--- title: "Need for excitation and improving dynamic response" subject: "Power Plant Equipment" chapter: 3 key: "semester_7/power_plant_equipment:excitation-dynamic-improvement" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/excitation-dynamic-improvement site: BCT Engineering --- # Need for excitation and improving dynamic response Power Plant Equipment · Chapter 3: Var/Voltage Control in Hydrogenerating Systems - Asked in: 2073 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/excitation-dynamic-improvement - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-3 ## Question Why do we need excitation system in power plants? How the dynamic response of excitation system can be improved? Explain with appropriate diagram and mathematical deduction. ## Short answer Excitation maintains terminal voltage and enhances transient stability. Dynamic response is improved by rate-feedback from field voltage to damp oscillations, feedback around the exciter to reduce its time constant, fast static exciters, and a power system stabilizer. ## Answer #### Need of excitation system A synchronous generator needs DC field current to build up its flux. The excitation system supplies this current and controls it so that: - terminal voltage stays constant from no load to full load; - reactive power is shared properly among parallel machines; - the field is "forced" quickly during faults to keep the machine in step (better transient stability); - field current stays inside the machine's capability limits. #### Basic loop and its limitation ```text Vref +--------+ +--------+ Efd +--------+ Vt -->(Σ)->| K_A |->| K_E |---->| K_G |--+--> ^- |1+sT_A | |1+sT_E | |1+sT_G | | | +--------+ +--------+ +--------+ | +-------------------------------------------+ ``` With loop gain $K = K_AK_EK_G$, the steady-state error for a step in $V_{ref}$ is $$ e_{ss} = \frac{1}{1+K} $$ Raising $K$ lowers the error but, with three lags in the loop, the response becomes oscillatory and finally unstable. The exciter time constant $T_E$ (about 0.5 to 1 s for a DC exciter) is the main cause of slow response. #### Methods to improve the dynamic response **1. Rate (derivative) feedback via a stabilizing transformer.** Feed back $H_F = \dfrac{sK_F}{1+sT_F}$ from $E_{fd}$ to the amplifier input. The closed loop becomes $$ \frac{V_t}{V_{ref}} = \frac{G_AG_EG_G}{1+G_AG_EH_F+G_AG_EG_G} $$ The extra term $G_AG_EH_F$ contains $s$, so it adds damping only during transients; at steady state ($s\to0$) it vanishes and the low error is kept. Hence high gain and good damping are obtained together. **2. Reducing the effective exciter time constant by feedback.** Put a proportional feedback $K_f$ around the exciter: $$ \begin{aligned} \frac{E_{fd}}{V_R} &= \frac{\dfrac{K_E}{1+sT_E}}{1+\dfrac{K_EK_f}{1+sT_E}} = \frac{K_E}{1+K_EK_f+sT_E} \\ &= \frac{K_E/(1+K_EK_f)}{1+s\,\dfrac{T_E}{1+K_EK_f}} \end{aligned} $$ The time constant falls from $T_E$ to $T_E/(1+K_EK_f)$: the exciter responds $(1+K_EK_f)$ times faster. The lost gain is made up in the amplifier. **3. Fast exciters.** Replace rotating DC exciters by static (thyristor) or brushless high-initial-response exciters with small time constants and high **ceiling voltage** (2 to 3 times rated), so field current rises quickly. **4. Power system stabilizer (PSS).** Adds a signal from rotor speed or electrical power through lead–lag blocks, giving positive damping to low-frequency rotor oscillations that a fast AVR may otherwise make worse. **5. Lead compensation** in the regulator to cancel the largest lag. Together these give a fast rise, small overshoot and small steady-state error. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is an excitation system needed in a power plant? A. To supply field current and keep terminal voltage constant and share reactive power B. To increase turbine speed C. To generate steam D. To control frequency Answer: A. The generator needs DC field current, which is controlled to regulate voltage and VAr. (Key point: Why needed) 2. What is the main cause of slow response in a DC excitation loop? A. The sensor time constant B. The exciter time constant T_E C. The amplifier gain D. The PSS Answer: B. T_E (about 0.5-1 s) is the largest lag. (Key point: Slow response cause) 3. With proportional feedback Kf around the exciter, the effective time constant becomes: A. T_E (1 + K_E K_f) B. T_E/(1 + K_E K_f) C. T_E K_E K_f D. T_E - K_f Answer: B. Efd/VR = KE/(1+KEKf+sTE), so the time constant is TE/(1+KEKf), making it faster. (Key point: Feedback around exciter) 4. What gives a quicker field current rise in modern exciters? A. Small time constant and high ceiling voltage (static or brushless) B. Large exciter inductance C. Lower ceiling voltage D. Slip rings Answer: A. Static or brushless exciters with small time constants and high ceiling (2-3 times rated) force field current quickly. 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