 | -i1->   |
  E(~)        |               E(~)        |
    |   C1/   |                 |  C <-i1-| C = C1 +(Ct^-1 +Cda^-1)^-1
    +---||----+                 +--||-----+
    |  /      |            -    | / -i2-> |  Cdb = Cd-Cda
    |     /   | /          -    |         |    /
    +---||---|||--+---->   -    |         +--||--+----> Detector
    |  /Ct   /Cd  |             |           /Cdb |
    |             |             | <-i2-          |
    +------UUUUU--+             +------UUUUU-----+
    |      =====                |      =====  <-i3- 
    |      Lrfc                 |      Lrfc
    +------------------<        +---------------------<

Note how the Hobbydyne is a capacitive transformer.⁶⁾

   +---------------->|--+
   |                    |
   |      +-UUUU-+      +-----VVV
   |      |      |      |      |
   +--||--+--||--+--||--+--||--+-----o
   |             |             |
   |             |             |
   +--UUUUUU-----+UUUUUUUUUUUUU+UU---o

Hobbydyne Eigenvalue Analysis

The Hobbydyne circuit has three current loops.

(E)→(L₁)→(C)
(L_rfc)→(C)→(C_db)
(E)→(L₁)→(C_db)→(L_rfc)

These can be represented by three equations setting the sum of the voltage drops around each loop to zero:

$E -L_1({{di_1}/{dt}} + {{di_3}/{dt}}) - 1/C int{}{}{}(i_1-i_2)dt =0$ ,
$L_{rfc}({{di_2}/{dt}} + {{di_3}/{dt}}) +1/C int{}{}{}(i_2-i_1)dt +1/{C_{db}}int{}{}{}(i_2+i_3)dt =0$ ,
$E -L_1({{di_1}/{dt}} + {{di_3}/{dt}}) -1/{C_{db}} int{}{}{}(i_2+i_3)dt -L_{rfc}({{di_2}/{dt}} + {{di_3}/{dt}}) =0$ .

These equations can be differentiated and rearranged,…

${dE}/{dt} = L_1{{d^2 i_1}/{dt^2}} +L_1{{d^2 i_3}/{dt^2}} +{1/C}i_1 -{1/C}i_2$ ,
$L_{rfc}{{d^2 i_2}/{dt^2}} +L_{rfc}{{d^2 i_3}/{dt^2}} -{1/C}i_1 +({1/C} +{1/C_{db}})i_2 +{1/{C_{db}}}i_3 =0$ ,
${dE}/{dt} = L_1{{d^2 i_1}/{dt^2}} +L_1{{d^2 3_1}/{dt^2}} +L_{rfc}{{d^2 i_2}/{dt^2}} +L_{rfc}{{d^2 i_3}/{dt^2}} +{1/{C_{db}}}i_2 +{1/{C_{db}}}i_3$ ,

…the currents assumed to be sine waves,…

${dE}/{dt} = -L_1 I_1 omega^2 sin(omega t + phi) -L_1 I_3 omega^2 sin(omega t + phi) +{1/C}I_1 sin(omega t + phi) -{1/C}I_2 sin(omega t + phi)$ ,
$-L_{rfc} I_2 omega^2 sin(omega t + phi) -L_{rfc} I_3 omega^2 sin(omega t + phi) -{1/C}I_1 sin(omega t + phi) +({1/C}+{1/C_{db}})I_2 sin(omega t + phi) +{1/{C_{db}}}I_3 sin(omega t + phi) =0$ ,
${dE}/{dt} = -L_1 I_1 omega^2 sin(omega t + phi) -(L_1+L_{rfc}) I_3 omega^2 sin(omega t + phi) -L_{rfc} I_2 omega^2 sin(omega t + phi) +{1/{C_{db}}}I_2 sin(omega t + phi) +{1/{C_{db}}}I_3 sin(omega t + phi)$ ,

…cancellations done and dE/dt set to 0 so ω represents the resonant frequencies,…

$sin^{-1}(omega t + phi){dE}/{dt} = -L_1 I_1 omega^2 -L_1 I_3 omega^2 +{1/C}I_1 -{1/C}I_2 =0$ ,
$-L_{rfc} I_2 omega^2 -L_{rfc} I_3 omega^2 -{1/C}I_1 +({1/C}+{1/C_{db}})I_2 +{1/{C_{db}}}I_3 =0$ ,
$sin^{-1}(omega t + phi){dE}/{dt} = -L_1 I_1 omega^2 -(L_1+L_{rfc}) I_3 omega^2 -L_{rfc} I_2 omega^2 +{1/{C_{db}}}I_2 +{1/{C_{db}}}I_3 =0$ ,

…and the result rearranged as a general eigenvalue problem of the form delim{[}{A}{]} vec{X} = lambda delim{[}{B}{]} vec{X} …

$delim{[}{C}{]} vec{I} =omega^2 delim{[}{L}{]} vec{I}$ ≡ $delim{[}{ matrix{3}{3}{ {1/C} {-1/C} 0 {-1/C} (1/C + 1/{C_{db}}) {1/C_{db}} 0 {1/C_{db}} {1/C_{db}} } } {]} delim{lbrace}{ matrix{3}{1}{I_1 I_2 I_3} }{rbrace} = omega^2 delim{[}{matrix{3}{3}{L_1 0 L_1 0 L_{rfc} L_{rfc} L_1 L_{rfc} (L_1 +L_{rfc}) } }{]} delim{lbrace}{ matrix{3}{1}{I_1 I_2 I_3} }{rbrace}$ ,

…or as a standard eigenvalue problem of the form delim{[}{A}{]} vec{X} = lambda vec{X} …

$delim{[}{ delim{[}{L}{]}^-1 delim{[}{C}{]} }{]} vec{I} =omega^2 vec{I}$ .

Note this is not the most efficient way to convert a general to a standard eigenvalue problem to solve it, but these are small matrices so performance is not an issue. However, there is an issue with these being singular matrices because they include the third loop, which is just the sum of loops one and two. There are ways to solve this, but conversion to a standard eigenvalue problem as shown above is not a valid method.

Besides using the above to work with resonant frequency response, by assuming

E = -I cos(ωt + φ) so
E' = dE/dt = I ω sin(ωt + φ),

the following can be used to characterize the response of the Hobbydyne circuit at any frequency:

$omega delim{lbrace}{ matrix{3}{1}{ {I_{i n}} 0 {I_{i n}} } }{rbrace} + omega^2 delim{[}{matrix{3}{3}{L_1 0 L_1 0 L_{rfc} L_{rfc} L_1 L_{rfc} (L_1 +L_{rfc}) } }{]} delim{lbrace}{ matrix{3}{1}{I_1 I_2 I_3} }{rbrace} = delim{[}{ matrix{3}{3}{ {1/C} {-1/C} 0 {-1/C} (1/C + 1/{C_{db}}) {1/C_{db}} 0 {1/C_{db}} {1/C_{db}} } } {]} delim{lbrace}{ matrix{3}{1}{I_1 I_2 I_3} }{rbrace}$ or
$omega vec{E prime} +delim{[}{ omega^2 delim{[}{L}{]} -delim{[}{C}{]} }{]} vec{I} = delim{[}{0}{]}$ .

In these equations, I₂ + I₃ represents the current through L_rfc. A sufficiently large impedance (Z_det » X_rfc) across L_rfc would load the circuit lightly enough to indicate the detector voltage potential without unduly effecting the tank and Hobbydyne response.

$V_{det} approx (I_2 + I_3) X_{rfc} = (I_2 + I_3) omega L_{rfc}$

Capacitive Transformers

                       Rout
          +---------o--VV--+
          |                |
o--+--||--+--||--+         |
   |  C2     C1  |         |
   |             |         |
   +----UUUUU----+         |
                 |         |
0----------------+--o------+
RT=40k   L

For 40kΩ input impedance to R_out=16Ω output centered at 1500 Hz with a bandwidth of 1000 Hz.

Analysis and Applications of the Capacitive Transformer by Ramon Vargas Patron.

The bandwidth is inversely proportional to the filter's Q: $Delta f = f_0/Q right Q={f_0}/{Delta f}=1500/1000=$ 1.5.
$C_1+C_2={Q_E}/{2 pi f_0 R_out}={Q_E}/{2 pi f_0 R_out}=1.5/{2*pi*1500*16}=9.947*10^{-6}$
0.02.
$C_2 = n(C_1+C_2)=0.02*9.947*10^{-6}=0.199*10^{-6} approx$ 0.2 μF.
$C_1 = C_1+C_2 - C_2=9.947x10^{-6}-0.199*10^{-6}=9.748*10^{-6} approx$ 10 μF.
$C_{i n}={C_1 C_2}/{C_1 + C_2}$ .
Input Q: $Q_T=Q_E({1-n}/{n})$ .

${omega_0}^2={C_1 + C_2}/{C_1 C_2 L} right L={C_1+C_2}/{C_1 C_2 4 pi^2 {f_0}^2}={10.146*10^-6}/{4*pi^2*1500*9.947*0.199*10^-12}=86.6approx$ 87 H ⁷⁾

o-------------+
              |
   +----||----+--||---+---o
   | C+(Ct&Cda)  Cdb  |
   |                  |
   +-------UUUUU------+
   |       L
o--+----------------------o

Hobbydyne

Check out the Hobbydyne. It's also backwards. Take f0≈1 MHz, C1=C+(Ct&Cda)≈95 pF, C2=Cdb≈5pF, R≈40kΩ.

$Q_E=2 pi f_0 (C_1 + C_2)R=2*pi*10^6*100*10^{-12}*40000=$ 25.13
40 kHz
$n = {C_2}/{C_1+C_2} = 5/100=$ 0.05
477

${omega_0}^2={C_1 + C_2}/{C_1 C_2 L} right L={C_1+C_2}/{C_1 C_2 4 pi^2 {f_0}^2} = {100*10^{-12}}/{4*pi^2*10^12*95*5*10^{-24}}=0.00533 approx$ 5 mH

References

Applied Numerical Methods for Engineers and Scientists, by Singiresu S. Rao, Prentice-Hall, Inc., 2002. Pages 276-277 have the circuit analysis used as a model for the Hobbydyne Eigenvalue Analysis.

¹⁾

and similarly for the next equation.

²⁾

$f(i)=h(g(i)), g(i)={i/{I_s}}+1 , h(i)= n V_T ln(i);$
$f prime(i)=h prime(g(i))g prime(i)= ({n V_T}/{g(i)}) ({1/{I_s}}) = ({n V_T}/{i/{I_s}+1}) ({1/{I_s}}) = {n V_T}/{i+{I_s}}$
$dv_D/dt=({n V_T}/{i+{I_s}}) di/dt = ({n V_T}/{I sin(omega t + phi)+{I_s}})I omega cos(omega t + phi) =$ $({n V_T}/{I tan(omega t + phi)+{I_s}cos^{-1}(omega t + phi)})I omega =$ $({n V_T}/{tan(omega t + phi)+{I_s/I}cos^{-1}(omega t + phi)})omega$