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Need for excitation and improving dynamic response

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.

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.

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

Vref    +--------+  +--------+ Efd +--------+  Vt
 -->(Σ)->|  K_A   |->|  K_E   |---->|  K_G   |--+-->
    ^-   |1+sT_A  |  |1+sT_E  |     |1+sT_G  |  |
    |    +--------+  +--------+     +--------+  |
    +-------------------------------------------+

With loop gain K=KAKEKGK = K_AK_EK_G, the steady-state error for a step in VrefV_{ref} is

ess=11+Ke_{ss} = \frac{1}{1+K}

Raising KK lowers the error but, with three lags in the loop, the response becomes oscillatory and finally unstable. The exciter time constant TET_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 HF=sKF1+sTFH_F = \dfrac{sK_F}{1+sT_F} from EfdE_{fd} to the amplifier input. The closed loop becomes

VtVref=GAGEGG1+GAGEHF+GAGEGG\frac{V_t}{V_{ref}} = \frac{G_AG_EG_G}{1+G_AG_EH_F+G_AG_EG_G}

The extra term GAGEHFG_AG_EH_F contains ss, so it adds damping only during transients; at steady state (s→0s\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 KfK_f around the exciter:

EfdVR=KE1+sTE1+KEKf1+sTE=KE1+KEKf+sTE=KE/(1+KEKf)1+s TE1+KEKf\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 TET_E to TE/(1+KEKf)T_E/(1+K_EK_f): the exciter responds (1+KEKf)(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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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