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Formula sheet
The formulas an operator uses.
The formulas an operator uses.
Every formula an operator actually reaches for, grouped by job, each with its units and a worked example. Units are base units unless noted: volts (V), amps (A), ohms (Ω), watts (W), hertz (Hz), farads (F), henries (H), seconds (s) and meters (m). Where a formula uses MHz, feet, or similar, the table says so. The worked examples are all computed.
| Quantity | Formula | Example |
|---|---|---|
| Ohm's law (E volts, I amps, R ohms) | E = I × R; I = E ÷ R; R = E ÷ I | 12 V across 4 Ω gives 3 A |
| Power (W) | P = E × I | 12 V × 3 A = 36 W |
| Power from I and R | P = I² × R | 3² × 4 = 36 W |
| Power from E and R | P = E² ÷ R | 12² ÷ 4 = 36 W |
| Series resistors | R = R1 + R2 + R3 | 100 + 200 = 300 Ω |
| Parallel resistors | 1/R = 1/R1 + 1/R2 | 100 and 300 in parallel give 75 Ω |
| Two in parallel | R = (R1 × R2) ÷ (R1 + R2) | (100 × 300) ÷ 400 = 75 Ω |
| N equal resistors in parallel | R = R1 ÷ N | four 200 Ω in parallel give 50 Ω |
| Capacitors | parallel: C = C1 + C2; series: 1/C = 1/C1 + 1/C2 | the reverse of resistors |
| Inductors | series: L = L1 + L2; parallel: 1/L = 1/L1 + 1/L2 | the same as resistors |
| Voltage divider | Eout = Ein × R2 ÷ (R1 + R2) | 12 V, 1 kΩ and 2 kΩ: 8 V across the 2 kΩ |
| Energy (Wh, kWh) | energy = P × time | 100 W for 5 h = 500 Wh |
| Efficiency | η = Pout ÷ Pin; heat = Pin − Pout | 100 W out for 160 W in: 62.5 %, 60 W of heat |
| Wire resistance | R = ρ × L ÷ A (copper ρ ≈ 1.72 × 10⁻⁸ Ω·m) | see Wire gauge and fusing |
| Battery runtime | hours = amp-hours ÷ amps (ideal) | 7 Ah at 2 A = 3.5 h at best |
| Quantity | Formula | Example |
|---|---|---|
| Frequency and period | f = 1 ÷ T; T = 1 ÷ f | 7.15 MHz has a period of about 140 ns |
| Sine waves | Erms = 0.707 × Epeak; Epp = 2 × Epeak | 170 V peak is about 120 V rms |
| PEP (SSB, into a load) | PEP = (0.707 × Epeak)² ÷ R | 100 V peak-to-peak into 50 Ω = 25 W |
| Inductive reactance | XL = 2π × f × L | 10 µH at 7.1 MHz = 446 Ω |
| Capacitive reactance | XC = 1 ÷ (2π × f × C) | 100 pF at 7.1 MHz = 224 Ω |
| Impedance (series R and X) | Z = √(R² + X²); X = XL − XC | R = 30 Ω, X = 40 Ω gives 50 Ω |
| Phase angle | θ = arctan(X ÷ R) | arctan(40 ÷ 30) = 53.1° |
| Resonant frequency | f = 1 ÷ (2π × √(L × C)) | 10 µH and 100 pF: 5.03 MHz |
| Q of a coil | Q = XL ÷ R | |
| Bandwidth from Q | BW = f0 ÷ Q | 7.1 MHz at Q = 71 gives 100 kHz |
| Transformer ratios | Ep ÷ Es = Np ÷ Ns; Ip ÷ Is = Ns ÷ Np | |
| Impedance transformation | Zp ÷ Zs = (Np ÷ Ns)² | a 2:1 turns ratio gives 4:1 impedance |
| RC time constant | τ = R × C (s) | 1 kΩ and 100 µF: 0.1 s |
| RL time constant | τ = L ÷ R (s) | 10 mH and 100 Ω: 100 µs |
A time constant τ is how fast a charge or decay proceeds. After 1τ a charging capacitor is at 63.2 % of the supply; after 2τ, 86.5 %; 3τ, 95.0 %; 4τ, 98.2 %; 5τ, 99.3 %. A discharging one is left with 36.8 %, 13.5 %, 5.0 %, 1.8 % and 0.7 %. See Time constants and Resonance and Q.
| Quantity | Formula | Example |
|---|---|---|
| Wave speed | v = f × λ (c ≈ 3 × 10⁸ m/s) | |
| Wavelength (m) | λ = 300 ÷ f (MHz) | 146 MHz: 2.05 m |
| Half-wave dipole (ft) | L = 468 ÷ f (MHz) | 7.15 MHz: 65.5 ft |
| Quarter-wave (ft) | L = 234 ÷ f (MHz) | 14.2 MHz: 16.5 ft |
| Full-wave loop (ft) | L = 1005 ÷ f (MHz) | 14.2 MHz: 70.8 ft |
| Line length with velocity factor | physical length = electrical length × VF | quarter-wave at 146 MHz, VF 0.66: 1.06 ft |
| Velocity factor | VF = 1 ÷ √εr | |
| dBi and dBd | dBi = dBd + 2.15 | a 6 dBd antenna is 8.15 dBi |
| ERP and EIRP | ERP = Ptx × G (G versus a dipole); EIRP = ERP × 1.64 | 100 W and 6 dBd (×3.98): ERP 398 W, EIRP 653 W |
| Radio horizon (miles) | d ≈ √(2 × h), h in feet | 100 ft: about 14 miles |
| Free-space path loss (dB) | FSPL = 32.44 + 20 log(d km) + 20 log(f MHz) | 146 MHz, 1 km: 75.7 dB |
| Doppler shift | Δf = f × v ÷ c (v is the speed along the line of sight) | 435 MHz, 7.5 km/s: up to about 10.9 kHz |
The 468 comes from 492 (a true half wave) less about 5 % for end effect and wire; it is a first estimate, then trim for the SWR. See The half-wave dipole.
| Quantity | Formula | Example |
|---|---|---|
| Coax impedance | Z0 = (138 ÷ √εr) × log(D ÷ d) | D/d = 3.5, εr = 2.25: 50.1 Ω |
| Line impedance | Z0 = √(L ÷ C) (per unit length) | |
| Reflection coefficient | Γ = (ZL − Z0) ÷ (ZL + Z0) | 100 Ω on 50 Ω line: Γ = 0.333 |
| SWR | SWR = (1 + |Γ|) ÷ (1 − |Γ|) | Γ = 0.333: SWR 2.0 |
| SWR, resistive load | SWR = ZL ÷ Z0 (or Z0 ÷ ZL, whichever is larger) | 100 Ω on 50 Ω line: 2:1 |
| Reflected power | Pr ÷ Pf = Γ² | SWR 3: Γ = 0.5, so 25 % reflected |
| Return loss (dB) | RL = −20 log(|Γ|) | Γ = 0.333: 9.5 dB |
| Mismatch loss (dB) | ML = −10 log(1 − Γ²) | SWR 2: 0.51 dB |
| Cable loss | total dB = dB per 100 ft × length ÷ 100 | 2.4 dB/100 ft, 150 ft: 3.6 dB |
| Quarter-wave transformer | Zmatch = √(Z1 × Z2) | 50 Ω to 200 Ω: 100 Ω line |
SWR is how a mismatch is measured; see SWR and reflections. On a lossless line the SWR is the same everywhere along it; with real cable loss, the reading at the radio looks better than the SWR at the antenna.
| Quantity | Formula | Example |
|---|---|---|
| Power ratio in dB | dB = 10 log(P2 ÷ P1) | 100 W to 50 W: −3 dB |
| Voltage ratio in dB | dB = 20 log(E2 ÷ E1) (same impedance) | doubling the voltage: +6 dB |
| Ratio from dB | P2 = P1 × 10^(dB ÷ 10) | 10 W with +6 dB: about 40 W |
| dBm | dBm = 10 log(P ÷ 1 mW) | 1 W = 30 dBm; 5 W = 37 dBm |
| S meter | 1 S unit = 6 dB; S9 = −73 dBm on HF | S5 is 24 dB below S9 |
| Noise floor | −174 dBm/Hz + 10 log(bandwidth in Hz) + noise figure (dB) | 2.4 kHz, NF 10 dB: −130 dBm |
| Link margin | received = transmitted + gains − losses (all in dB) |
Handy anchors: 3 dB is double (or half) the power, 6 dB is four times (voltage doubles), 10 dB is ten times, and 20 dB is a hundred times. Add dB for cascaded gains and losses. See Decibels.
| Quantity | Formula | Example |
|---|---|---|
| Mixer output | fout = f1 + f2 and f1 − f2 | 7.1 MHz mixed with 5.0 MHz gives 12.1 and 2.1 |
| AM bandwidth | BW = 2 × highest audio frequency | 3 kHz audio: 6 kHz |
| FM modulation index | m = deviation ÷ modulating frequency | 5 kHz ÷ 1 kHz = 5 |
| Carson's rule (FM) | BW ≈ 2 × (deviation + highest audio) | 5 kHz, 3 kHz: 16 kHz |
| Nyquist | sampling rate must exceed 2 × the highest frequency | |
| Bit rate | bit rate = baud × bits per symbol | |
| Shannon limit | C = B × log2(1 + SNR) (SNR as a ratio) | 2.5 kHz, 20 dB: 16.6 kbit/s |
| CW speed | dit length = 1.2 ÷ WPM seconds | 20 WPM: 60 ms |
| Duty cycle | average power = peak power × fraction of time transmitting | |
| Rectifier ripple | full wave: 2 × line frequency; half wave: line frequency | 120 Hz at 60 Hz |