6.3 Power Factor, Transformer Ratios & Resistive Voltage Dividers

6.3 Power Factor, Transformer Ratios & Resistive Voltage Dividers

In pure direct current (DC) circuits, calculating electrical power is straightforward: electrical power in watts is simply the product of voltage and current ($P = V \cdot I$). In alternating current (AC) and radio frequency (RF) circuits, however, the presence of inductors and capacitors creates phase shifts between voltage and current. These phase displacements cause a portion of the supplied energy to oscillate back and forth between the source and reactive magnetic/electric fields without performing real work, introducing the concepts of Real Power, Reactive Power, Apparent Power, and Power Factor ($\text{PF}$).

Additionally, amateur operators frequently configure resistive networks for bias supplies and attenuators using voltage dividers, and utilize broadband transformers and baluns to match feedline impedances ($50,\Omega$) to varied antenna loads ($200,\Omega$, $450,\Omega$, $800,\Omega$). This section presents the theoretical equations and practical calculations governing power factor, resistive dividers, and transformer impedance transformation.

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1. AC Power Concepts: Real, Reactive & Apparent Power

When AC voltage and AC current are not in phase ($\theta \neq 0^\circ$), the total power supplied by the source separates into three distinct mathematical quantities forming the AC Power Triangle:

                               + (Apparent Power: S, in VA)
                              /|
                             / |
                            /  | (Reactive Power: Q, in VAR)
                           /   | Q = I² * X = V * I * sin(θ)
                          /    |
                         / θ   |
     (Reference) 0 -----+------+---> (Real Power: P, in Watts)
                                     P = I² * R = V * I * cos(θ)

1. Real Power (True Power, $P$)

• Definition: The actual power dissipated in resistive elements and converted into heat, light, mechanical motion, or radiated as electromagnetic RF waves.
• Unit: Watts (W).
• Formula: $$P = I_{\text{RMS}}^2 \cdot R = V_{\text{RMS}} \cdot I_{\text{RMS}} \cdot \cos(\theta)$$

2. Reactive Power (Imaginary / Wattless Power, $Q$)

• Definition: The power that oscillates back and forth between the generator and the reactive components (stored in the magnetic field of coils and electric field of capacitors). It performs zero net physical work over a full AC cycle.
• Unit: Volt-Amperes Reactive (VAR).
• Formula: $$Q = I_{\text{RMS}}^2 \cdot X = V_{\text{RMS}} \cdot I_{\text{RMS}} \cdot \sin(\theta)$$

3. Apparent Power ($S$)

• Definition: The simple product of measured RMS voltage and RMS current supplied by the power source, without regard to phase angle. It represents the total capacity that power lines, transformers, and generators must handle.
• Unit: Volt-Amperes (VA).
• Formula: $$S = V_{\text{RMS}} \cdot I_{\text{RMS}} = \sqrt{P^2 + Q^2}$$

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2. Power Factor ($\text{PF}$) & Circuit Efficiency

The Power Factor ($\text{PF}$) is the ratio of Real Power dissipated to Apparent Power supplied:

$$\text{PF} = \frac{\text{Real Power (Watts)}}{\text{Apparent Power (VA)}} = \frac{P}{S} = \cos(\theta)$$

For any series AC circuit with resistance $R$ and total impedance $Z$:

$$\text{PF} = \frac{R}{Z}$$

+-----------------------------------------------------------------------------------------+
|                              POWER FACTOR OPERATIONAL MATRIX                            |
|                                                                                         |
| Circuit Condition    Phase Angle (θ)  Power Factor (PF)  Real vs. Apparent Power        |
| --------------------------------------------------------------------------------------- |
| Purely Resistive     θ = 0°           PF = 1.0 (Unity)   Real Power = Apparent Power    |
| (or at Resonance)                                        (All current performs work)    |
| Purely Inductive     θ = +90°         PF = 0.0           Real Power = 0 Watts           |
|                                                          (100% reactive energy storage) |
| Purely Capacitive    θ = -90°         PF = 0.0           Real Power = 0 Watts           |
|                                                          (100% reactive energy storage) |
| Typical Mixed AC     0° < |θ| < 90°   0.0 < PF < 1.0     Real Power < Apparent Power    |
+-----------------------------------------------------------------------------------------+

Operational Significance of Power Factor

When power factor is low (e.g., $\text{PF} = 0.5$), the power source must deliver twice as much RMS current to supply the same amount of real power to the load ($I = P / (V \cdot \text{PF})$). This excess circulating current causes excessive $I^2 R$ heat loss in station wiring, power cords, and supply transformers without delivering useful power.

Step-by-Step Worked Example: Power Factor Calculation

Problem: An AC linear power supply connected to a $120\text{ V AC}$ mains line draws $5.0\text{ Amperes RMS}$. An active wattmeter measures a true real power consumption of $480\text{ Watts}$. What is the apparent power and power factor?

1. Calculate Apparent Power ($S$): $$S = V_{\text{RMS}} \times I_{\text{RMS}} = 120\text{ V} \times 5.0\text{ A} = \mathbf{600\text{ VA}}$$
2. Calculate Power Factor ($\text{PF}$): $$\text{PF} = \frac{\text{Real Power}}{\text{Apparent Power}} = \frac{480\text{ W}}{600\text{ VA}} = \mathbf{0.80} \quad (80\%)$$
3. Calculate Phase Angle ($\theta$): $$\theta = \arccos(0.80) \approx 36.87^\circ$$

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3. Resistor Networks & Voltage Dividers

Resistor Combination Fundamentals

• Series Resistors: Total resistance is the direct algebraic sum. Current is identical through each resistor. $$R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$$
• Parallel Resistors: Total resistance is always less than the smallest individual branch resistance. Voltage is identical across each branch. $$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$$
  • Two Resistors in Parallel: $R_{\text{total}} = \frac{R_1 \cdot R_2}{R_1 + R_2}$
  • $N$ Identical Resistors in Parallel: $R_{\text{total}} = \frac{R}{n}$ (e.g., four $100,\Omega$ resistors in parallel equal $25,\Omega$; two $100,\Omega$ resistors in parallel equal $50,\Omega$).

                    +Vin
                      |
                      +---[ R1 ]---+
                                   |
                                   +--------> +Vout
                                   |
                                 [ R2 ]
                                   |
                                  GND

The Resistive Voltage Divider Formula

A voltage divider uses two series resistors to produce an output voltage ($V_{\text{out}}$) that is a predetermined fraction of the input voltage ($V_{\text{in}}$):

$$V_{\text{out}} = V_{\text{in}} \left(\frac{R_2}{R_1 + R_2}\right)$$

Worked Example: Voltage Divider

Problem: A $24\text{ V DC}$ power source is applied across a voltage divider consisting of $R_1 = 90,\Omega$ and $R_2 = 30,\Omega$. What is the unloaded output voltage measured across $R_2$?

$$V_{\text{out}} = 24\text{ V} \times \left(\frac{30\,\Omega}{90\,\Omega + 30\,\Omega}\right) = 24\text{ V} \times \left(\frac{30}{120}\right) = 24\text{ V} \times 0.25 = \mathbf{6.0\text{ Volts}}$$

[!WARNING] The Voltage Divider Loading Effect: The standard divider formula assumes no current is drawn by an external load (unloaded condition). Connecting an external load resistor ($R_L$) in parallel with $R_2$ reduces the effective lower resistance to $R_2' = (R_2 \cdot R_L) / (R_2 + R_L)$, causing $V_{\text{out}}$ to drop below the calculated unloaded value.

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4. Ideal Transformers & Impedance Transformation

A transformer consists of two or more electrically isolated coils of wire wound around a common magnetic core (typically laminated iron for audio/power frequencies or ferrite/powdered iron for RF). By mutual electromagnetic induction, alternating current in the primary winding ($N_p$) creates an oscillating magnetic flux that induces an AC voltage across the secondary winding ($N_s$).

       Primary Winding (Np)            Magnetic Core          Secondary Winding (Ns)
            +---*                                                  *---+
            |   |                                                  |   |
   Vp, Ip   |  ( )                                                ( )  |  Vs, Is
   (Zp)     |  ( )================================================( )  |  (Zs)
            |  ( )                                                ( )  |  Load
            |   |                                                  |   |
            +---+                                                  +---+

Fundamental Transformer Ratio Relationships

In an ideal lossless transformer ($100%$ efficiency, where Real Power in = Real Power out, $V_p \cdot I_p = V_s \cdot I_s$):

+-----------------------------------------------------------------------------------------+
|                          IDEAL TRANSFORMER FORMULA MATRIX                               |
|                                                                                         |
| Parameter              Formula                                  Description             |
| --------------------------------------------------------------------------------------- |
| Voltage Ratio          V_p / V_s = N_p / N_s                    Directly proportional   |
|                        V_s = V_p * (N_s / N_p)                  to turns ratio          |
| --------------------------------------------------------------------------------------- |
| Current Ratio          I_p / I_s = N_s / N_p                    Inversely proportional  |
|                        I_s = I_p * (N_p / N_s)                  to turns ratio          |
| --------------------------------------------------------------------------------------- |
| Impedance Ratio        Z_p / Z_s = (N_p / N_s)²                 Proportional to SQUARE  |
|                        N_p / N_s = √(Z_p / Z_s)                 of turns ratio          |
+-----------------------------------------------------------------------------------------+

Why Impedance Transforms as the Square of the Turns Ratio

Impedance is the ratio of voltage to current ($Z = V / I$). When a transformer steps voltage up by a factor of $n = N_s / N_p$, it simultaneously steps current down by the exact same factor $n$:

$$Z_s = \frac{V_s}{I_s} = \frac{V_p \cdot n}{I_p / n} = \left(\frac{V_p}{I_p}\right) \cdot n^2 = Z_p \cdot \left(\frac{N_s}{N_p}\right)^2$$ $$\frac{Z_p}{Z_s} = \left(\frac{N_p}{N_s}\right)^2 \iff \frac{N_p}{N_s} = \sqrt{\frac{Z_p}{Z_s}}$$

Step-Up vs. Step-Down Transformers

• Step-Up Transformer ($N_s > N_p$): Secondary voltage is higher than primary voltage, secondary current is lower than primary current, and load impedance is transformed to a lower value looking back into the primary.
• Step-Down Transformer ($N_s < N_p$): Secondary voltage is lower than primary voltage, secondary current is higher than primary current, and load impedance is transformed to a higher value looking back into the primary.

RF Impedance Transformation in Amateur Radio

Amateur radio operators use RF transformers, ununs (unbalanced-to-unbalanced), and baluns (balanced-to-unbalanced) to match standard $50,\Omega$ transceivers to various antenna systems:

Step-by-Step Worked Transformer Problems

Problem 1: Power Supply Step-Down Transformer

• An ideal transformer connected to $120\text{ V AC}$ primary delivers $12\text{ V AC}$ to a secondary rectifier circuit.
• What is the required turns ratio? $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{120}{12} = \mathbf{10:1}$.
• If the secondary load draws $10\text{ Amperes}$, what is the primary current? $I_p = I_s \times (N_s / N_p) = 10\text{ A} \times (1/10) = \mathbf{1.0\text{ Ampere}}$.

Problem 2: Matching a $50,\Omega$ Transceiver to an $800,\Omega$ Beverage Antenna

• An operator must match an $800,\Omega$ beverage antenna wire to a $50,\Omega$ coaxial feedline.
• Calculate the impedance transformation ratio: $\frac{Z_p}{Z_s} = \frac{50,\Omega}{800,\Omega} = \frac{1}{16}$.
• Calculate the required primary-to-secondary turns ratio: $$\frac{N_p}{N_s} = \sqrt{\frac{Z_p}{Z_s}} = \sqrt{\frac{50}{800}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = \mathbf{1:4}$$ The secondary winding connected to the antenna must have $4$ times as many turns as the primary winding connected to the coaxial feedline.