7.3 Power Relationships & Decibels

7.3 Power Relationships & Decibels

Quick Answer: P = VI = I²R = V²/R. PEP is average power at the envelope crest, not the long-term average of a keyed/modulated signal. dB (power) = 10 log(P2/P1); dB (voltage) = 20 log(V2/V1) at constant Z. +3 dB ≈ double power; +6 dB ≈ double voltage.

Topic 3-B power questions mix bench DC formulas with RF concepts (PEP) and logarithmic ratios (dB). GROL techs use all three daily: supply loading, transmitter output, and gain/loss budgets.

Three forms of electrical power

For a resistance R carrying current I with voltage V across it:

$$P = V I = I^2 R = \frac{V^2}{R}$$

DC circuit pool form: the formula for power is P = V × I (among the three equivalents). On pure resistance, all three agree.

Worked example — resistor dissipation

A 50 Ω dummy load shows 100 V RMS RF (sine, properly metered):

$$P = \frac{V^2}{R} = \frac{100^2}{50} = 200\,\mathrm{W}$$

Alternatively, $I = V/R = 2\,\mathrm{A}$, so $P = VI = 200\,\mathrm{W}$ or $P = I^2 R = 4 \times 50 = 200\,\mathrm{W}$.

Worked example — series power split

12 V across series R1 = 4 Ω and R2 = 8 Ω:

1. $R_T = 12\,\Omega$, $I = 1\,\mathrm{A}$
2. $P_1 = I^2 R_1 = 4\,\mathrm{W}$, $P_2 = 8\,\mathrm{W}$, total 12 W (also $P = VI = 12 \times 1$)

Larger series R dissipates more power at the same current.

Apparent, true, and reactive power (callback)

From Chapter 6: on AC with phase shift, V_RMS × I_RMS is apparent power (VA); true power (W) is $S \times \mathrm{PF}$ or $I^2 R$ in resistive parts; reactive power (VAR) is the out-of-phase L/C component. Section 7.3’s P = VI for DC/resistive sine heating still holds for true power when V and I are the resistive-component values or when PF is included.

Peak Envelope Power (PEP) vs average power

Peak Envelope Power (PEP) is the average power supplied to the antenna transmission line by a transmitter during one RF cycle at the crest of the modulation envelope. Read that carefully:

• It is not the instantaneous peak of a single RF sine sample without the “average over one RF cycle” idea.
• It is not automatically the long-term average power of SSB speech or CW keying.
• For a steady unmodulated carrier (CW key-down, constant envelope), PEP equals the continuous average power.
• For SSB voice, PEP is set by the loudest envelope peaks; long-term average power is lower (depends on voice, compression, and duty of speech).

Worked example — pulse average from PEP

Transmitter PEP = 100 W during a pulse; duty cycle 20%; off power negligible:

$$P_{\mathrm{avg}} \approx 100 \times 0.20 = 20\,\mathrm{W}$$

That average matters for power-supply sizing and thermal design even when the plate/license discussion quotes PEP.

Decibels — power ratios

The decibel expresses a ratio, not an absolute unit by itself (unless referenced: dBm, dBW, etc.).

Power ratio:

$$\mathrm{dB} = 10 \log_{10}\left(\frac{P_2}{P_1}\right)$$

Worked example — amplifier gain (pool style)

Input 0.5 mW, output 50 mW:

$$\frac{P_2}{P_1} = \frac{50}{0.5} = 100 \Rightarrow 10 \log_{10}(100) = 20\,\mathrm{dB}$$

Trap answers: 10 dB (forgot the ratio is 100, not 10), 100 dB (confused linear ratio with dB), 17 dB (mis-log).

Worked example — half power

A filter’s cutoff is often specified at −3 dB, meaning output power is about half the midband power (−3 dB ≈ 10 log(0.5)).

Decibels — voltage and current ratios

When comparing voltages (or currents) across the same impedance:

$$\mathrm{dB} = 20 \log_{10}\left(\frac{V_2}{V_1}\right) = 20 \log_{10}\left(\frac{I_2}{I_1}\right)$$

Why 20? Because power scales with V² (or I²) at fixed R, so 10 log(V² ratio) = 20 log(V ratio).

Memory hooks Element 3 loves:

• 3 dB ↔ factor-of-two power
• 6 dB ↔ factor-of-two voltage (and four times power)
• 10 dB ↔ ×10 power
• 20 dB ↔ ×100 power or ×10 voltage

Worked example — voltage gain to dB

Amplifier voltage gain ×10 into matched Z:

$$20 \log_{10}(10) = 20\,\mathrm{dB}$$

Power gain is also ×100 → 20 dB when Z is the same—consistent story.

Cascaded gains and losses

In decibels, stage gains and losses add:

$$G_{\mathrm{total,dB}} = G_1 + G_2 + G_3 + \cdots$$

(Negative numbers for attenuators and line loss.)

In linear ratios, you multiply.

Worked example — RF chain budget

Total: $12 - 2 - 3 + 20 = +27\,\mathrm{dB}$.

If Pin = 1 mW (0 dBm), Pout ≈ 27 dBm = 0.5 W (because 30 dBm = 1 W, 27 dBm is 3 dB less → half a watt).

Attenuators

An RF attenuator reduces signal power by a known amount (fixed pad or step attenuator). Purpose is level control, protection, and measurement—not frequency conversion or demodulation. A 10 dB pad multiplies power by 0.1 and multiplies voltage (same Z) by ≈ 0.316.

Absolute decibel references (useful, not always tested)

Conversion example: 30 dBm = 1 W = 0 dBW. 0 dBm = 1 mW.

Exam-day checklist for §7.3

1. Pick the convenient P form: VI, I²R, or V²/R.
2. PEP = average power of one RF cycle at the envelope peak.
3. Power dB → 10 log; voltage dB → 20 log (same Z).
4. Memorize 3 / 6 / 10 / 20 dB anchors cold.
5. Cascade: add dB; multiply linear ratios.
6. Attenuator: known power reduction, not a frequency mixer.

Next, RC/RL time constants and impedance networks complete Topic 3-B’s quantitative toolkit.