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 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.

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.

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. 

Other Formulas