## Xtal Set Society, Inc

**Formula Index:**

Updated 11/22/13

Formulas this page:

Calculators are provided for some of the formulas listed. These process by using your computer CPU in conjunction with a JavaScript loaded with the webpage. Most times you see a form on a webpage, it's powered by a script. Most browsers now support scripting, including Explorer 4.5 up, Netscape 4.6 and up, and the current Mozilla Foxfire. You may need to "enable" scripting. Look in the HELP menu of your browser's toolbar.

**The calculators we've coded so far are easy to use** and start with a typcial solution displayed. RESET will take you back to the typical initial solution. If a formula calls for N parameters, enter all N or N-1 and click on the round parameter button you desire calculated/updated. If interest develops for these, we'll add more. Let us know what you'd like to see added/changed. Constructive suggestions and comments are prized and appreciated!

**Resonant Frequency of an LC Circuit** - with calculator

where *f* is the frequency in hertz, L inductance in henrys, and C capacitance in farads.

For the AM broadcast band, typical inductance (coil) values range from 250 uH down. Capacitance values vary from tens of pfs to several hundred pfs. 365 pf air variable caps are common.

**Inductive Reactance of a Coil**

where f is the frequency in hertz and L is the coil inductance in henrys. X is the AC impedance of the coil. Example: At 1 MHz with L equal to 250 uH, X = 2*3.14*1*exp(6)*250*exp(-6)= 1,570 ohms. exp(6) means 10 to the 6th power, or 1,000,000.

**Capacitive Reactance of a Capacitor**

where f is the frequency in hertz and C is the capacitance in farads. X is the AC impedance of the cap. Example: At 1 MHz with C equal to 105 pf, X = 1/(2*3.14*1*exp(6)*105*exp(-12) = 1,516 ohms.

**Inductance of a Single-Layer Cylindrical (Coil)** - with calculator

where L is in uH, N is turns, length* "l"* in inches, coil radius r in inches, and pitch, p, in inches. Pitch is the distance between the center of one turn and the next. Np is the coil length. The calculator below uses the second formula. The first is attributed to Wheeler; the second a slight adaptation. The reset example shown is for a coil with L = 250 uH, radius of 1.75 inches, with 61 turns, and winding spacing of 0.05 inches.

**Inductance of a Spider Web (Spiral) Coil**

The Spider (spiral) coil will produce a stronger signal than its cousin, the standard close wound cylindrical. This is accomplished by separating adjacent windings from each other, thus reducing the proximity effect between wires, wherein the closer they are a current in one will increase the resistance of the other. The separation of adjacent turns is accomplished by breaking up each full turn into an odd number of segments wherein each addition segment is wound on the opposite side of a flat form. This is enabled by cutting an odd-number of radial slits extending from the outer edge of the center support of the form. Windings of the first few turns are displayed in picture 1. The winding starts from the bottom of the interior ring of the form; hence the first and all odd segments of the first turn will show on the top as noted by "1." The last segment of turn 1, segment 9, comes out on top and zig-zags over to become turn 2, wherein it goes underneath the form again. The following segment of turn 2 is then on the top as noted, and so on. The inductance of the spider coil is determined by its inner and outer radiuses and the diameter and spacing of the wire used. Here's the equation for the Spider Web coil, derived from the original Wheeler formula for cylindrical coils:

For our coil formula the inner diameter, Di, is the opening at the center of the coil and the outer diamater measures the full length of the coil. For example, using 56 turns of 150/45 Litz wire, an inner diameter of 1.6 nchs and and an outer diameter of 4.2 inches, the formula says the coil inductance is 254 uH. Installed in the crystal set, that seemed about right, given that the 365 pf cap tuned the band from about 550 to 1500 kHz. See the download files page for more detail.

**Q of a Coil**

where omega is 2(pi)f, f is in hertz, L in heneries, and (AC) r in ohms. Example: At 1 MHz, a 250 uH coil with 6 ohms of AC resistance has a Q = 2*3.14*1*exp(6)*250*exp(-6)/6 = 261. Q is proportional to coil form diameter and inversely proportional to the AC resistance in the coil. Hence, a 3.5 inch dia coil with Litz wire (low loss multi-strand wire) will have a much higher Q than a coil with hookup wire wound on a 2 inch form. Selectivity of a crystal se is inversely proportional to Q; that is, the bandwidth of a parallel tuned circuit at near resonance is it's frequency divided by the effective bandwidth, BW.

**Measurement for Coil Self-Capacitance** - using the Double-Frequency method

where Co is the self-capacitance, and C2 and C1 are measurement values. The coil in question is resonated at frequency f2, and the capacitance, C2, is measured. (Perhaps using a capacitance meter). The coil is then resonated at f1, which is set to twice f2, and the capacitance needed, C1, is measured. The self-capacitance of the coil is then calculated using the above formula. Example: A 90 uH coil is resonated at 909 kHz with a 340 pf capacitor. Then coil is again resonated at twice that, 1818 kHz with an 80 pf capacitor. Self capacitance is therefore = (340 - 4*80)/3 = 6.6 pf. For a derivation of the formula, see the September 2007 issue of the XSS Newsletter.

**Series to Parallel Circuit Conversion** - at a given frequency

where Rp and Xp are the parallel equivalent values at a particular frequency corresponding to the series circuit. See Article 1: Equivalent Series and Parallel Circuits for details. Example: A 50 ohm resistor in series with a 100 pf capacitor at 1 MHz has an equivalent parallel circuit at that frequency of a 50.7 K resistor in parallel with about 100 pf of capacitance.

**AC Resistance/Foot of Wire**

where R is the AC resistance of the wire in ohms/foot, f is in Hz, and D is the wire diameter in inches. Example: The AC resistance/foot of #24 hookup wire, assuming copper, at 1 MHz is = exp(-6)sqrt(exp(6)/0.02 = 0.05 ohms/foot. The total AC resistance therefore of a 250 uH coil with 61 turns and 3.5 inch diameter form would be = 0.05*2*3.14*3.5/2*61/12 = 2.79 ohms!

**Skin Depth for Copper**

where the skin depth (of copper) is in mils and f is the frequency in MHz. Example: The skin depth of AC current at 1 MHz in copper is = 2.6/sqrt(exp(6)) = 2.6exp(-3) mils or 2.6 micro-inches.

**Diode Current and Dynamic Resistance** - with calculator

where iD is the diode current, Io is the diode saturation current, m is the quality factor [from 1 to 2, use 1.2], and vd is the voltage across the diode. Also, rd is the forward dynamic resistance of the diode at the operating point Q.