1.2 Power Rectification & Filtering Circuits

1.2 Power Rectification & Filtering Circuits

Aircraft electrical systems are among the most sophisticated power distribution networks in engineering. Modern commercial and military aircraft generate primary alternating current (AC) at 115 V AC, 400 Hz (or 230 V AC variable frequency on modern twin-aisle jets) via engine-driven Integrated Drive Generators (IDGs). However, the vast majority of critical avionics—including flight management computers, electronic flight instrument system (EFIS) displays, flight control actuators, cockpit voice recorders, and DC battery backup buses—require clean, highly regulated 28 V DC. The conversion of AC to DC is performed by Transformer Rectifier Units (TRUs) and discrete rectification circuits.

---

Half-Wave Rectification

The half-wave rectifier is the simplest AC-to-DC conversion circuit, consisting of a single semiconductor diode connected in series between the AC voltage source and the load resistor ($R_L$).

          +--------|>|--------+  
          |      Diode        |  
  AC o----+                   +----o + 
  Input   |                   |      V_out
  (Vm)    +------[ RL ]-------+----o - 

Operational Dynamics

• Positive Half-Cycle ($0$ to $\pi$ radians): The AC voltage makes the diode anode positive relative to its cathode. Once the input exceeds the forward threshold ($0.7\text{ V}$ for silicon), the diode turns on and conducts. The output voltage across $R_L$ tracks the input wave minus the diode drop: $v_L(t) = V_m \sin(\omega t) - V_D$.
• Negative Half-Cycle ($\pi$ to $2\pi$ radians): The AC voltage reverses, making the anode negative with respect to the cathode. The diode is reverse-biased, presenting an open circuit. No current flows, and output voltage is zero ($v_L(t) = 0$).
• Conduction Angle: The diode conducts for approximately $180^\circ$ of the complete $360^\circ$ AC cycle.

Mathematical Characteristics

1. Peak Output Voltage: $$V_{peak} = V_m - V_D$$
2. Average (DC) Output Voltage: $$V_{DC} = \frac{1}{2\pi} \int_0^\pi (V_m \sin\theta - V_D) d\theta = \frac{V_m - V_D}{\pi} \approx 0.318(V_m - V_D)$$
3. RMS Output Voltage: $$V_{RMS} = \frac{V_m}{2}$$
4. Ripple Frequency: The output pulses once per input cycle, so ripple frequency equals input frequency: $$f_{ripple} = f_{in}$$
5. Peak Inverse Voltage (PIV): During the negative half-cycle, the diode blocks the entire peak negative potential: $$\text{PIV} = V_m$$

Severe Aerospace Limitations

Half-wave rectifiers exhibit a high ripple factor ($\gamma \approx 1.21$ or $121\%$), a very low theoretical maximum rectification efficiency of only $40.6\%$, and draw unidirectional DC pulses from the transformer. This unidirectional current causes direct-current saturation of the transformer core, resulting in heavy core losses and overheating. Consequently, half-wave rectifiers are never utilized for primary aircraft DC power distribution.

---

Full-Wave Center-Tapped Rectification

The center-tapped full-wave rectifier uses a step-down transformer with a center-tapped secondary winding and two diodes ($D_1$ and $D_2$). The center tap serves as the common zero-volt ground reference for the DC output.

           +-------|>|-------+ (D1)
           |                 |
      +----+ Secondary (Vm)  +----+-----o + V_out
      |    |                 |    |
  AC  |    o CT (GND)        |   [RL]
 Input|    |                 |    |
      +----+ Secondary (Vm)  +----+-----o GND (0V)
           |                 |
           +-------|>|-------+ (D2)

Operational Dynamics

• Positive Half-Cycle: The top of the secondary winding is positive relative to the center tap, while the bottom is negative. Diode $D_1$ is forward-biased and conducts, delivering current through $R_L$ into the center tap. Diode $D_2$ is reverse-biased.
• Negative Half-Cycle: The AC polarity reverses: the top winding becomes negative and the bottom winding becomes positive relative to the center tap. Diode $D_2$ is forward-biased and conducts current through $R_L$ in the exact same direction as $D_1$ did. Diode $D_1$ is reverse-biased.
• Conduction: Each diode conducts for $180^\circ$, resulting in continuous $360^\circ$ conduction into the load.

Mathematical Characteristics

1. Average DC Output Voltage: $$V_{DC} = \frac{2(V_m - V_D)}{\pi} \approx 0.636(V_m - V_D)$$ (where $V_m$ is the peak AC voltage of half the secondary winding)
2. Ripple Frequency: The output pulses twice per AC cycle: $$f_{ripple} = 2 f_{in}$$
3. Peak Inverse Voltage (PIV) Stress: When $D_1$ conducts at peak positive voltage $+V_m$, the common cathode line sits at $(V_m - V_D)$. At that same instant, the anode of $D_2$ is connected to the bottom of the secondary winding at $-V_m$. Therefore, the total reverse voltage across $D_2$ is: $$\text{PIV} = V_m - (-V_m) - V_D = 2V_m - V_D \approx 2V_m$$

---

Full-Wave Bridge Rectification (Graetz Bridge)

The full-wave bridge rectifier is the premier rectification topology in aviation power conversion. It employs four diodes ($D_1, D_2, D_3, D_4$) arranged in a bridge diamond, requiring only a standard two-wire secondary transformer winding.

Loading diagram...

Operational Dynamics

• Positive Half-Cycle: The top terminal is positive with respect to the bottom terminal. Current flows through diode $D_1$, downward through load resistor $R_L$, returns through diode $D_2$, and completes the circuit to the transformer bottom terminal. Diodes $D_3$ and $D_4$ are reverse-biased.
• Negative Half-Cycle: The bottom terminal is positive with respect to the top terminal. Current flows through diode $D_3$, downward through load resistor $R_L$ (maintaining identical polarity), returns through diode $D_4$, and enters the transformer top terminal. Diodes $D_1$ and $D_2$ are reverse-biased.

Mathematical Characteristics

1. Peak Rectified Voltage: Because current flows through two forward-biased diodes in series during each half-cycle: $$V_{peak} = V_m - 2V_D = V_m - 1.4\text{ V (for silicon)}$$
2. Average DC Output Voltage: $$V_{DC} = \frac{2(V_m - 2V_D)}{\pi} \approx 0.636(V_m - 2V_D)$$
3. Ripple Frequency: $$f_{ripple} = 2 f_{in}$$
4. Peak Inverse Voltage (PIV): When $D_1$ and $D_2$ are conducting, the non-conducting diodes ($D_3$ and $D_4$) are clamped across the peak AC source minus one diode drop: $$\text{PIV} = V_m - V_D \approx V_m$$

Bridge Rectifier Engineering Advantages

• Half the PIV Rating: Diodes require only half the PIV rating ($V_m$) compared to the center-tapped topology ($2V_m$).
• No Center-Tap Required: Eliminates the bulk, weight, and copper losses of a center-tapped transformer.
• Zero DC Core Saturation: Symmetrical current draws during alternate half-cycles ensure net zero DC flux in the transformer core, preventing saturation.

---

The Critical 400 Hz Aircraft Advantage

Commercial and military aviation standardizes on 400 Hz AC power (governed by MIL-STD-704 and RTCA DO-160) rather than 50 Hz or 60 Hz ground utility power. The mathematical and physical justification is profound:

1. Transformer & Core Weight: By Faraday's Law of Electromagnetic Induction ($E = 4.44 f N \Phi_m$), the magnetic core cross-sectional area and volume required to support a given voltage are inversely proportional to frequency ($A_{core} \propto 1/f$). Operating at 400 Hz reduces transformer magnetic iron weight by approximately $70\%\text{ to } 80\%$ compared to 50/60 Hz transformers of equivalent kVA rating.
2. Output Ripple Frequency:
   • On $50\text{ Hz}$ mains, full-wave ripple frequency is $2 \times 50 = 100\text{ Hz}$ (period $T_r = 10.0\text{ ms}$).
   • On $60\text{ Hz}$ mains, full-wave ripple frequency is $2 \times 60 = 120\text{ Hz}$ (period $T_r = 8.33\text{ ms}$).
   • On $400\text{ Hz}$ aircraft AC, full-wave ripple frequency is $2 \times 400 = 800\text{ Hz}$ (period $T_r = 1.25\text{ ms}$).

---

Reservoir Capacitor Smoothing & Filter Networks

The raw output of a full-wave rectifier consists of pulsating DC that drops to zero volts twice per cycle. To provide smooth DC for avionics, a reservoir capacitor ($C$) is connected directly in parallel across the load resistor ($R_L$).

  Rectifier o-------+-------------+------o + DC Output
  Output            |             |
                   === C         [RL] Load
                    |             |
  GND       o-------+-------------+------o - GND

Smoothing Mechanism

1. Charging Cycle: When the rectified transformer voltage rises above the capacitor voltage, the diodes conduct, rapidly charging the capacitor to peak voltage $V_{peak} = V_m - 2V_D$ through the very low forward resistance of the diodes.
2. Discharging Cycle: As the transformer AC wave falls below the peak, the diodes become reverse-biased and turn off. The capacitor discharges its stored charge through $R_L$, sustaining the load voltage until the next AC pulse arrives.

Peak-to-Peak Ripple Voltage ($V_{r(p-p)}$)

Assuming the discharge is approximately linear across the short ripple period:

$$Q = I_{load} \cdot \Delta t = C \cdot V_{r(p-p)}$$

Since $\Delta t \approx \frac{1}{f_{ripple}}$:

$$V_{r(p-p)} = \frac{I_{load}}{f_{ripple} \cdot C} = \frac{V_{DC}}{f_{ripple} \cdot R_L \cdot C}$$

The resulting average DC output voltage is:

$$V_{DC} = V_{peak} - \frac{V_{r(p-p)}}{2}$$

The ripple factor ($\gamma$) is defined as the ratio of the RMS ripple voltage to the DC output voltage:

$$\gamma = \frac{V_{r(rms)}}{V_{DC}} = \frac{V_{r(p-p)}}{2\sqrt{3} \cdot V_{DC}} = \frac{1}{2\sqrt{3} \cdot f_{ripple} \cdot R_L \cdot C}$$

Filter Network Topologies & Bleeder Resistors

• RC Pi-Filter: Placing a series resistor $R$ between two shunt capacitors ($C_1$ and $C_2$) provides two stages of ripple attenuation. However, the series resistor causes a significant DC voltage drop ($I_{load} R$) and $I^2 R$ heat dissipation, making it suitable only for light, constant avionics loads.
• LC Choke-Input / Pi-Filter: Replaces the series resistor with an inductor (filter choke, $L$). The inductor presents high AC impedance ($X_L = 2\pi f L$) to the 800 Hz ripple while presenting virtually zero resistance ($R_{DC} \approx 0$) to direct current. At 400 Hz/800 Hz, a compact 10 mH choke provides substantial ripple rejection without wasting DC power.
• Bleeder Resistor ($R_B$): A high-value power resistor permanently connected across the filter capacitor terminals:
  1. Safety Discharge: Safely discharges the stored capacitive charge within a specified time (typically $< 5\text{ seconds}$ per EASA/FAA specifications) once aircraft power is removed, protecting technicians from electrical shock during maintenance.
  2. Voltage Regulation Stabilization: Draws a small, constant bleed current, preventing the output voltage from floating up to the peak AC voltage ($V_{peak}$) when the external avionics load is disconnected.

---

Comparison of Rectifier Topologies

---

Worked Numerical Calculation: Aircraft 400 Hz TRU Bridge Rectifier Sizing

An aircraft Transformer Rectifier Unit (TRU) stage is energized from a single-phase secondary winding delivering $115\text{ V RMS}$ at $400\text{ Hz}$. The circuit utilizes a full-wave silicon bridge rectifier ($V_D = 0.70\text{ V}$ per diode) supplying a continuous avionics load of $I_{load} = 2.50\text{ A}$.

Step 1: Calculate Peak AC Input Voltage ($V_m$)

$$V_m = V_{RMS} \cdot \sqrt{2} = 115\text{ V} \times 1.4142 = 162.63\text{ V}$$

Step 2: Determine Peak Rectified DC Voltage ($V_{peak}$)

Accounting for two conducting silicon diodes in series:

$$V_{peak} = V_m - 2V_D = 162.63\text{ V} - (2 \times 0.70\text{ V}) = 161.23\text{ V}$$

Step 3: Determine Output Ripple Frequency ($f_{ripple}$)

For a full-wave bridge rectifier on a 400 Hz aircraft supply:

$$f_{ripple} = 2 \cdot f_{in} = 2 \times 400\text{ Hz} = 800\text{ Hz}$$

Step 4: Size the Reservoir Capacitor ($C$) for $V_{r(p-p)} = 5.0\text{ V}$

The avionics load specification mandates that the peak-to-peak ripple voltage must not exceed $5.0\text{ V}$:

$$V_{r(p-p)} = \frac{I_{load}}{f_{ripple} \cdot C} \implies C = \frac{I_{load}}{f_{ripple} \cdot V_{r(p-p)}}$$ $$C = \frac{2.50\text{ A}}{800\text{ Hz} \times 5.0\text{ V}} = \frac{2.50}{4000} = 6.25 \times 10^{-4}\text{ F} = 625\ \mu\text{F}$$

Selection: An aerospace engineer selects a standard military-grade $680\ \mu\text{F}$ or $1,000\ \mu\text{F}$ capacitor.

Step 5: Calculate Average DC Output Voltage ($V_{DC}$)

$$V_{DC} = V_{peak} - \frac{V_{r(p-p)}}{2} = 161.23\text{ V} - \frac{5.0\text{ V}}{2} = 158.73\text{ V}$$

Step 6: Determine Minimum Diode PIV Rating

$$\text{PIV} = V_m - V_D = 162.63\text{ V} - 0.70\text{ V} = 161.93\text{ V}$$

Aerospace Safety Derating: Applying the standard 50% derating guideline for airborne systems, the technician ensures diodes with a minimum PIV rating of at least $400\text{ V}$ to $600\text{ V}$ are installed.