1.3 Power, Energy, and Resistance Calculations

1.3 Power, Energy, and Resistance Calculations

In amateur radio engineering, calculating electrical power and signal level changes is essential. Radio transmitters convert DC electrical power into high-frequency RF power, while receivers amplify microvolt-level signals over many orders of magnitude. Master candidate proficiency in power dissipation, internal battery resistance, decibel ratios, and absolute power units (dBm, dBW) is directly assessed on the ACMA Standard Theory examination.

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1. Electrical Power and Energy Fundamentals

Electrical Power ($P$)

Electrical Power ($P$) is defined as the rate at which electrical energy is transferred, converted, or dissipated per unit of time. The SI unit of power is the Watt (W), named after James Watt. One Watt represents an energy transfer rate of one Joule per second ($1\text{ W} = 1\text{ J/s}$).

By combining the basic definition of power ($P = V \times I$) with Ohm's Law ($V = I \times R$ and $I = V / R$), three equivalent power formulas are derived:

$\text{Power Formulas: } \quad P = V \times I \quad \vert \quad P = I^2 \times R \quad \vert \quad P = \frac{V^2}{R}$

Where:

• $P$ is power in Watts (W)
• $V$ is potential difference in Volts (V)
• $I$ is current in Amperes (A)
• $R$ is resistance in Ohms ($\Omega$)

Electrical Energy ($E$)

Energy ($E$) measures total work performed over a given duration: $E = P \times t$

While the SI unit of energy is the Joule (J), electrical energy is routinely measured in Watt-hours (Wh) or Kilowatt-hours (kWh):

• $1\text{ Watt-hour (Wh)} = 1\text{ W} \times 3600\text{ s} = 3,600\text{ Joules (J)}$
• $1\text{ Kilowatt-hour (kWh)} = 1,000\text{ Wh} = 3.6 \times 10^6\text{ Joules}$

Worked Example 1.3: Battery Operating Capacity

A $12\text{ V}$, $20\text{ Ampere-hour (Ah)}$ lithium iron phosphate ($\text{LiFePO}_4$) battery powers a portable 50W VHF transmitter. On transmit, the radio draws $9.0\text{ A}$ from the battery. Calculate:

1. Total energy stored in the battery in Watt-hours.
2. Theoretical continuous transmit operating time in hours.

Solution: $E = V \times Q = 12\text{ V} \times 20\text{ Ah} = 240\text{ Wh}$ $\text{Transmit Power Drawn } P = 12\text{ V} \times 9.0\text{ A} = 108\text{ W}$ $\text{Operating Time } t = \frac{E}{P} = \frac{240\text{ Wh}}{108\text{ W}} \approx 2.22\text{ hours (2 hours 13 minutes)}$

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2. Resistor Power Ratings and Thermal Dissipation

Every physical resistor converts electrical energy into heat energy. Resistors carry a maximum power rating (e.g., $0.25\text{ W}$, $0.5\text{ W}$, $2\text{ W}$, $50\text{ W}$) indicating the maximum power the device can continuously dissipate without structural degradation, resistance value shift, or fire risk.

When designing or repairing amateur radio equipment, designers must select resistors with a power rating significantly higher than the calculated actual dissipation—typically applying a safety factor of $2\times$ (derating).

If a $100\ \Omega$ resistor carries a current of $0.1\text{ A}$, the power dissipated is: $P = I^2 \times R = (0.1)^2 \times 100 = 0.01 \times 100 = 1.0\text{ Watt}$ In this case, a $2\text{ W}$ or $5\text{ W}$ rated power resistor must be used.

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3. Internal Resistance of Power Sources & Maximum Power Transfer

Real-world voltage sources (batteries, power supplies, solar panels) are not ideal. They possess an internal resistance ($r$) in series with their ideal electromotive force ($E_{\text{emf}}$).

+---[ Ideal Source E ]---+---[ Internal Resistance r ]---+--- ( Positive Terminal )
                                                         |
                                                     [ Load R_L ]
                                                         |
---------------------------------------------------------+--- ( Negative Terminal )

When current $I$ is drawn by an external load $R_L$, a voltage drop occurs across the internal resistance: $V_{\text{load}} = E_{\text{emf}} - (I \times r)$ Where: $I = \frac{E_{\text{emf}}}{r + R_L}$

Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem states that maximum power is transferred from a power source to an external load when the load resistance equals the internal resistance of the source ($R_L = r$).

Important implications for ACMA exams:

1. When $R_L = r$, power transferred to the load is at its maximum absolute value.
2. At maximum power transfer, electrical efficiency is exactly 50%, as equal power is dissipated inside the source ($I^2 r$) and inside the load ($I^2 R_L$).
3. In RF antenna systems, matching generator output impedance to transmission line and load impedance (e.g., $50\ \Omega$) maximises RF energy transfer.

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4. Decibel (dB) Calculations

The Decibel (dB) is a logarithmic unit used to express ratios of power, voltage, or current. Logarithmic scales are essential in radio communications because signal power ranges from fractions of a picowatt at receiver antennas to kilowatts at transmitter output stages.

Power Ratio Formula

$\text{Gain/Loss in dB} = 10 \log_{10} \left( \frac{P_2}{P_1} \right)$ Where $P_1$ is reference power and $P_2$ is measured power.

Voltage Ratio Formula (Equal Impedance Circuits)

$\text{Gain/Loss in dB} = 20 \log_{10} \left( \frac{V_2}{V_1} \right)$

Common Decibel Rules of Thumb (Must Memorize):

Absolute Power Units: dBm and dBW

• $\text{dBm}$: Power relative to $1\text{ Milliwatt } (1\text{ mW} = 10^{-3}\text{ W})$. $P_{\text{dBm}} = 10 \log_{10} \left( \frac{P_{\text{watts}}}{0.001\text{ W}} \right)$
• $\text{dBW}$: Power relative to $1\text{ Watt } (1\text{ W})$. $P_{\text{dBW}} = 10 \log_{10} \left( \frac{P_{\text{watts}}}{1\text{ W}} \right)$

Key Conversions:

• $0\text{ dBm} = 1\text{ mW}$
• $30\text{ dBm} = 1000\text{ mW} = 1\text{ W} = 0\text{ dBW}$
• $50\text{ dBm} = 100\text{ W} = +20\text{ dBW}$ (the HF limit for an amateur standard station using emission mode J3E or R3E; 30 W pY mean power applies in any other mode).

Worked Example 1.4: Multi-Stage RF System Decibel Calculation

An RF signal generator outputs $2\text{ mW}$ ($+3\text{ dBm}$). The signal passes through a $+13\text{ dB}$ preamplifier, a $-4\text{ dB}$ bandpass filter attenuator, and a $+20\text{ dB}$ power amplifier. Calculate total system gain and final output power in Watts.

Solution: $\text{Total System Gain} = +13\text{ dB} - 4\text{ dB} + 20\text{ dB} = +29\text{ dB}$ $\text{Output Power in dBm} = +3\text{ dBm} + 29\text{ dB} = +32\text{ dBm}$ $\text{Converting +32 dBm to Watts}: \quad 32 = 10 \log_{10}\left(\frac{P}{0.001}\right) \implies 10^{3.2} \times 0.001 = 1585 \times 0.001 \approx 1.585\text{ Watts}$