1.1 DC Circuit Analysis, Ohm's Law, and Kirchhoff's Laws

1.1 DC Circuit Analysis, Ohm's Law, and Kirchhoff's Laws

Direct current (DC) circuit theory provides the foundational bedrock for all electrical engineering and amateur radio communication systems. Whether analysing the biasing networks of a bipolar junction transistor power amplifier, sizing power cabling for an HF transceiver drawing tens of amperes, or constructing high-power non-inductive dummy loads, a rigorous understanding of charge flow, potential difference, and energy conservation is indispensable.

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1. Fundamental Electrical Units and Quantities

All electrical phenomena in direct current circuits arise from the displacement and interaction of subatomic charges. In the International System of Units (SI), electrical quantities are strictly defined in terms of fundamental mechanical and electromagnetic dimensions:

• Electric Charge ($Q$): Measured in Coulombs ($\text{C}$). One Coulomb represents the aggregate charge carried by approximately $6.2415 \times 10^{18}$ electrons. The elementary charge of a single electron is $e \approx -1.6022 \times 10^{-19}\text{ C}$.
• Electric Current ($I$): Measured in Amperes ($\text{A}$). Current defines the time rate of charge displacement through a specified cross-sectional boundary:

$$I = \frac{dQ}{dt}$$

One Ampere corresponds to the transfer of one Coulomb of charge per second ($1\text{ A} = 1\text{ C/s}$). By international convention, conventional current flows from positive potential to negative potential, opposite to the physical drift of free electrons.

• Potential Difference and Electromotive Force ($V$ or $E$): Measured in Volts ($\text{V}$). Potential difference represents the work ($W$) required to displace a unit charge between two points in an electric field:

$$V = \frac{dW}{dQ}$$

One Volt represents one Joule of energy expended per Coulomb of charge displaced ($1\text{ V} = 1\text{ J/C}$).

• Electrical Resistance ($R$): Measured in Ohms ($\Omega$). Resistance quantifies the opposition presented by a material to the uniform flow of charge carriers, converting electrical potential energy irreversibly into thermal kinetic energy (Joule heating). One Ohm is the resistance across which a potential difference of one Volt produces a current of one Ampere ($1\ \Omega = 1\text{ V/A}$).
• Energy ($W$): Measured in Joules ($\text{J}$). Energy represents the total physical work performed over time. In electrical systems, one Joule equals one Watt-second ($1\text{ J} = 1\text{ W}\cdot\text{s}$).
• Power ($P$): Measured in Watts ($\text{W}$). Power represents the instantaneous rate of energy conversion or dissipation over time:

$$P = \frac{dW}{dt}$$

One Watt equals one Joule per second ($1\text{ W} = 1\text{ J/s}$).

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2. Ohm's Law and Joule's Power Formulations

Ohm's Law

Formulated by Georg Simon Ohm, this law states that for an ideal conductor operating at a constant temperature, the current flowing through the conductor is directly proportional to the potential difference across it and inversely proportional to its resistance:

$$V = I \cdot R, \quad I = \frac{V}{R}, \quad R = \frac{V}{I}$$

In linear (ohmic) materials, resistance remains constant regardless of applied voltage or current magnitude. Non-ohmic devices (such as semiconductor diodes, thermionic valves, and varistors) exhibit non-linear $V\text{-}I$ characteristics where dynamic resistance $r_d = dV/dI$ varies with operating point.

Joule's Law and Electrical Power

Joule's Law defines the thermal power generated when an electrical current traverses a resistive medium. By combining the fundamental definition of power ($P = V \cdot I$) with Ohm's Law substitutions, three interrelated power equations emerge:

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

These formulations are critical when calculating thermal dissipation in dropping resistors, dummy loads, and transmission lines.

Worked Example 1.1: DC Power Supply and Amplifier Load Analysis

An amateur transceiver draws a total of $18.5\text{ A}$ from a $13.8\text{ V}$ regulated supply during CW key-down. Treat these as measurements at the radio's DC input terminals.

1. Calculate the total DC input power supplied to the radio:

$$P_{\text{in}} = V \cdot I = 13.8\text{ V} \times 18.5\text{ A} = 255.3\text{ W}$$

1. Determine the equivalent DC resistance presented by the transceiver terminals:

$$R_{\text{eq}} = \frac{V}{I} = \frac{13.8\text{ V}}{18.5\text{ A}} \approx 0.7459\ \Omega$$

1. If the transmitter delivers $100\text{ W}$ of RF power into a matched $50\ \Omega$ antenna load, calculate the whole-radio heat dissipation and overall DC-to-RF efficiency:

$$P_{\text{heat}} = P_{\text{in}} - P_{\text{RF}} = 255.3\text{ W} - 100.0\text{ W} = 155.3\text{ W}$$ $$\eta = \frac{P_{\text{RF}}}{P_{\text{in}}} \times 100\% = \frac{100.0\text{ W}}{255.3\text{ W}} \times 100\% \approx 39.17\%$$

This calculation illustrates why high-power solid-state amplifiers require extensive extruded aluminium heat sinks and forced-air cooling fans to safely shed heat.

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3. Kirchhoff's Laws

Complex networks containing multiple interconnected branches, loops, and sources cannot be simplified by Ohm's Law alone. Gustav Kirchhoff formulated two governing circuit laws rooted directly in fundamental conservation laws of physics.

Kirchhoff's Current Law (KCL) — Conservation of Charge

Kirchhoff's Current Law states that the algebraic sum of all electric currents entering and exiting any electrical node (junction) is identically zero:

$$\sum_{k=1}^n I_k = 0 \quad \Longleftrightarrow \quad \sum I_{\text{entering}} = \sum I_{\text{leaving}}$$

Because electric charge can neither accumulate indefinitely at a junction nor be created out of nothing, the total rate at which charge arrives at a node must precisely equal the rate at which charge departs.

Kirchhoff's Voltage Law (KVL) — Conservation of Energy

Kirchhoff's Voltage Law states that the algebraic sum of all electrical potential differences (electromotive forces and resistive voltage drops) around any closed conductive loop or mesh is identically zero:

$$\sum_{k=1}^m V_k = 0 \quad \Longleftrightarrow \quad \sum E_{\text{sources}} = \sum (I \cdot R)_{\text{drops}}$$

Because the electrostatic field is conservative, displacing a test charge completely around any closed loop returns it to its original electrical potential, meaning no net work is performed on the charge over the full circuit.

Worked Example 1.2: DC Station Bus Distribution

A $13.8\text{ V}$ primary power supply feeds an amateur radio station distribution block. Three separate equipment branches connect in parallel to the central distribution node:

• Branch 1 (HF Transceiver): draws $16.0\text{ A}$.
• Branch 2 (VHF/UHF Repeater Link): draws $2.5\text{ A}$.
• Branch 3 (Antenna Tuner & Accessories): draws $0.7\text{ A}$.

The main feed cable connecting the power supply to the distribution block has an unavoidable loop lead resistance (positive plus negative conductors) of $0.025\ \Omega$.

1. By KCL, calculate total current leaving the distribution node:

$$I_{\text{total}} = I_1 + I_2 + I_3 = 16.0\text{ A} + 2.5\text{ A} + 0.7\text{ A} = 19.2\text{ A}$$

1. By Ohm's Law, calculate the voltage drop along the main feed cable:

$$V_{\text{cable}} = I_{\text{total}} \cdot R_{\text{cable}} = 19.2\text{ A} \times 0.025\ \Omega = 0.48\text{ V}$$

1. Applying KVL around the loop including the power supply, cable drop, and distribution node:

$$E_{\text{supply}} - V_{\text{cable}} - V_{\text{node}} = 0$$ $$V_{\text{node}} = 13.80\text{ V} - 0.48\text{ V} = 13.32\text{ V}$$

Each connected radio receives $13.32\text{ V}$ rather than the full $13.80\text{ V}$, demonstrating the practical importance of heavy-gauge power conductors in high-current amateur stations.

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4. Resistor Network Combinations

Resistors in Series

When resistors are connected end-to-end such that the same current traverses each element sequentially, the total equivalent resistance ($R_{\text{total}}$) is the direct arithmetic sum of individual resistances:

$$R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$$

Because current is identical through each element ($I_1 = I_2 = \dots = I$), the total voltage drop divides proportionally across the resistors according to their individual values:

$$V_k = V_{\text{total}} \cdot \frac{R_k}{R_{\text{total}}}$$

Resistors in Parallel

When resistors are connected across the same two common nodes, the identical potential difference appears across each branch. Total circuit current is the sum of the branch currents. The equivalent resistance is calculated from the sum of the conductances ($G = 1/R$):

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$$

For exactly two resistors in parallel, this simplifies to the product-over-sum rule:

$$R_{\text{total}} = \frac{R_1 \cdot R_2}{R_1 + R_2}$$

For $n$ identical resistors of value $R$ connected in parallel:

$$R_{\text{total}} = \frac{R}{n}$$

The total equivalent resistance of any parallel network is always strictly less than the smallest individual resistor in that network.

Practical Application: Constructing an RF Dummy Load

Wire-wound power resistors cannot be used at radio frequencies because their spiral construction introduces significant parasitic series inductance, ruining impedance matching. Instead, amateurs construct non-inductive dummy loads using parallel arrays of metal oxide or thick-film resistors.

To construct a $50\ \Omega$ non-inductive dummy load rated for $100\text{ W}$ continuous dissipation:

• Twenty $1.0\ \text{k}\Omega$ metal-oxide resistors ($R = 1000\ \Omega$) are connected in parallel.
• Equivalent resistance: $R_{\text{total}} = 1000\ \Omega / 20 = 50.0\ \Omega$.
• If each resistor has a continuous power rating of $5\text{ W}$, the aggregate rating is $20 \times 5\text{ W} = 100\text{ W}$.
• This configuration divides current equally across all twenty resistors ($I_k = I_{\text{total}} / 20$), ensuring even heat distribution.

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5. Voltage Dividers: Unloaded vs. Loaded Analysis

The Unloaded Voltage Divider

A series pair of resistors connected across a source potential $V_{\text{in}}$ constitutes a basic voltage divider. With no external load attached to the output terminals:

$$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$$ $$I_{\text{bleeder}} = \frac{V_{\text{in}}}{R_1 + R_2}$$

The Loaded Voltage Divider Effect

When an external load resistor ($R_L$) is connected across $R_2$, current is diverted away from $R_2$ into the load. The lower arm of the divider now consists of the parallel combination of $R_2$ and $R_L$:

$$R_{\text{eq}} = R_2 \parallel R_L = \frac{R_2 \cdot R_L}{R_2 + R_L}$$

Because $R_2 \parallel R_L$ is strictly less than $R_2$, the loaded output voltage drops:

$$V_{\text{out(loaded)}} = V_{\text{in}} \cdot \frac{R_2 \parallel R_L}{R_1 + (R_2 \parallel R_L)}$$ $$\text{Voltage Regulation (\%)} = \frac{V_{\text{no-load}} - V_{\text{full-load}}}{V_{\text{full-load}}} \times 100\%$$

Stiff Divider Design Rule

To maintain stable output voltage against variations in load current, a divider must be "stiff". A standard engineering rule of thumb dictates that the bleeder current flowing through $R_2$ should be at least ten times the full-load current ($I_{\text{bleeder}} \ge 10 \cdot I_L$), which is equivalent to designing $R_2 \le 0.1 \cdot R_L$. While this wastes some power in the divider resistors, it limits voltage sagging under load to under 10%.

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6. Internal Resistance and Maximum Power Transfer

Practical Voltage Sources and Terminal Voltage

A useful model for a practical voltage source is an ideal EMF $E$ in series with resistance $r_{\text{int}}$. It approximates terminal behavior over the range where $E$ and $r_{\text{int}}$ can be treated as fixed. A regulated supply entering current limit, for example, no longer follows that fixed-resistance model.

When connected to an external load $R_L$, the circuit current is:

$$I = \frac{E}{R_L + r_{\text{int}}}$$

Within this model, terminal voltage decreases linearly with load current:

$$V_{\text{terminal}} = E - I \cdot r_{\text{int}} = E \cdot \frac{R_L}{R_L + r_{\text{int}}}$$

The model predicts $I_{\text{sc}}=E/r_{\text{int}}$ at $R_L=0$. An actual source may instead enter current limit or have changing internal resistance; use its specified behavior when assessing fault current.

The Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem states that a source with a fixed internal resistance $r_{\text{int}}$ delivers maximum power to an external resistive load $R_L$ when the load resistance equals the source internal resistance:

$$R_L = r_{\text{int}}$$

To prove this mathematically, express power delivered to the load as a function of $R_L$:

$$P_L = I^2 \cdot R_L = \left( \frac{E}{R_L + r_{\text{int}}} \right)^2 R_L = \frac{E^2 \cdot R_L}{(R_L + r_{\text{int}})^2}$$

Differentiating $P_L$ with respect to $R_L$ and setting the derivative to zero yields:

$$\frac{dP_L}{dR_L} = E^2 \cdot \frac{(R_L + r_{\text{int}})^2 - 2R_L(R_L + r_{\text{int}})}{(R_L + r_{\text{int}})^4} = 0 \implies R_L = r_{\text{int}}$$

Under this matched condition, the maximum power delivered to the load is:

$$P_{L(\text{max})} = \frac{E^2}{4 \cdot r_{\text{int}}}$$

The Efficiency Trade-off

At the point of maximum power transfer ($R_L = r_{\text{int}}$), an identical amount of power is dissipated inside the source itself as heat ($P_{\text{internal}} = I^2 r_{\text{int}} = P_L$). Consequently, the electrical efficiency is exactly:

$$\eta = \frac{P_L}{P_{\text{total}}} = \frac{I^2 R_L}{I^2(R_L + r_{\text{int}})} = \frac{R_L}{2R_L} = 50\%$$

For an RF line, a load equal to its characteristic impedance produces no reflected wave: a $50\ \Omega$ line needs a $50\ \Omega$ termination. Matching only the transmitter end does not remove a load mismatch. This reflection condition does not require 50% DC-to-RF efficiency. For DC distribution, low source and cable resistance reduce voltage drop; a regulated supply's conversion efficiency must be assessed separately.

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Battery Capacity and Series/Parallel Sources

Battery capacity in ampere-hours (Ah) measures charge: $Q_{\text{Ah}}=I_{\text{A}}t_{\text{h}}$ for constant current, and 1 Ah = 3600 C. An ideal 20 Ah battery supplies 2 A for 10 hours. Actual usable time depends on discharge rate, temperature, battery condition and permitted depth of discharge. Energy also requires voltage: at approximately constant 12 V, 20 Ah corresponds to about 240 Wh.

For matched, compatible batteries:

• Series: Connect positive to negative. EMFs add algebraically and internal resistances add. Two 12 V, 20 Ah batteries give 24 V, 20 Ah; the same current passes through both.
• Parallel: Connect like polarities. Voltage remains 12 V while the two matched capacities add to 40 Ah. Identical internal resistances in parallel give half the individual resistance.

These are circuit models, not permission to parallel arbitrary batteries or regulated supplies. Follow the manufacturer’s compatibility, charging and protection instructions.

7. Practical Amateur Radio DC Applications

Ammeter Shunt Resistors

A standard moving-coil (d'Arsonval) galvanometer movement is inherently a current-sensitive device characterized by two values: its full-scale deflection current ($I_m$) and its internal coil resistance ($R_m$). Typical movements deflect fully at $1\text{ mA}$ or $50\ \mu\text{A}$.

To measure larger currents (e.g. $10\text{ A}$ power supply current), a precision low-resistance ammeter shunt ($R_{\text{shunt}}$) is connected directly in parallel with the movement:

$$R_{\text{shunt}} = \frac{I_m \cdot R_m}{I_{\text{total}} - I_m}$$

Worked Example 1.3: Sizing an Ammeter Shunt

An amateur wishes to construct a high-current station ammeter to monitor a $20.0\text{ A}$ linear amplifier power supply using a sensitive meter movement with $I_m = 1.0\text{ mA}$ ($0.001\text{ A}$) and internal coil resistance $R_m = 50.0\ \Omega$.

1. Calculate full-scale voltage drop across the meter movement:

$$V_m = I_m \cdot R_m = 0.001\text{ A} \times 50.0\ \Omega = 0.050\text{ V} = 50\text{ mV}$$

1. The shunt must carry all remaining current at full scale:

$$I_{\text{shunt}} = I_{\text{total}} - I_m = 20.0\text{ A} - 0.001\text{ A} = 19.999\text{ A}$$

1. Calculate required shunt resistance:

$$R_{\text{shunt}} = \frac{V_m}{I_{\text{shunt}}} = \frac{0.050\text{ V}}{19.999\text{ A}} \approx 0.0025001\ \Omega = 2.50\text{ m}\Omega$$

A heavy manganin wire or calibrated copper-strip shunt having a resistance of $2.50\text{ m}\Omega$ provides accurate full-scale measurement.

Voltmeter Multiplier Resistors

To convert the same movement into a voltmeter with full-scale range $V_{\text{fs}}$, a multiplier resistor ($R_{\text{mult}}$) is placed in series with the movement:

$$R_{\text{mult}} = \frac{V_{\text{fs}}}{I_m} - R_m$$

For a $30.0\text{ V}$ meter with $I_m = 1\text{ mA}$ and $R_m = 50\ \Omega$:

$$R_{\text{mult}} = \frac{30.0\text{ V}}{0.001\text{ A}} - 50\ \Omega = 30000\ \Omega - 50\ \Omega = 29950\ \Omega \approx 29.95\ \text{k}\Omega$$