G5B: Decibels and Power – Ham Radio General License Study Guide
G5B covers the mathematics of RF power measurement: the decibel scale, how to calculate electrical power using Ohm's law in its various forms, how to convert between peak, peak-to-peak, and RMS voltage values for sine waves, and how to calculate peak envelope power (PEP) for SSB transmissions.
The exam draws from topics including what dB change represents a factor of two in power, how total current relates to branch currents in parallel resistors, how to calculate power in watts from voltage and resistance values, how to calculate PEP from peak-to-peak voltage across a load, what value of AC signal produces the same power as DC, how to convert between RMS and peak-to-peak voltages, what percentage of power is lost with a 1 dB reduction, what the PEP-to-average power ratio is for an unmodulated carrier, and how to find RMS voltage from power and resistance.
The Decibel
The decibel (dB) is a logarithmic unit used to express power ratios. Because RF power levels span many orders of magnitude, a logarithmic scale makes comparisons practical. The decibel is defined as:
dB = 10 × log₁₀(P2 / P1)
Key decibel values to memorize:
| dB Change | Power Ratio | Effect |
|---|---|---|
| +3 dB | 2× more power | Double the power |
| −3 dB | ½ power | Half the power |
| +10 dB | 10× more power | Ten times the power |
| −1 dB | ~79.4% remaining | ~20.6% power loss |
A loss of 1 dB corresponds to approximately a 20.6% reduction in power. This is a frequently tested value.
For an unmodulated carrier (a constant-amplitude sine wave with no modulation), the peak envelope power (PEP) equals the average power — the ratio of PEP to average power is 1.00. The signal amplitude does not vary, so the average and peak are identical.
Electrical Power Calculations
Electrical power in DC and AC resistive circuits is calculated using three equivalent forms of the power formula, all derived from Ohm's law (V = IR):
| Formula | When to use it |
|---|---|
| P = I × V | When you know current (I) and voltage (V) |
| P = V² ÷ R | When you know voltage (V) and resistance (R) |
| P = I² × R | When you know current (I) and resistance (R) |
400 VDC into 800 Ω: P = 400² ÷ 800 = 160,000 ÷ 800 = 200 watts
12 VDC, 0.2 A bulb: P = 0.2 × 12 = 2.4 watts
7.0 mA through 1,250 Ω: P = (0.007)² × 1,250 = 0.000049 × 1,250 = ~61 milliwatts
Current in Parallel Circuits
In a circuit with parallel resistors, each resistor has the same voltage across it but carries a different current depending on its value. The total current drawn from the source equals the sum of the individual branch currents. Adding more parallel branches increases the total current and decreases total resistance.
RMS and Peak Voltage
For AC sine waves, three different voltage values are used to describe the waveform:
- Peak voltage (Vpeak) — the maximum amplitude reached by the sine wave
- Peak-to-peak voltage (Vpp) — the total swing from negative peak to positive peak: Vpp = 2 × Vpeak
- RMS voltage (Vrms) — the effective value: Vrms = Vpeak ÷ √2 ≈ Vpeak × 0.707
The RMS value produces the same power dissipation in a resistor as a DC voltage of the same value. This is why household AC is rated in RMS volts — 120 VAC RMS delivers the same heating effect to a resistor as 120 VDC.
17 V peak → RMS: 17 ÷ 1.414 = ~12 V RMS
120 V RMS → peak: 120 × 1.414 = 169.7 V peak → peak-to-peak: 169.7 × 2 = ~339.4 V peak-to-peak
To find RMS from power and resistance: Vrms = √(P × R) → √(1200 × 50) = √60,000 = ~245 V RMS
Peak Envelope Power (PEP)
Peak Envelope Power (PEP) is the average power at the crest of the modulation envelope — the highest instantaneous power during a transmission. For SSB, PEP is measured during a speech peak. For an unmodulated carrier, PEP equals average power.
To calculate PEP from a peak-to-peak voltage across a known load resistance:
- Find the peak voltage: Vpeak = Vpp ÷ 2
- Find the RMS voltage: Vrms = Vpeak ÷ √2
- Calculate power: PEP = Vrms² ÷ R
200 Vpp across 50 Ω:
Vpeak = 200 ÷ 2 = 100 V
Vrms = 100 ÷ √2 = 70.7 V
PEP = 70.7² ÷ 50 = 5,000 ÷ 50 = 100 watts
500 Vpp across 50 Ω:
Vpeak = 500 ÷ 2 = 250 V
Vrms = 250 ÷ √2 = 176.8 V
PEP = 176.8² ÷ 50 = 31,250 ÷ 50 = 625 watts
Unmodulated carrier, average power = 1060 W → PEP = 1060 watts (ratio 1.00)
G5B Practice Questions
Check Your Knowledge
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