Let's derive the **harmonic profile of the primary phase current** of a **12-pulse rectifier**. We will: 1. **Understand the system** 2. **Write the AC line current in terms of switching events** 3. **Expand it into a Fourier series** 4. **Identify the harmonic order equation** 5. **State the final answer** --- ## 1. **Understanding the 12-pulse rectifier** A **12-pulse rectifier** typically comprises two 6-pulse rectifiers fed by transformers with a 30° phase shift between their secondaries (Y-Y and Y-Δ). The input (primary) current is the **sum** of the two rectifier input currents, phase-shifted by 30°. --- ## 2. **Expression for AC Line Current** For a single 6-pulse rectifier, the AC line current is a quasi-square waveform with six pulses per cycle. For a 12-pulse rectifier, the **primary line current** is the sum of two 6-pulse currents, one shifted by 30°. Let’s denote: - \( i_{6,1}(t) \): 6-pulse current from Y-Y transformer (reference phase) - \( i_{6,2}(t) \): 6-pulse current from Y-Δ transformer (shifted by 30°) The total primary current is: \[ i_{12}(t) = i_{6,1}(t) + i_{6,2}(t - T/12) \] where \( T \) is the fundamental period. --- ## 3. **Fourier Series for 6-pulse Rectifier Current** A 6-pulse rectifier’s line current contains odd harmonics of order: \[ h = 6n \pm 1, \quad n = 1, 2, 3, \ldots \] That is: 5th, 7th, 11th, 13th, 17th, 19th, etc. The general Fourier expansion for a square wave with period \( T \) is: \[ i_{6}(t) = \sum_{k=1}^{\infty} I_k \sin(k\omega t + \phi_k) \] where \( I_k \) is the amplitude of the \( k \)-th harmonic. For a 6-pulse rectifier, only harmonics for \( h = 6n \pm 1 \) are present. --- ## 4. **Harmonic Cancellation in 12-pulse Rectifier** When you sum two 6-pulse rectifier currents shifted by 30°, certain harmonics cancel. Recall for phase-shifted addition: \[ I_{12,h} = I_{6,h} \left[ 1 + e^{-j h \cdot 30^{\circ}} \right] \] For harmonics where \( h \) is a multiple of 12, the two components **add**; for other harmonics, **partial or complete cancellation** occurs. For a 12-pulse rectifier, the remaining harmonics are: \[ h = 12n \pm 1, \quad n = 1,2,3,\ldots \] That is: 11th, 13th, 23rd, 25th, etc. **Lower-order harmonics (5th, 7th, etc.) are eliminated.** --- ## 5. **Final Equation and Harmonic Profile** ### **General Equation** The primary phase current \( i_{12}(t) \) is: \[ i_{12}(t) = \sum_{n=1}^{\infty} I_{h} \sin(h \omega t + \phi_{h}) \] where - \( h = 12n \pm 1 \) - \( I_{h} \) is the amplitude (attenuated compared to 6-pulse; see below) - \( \phi_{h} \) is the phase of each harmonic ### **Harmonic Orders** \[ \boxed{ h = 12n \pm 1, \;\;\; n=1,2,3,\ldots } \] **(i.e., 11th, 13th, 23rd, 25th, ... harmonics)** ### **Harmonic Amplitude** For a rectifier with pulse number \( p \), the amplitude of the \( h \)-th harmonic is: \[ I_{h} = \frac{4 I_{dc}}{h \pi} \] where \( I_{dc} \) is the output DC current, and \( h = p n \pm 1 \). So, for a 12-pulse rectifier: \[ I_{h} = \frac{4 I_{dc}}{h \pi}, \quad h=12n \pm 1 \] --- ## **Summary Table** | Rectifier Type | Harmonic Orders in AC line current | |-----------------|--------------------------------------| | 6-pulse | \( h=6n\pm1 \) (5,7,11,13,...) | | 12-pulse | \( h=12n\pm1 \) (11,13,23,25,...) | --- ## **Final Answer (Boxed):** \[ \boxed{ i_{12}(t) = \sum_{n=1}^{\infty} \frac{4 I_{dc}}{(12n \pm 1)\pi} \sin\left[(12n \pm 1)\omega t + \phi_{12n\pm1}\right] } \] **where:** - \( h = 12n \pm 1 \) (i.e., 11, 13, 23, 25, ...) - \( I_{dc} \) is the DC current - \( \omega \) is the supply angular frequency **Thus, the harmonic profile of the primary phase current of a 12-pulse rectifier contains only harmonics of order \( 12n \pm 1 \) (11th, 13th, 23rd, 25th, ...), with amplitudes \( \frac{4I_{dc}}{h\pi} \).**