3. Current harmonic caused by DC-link voltage ripple

section. The lowest trace in Figure 4 sketches the waveform of each stage, and the details are

56 Power System Harmonics - Analysis, Effects and Mitigation Solutions for Power Quality Improvement

1. S1 is the error between the current reference and the output current of the inverter,

2: S2 is the amplitude modulation (AM) ratio, S2 ¼ S1GPI, where the PI controller's transfer function is GPI ¼ kp þ ki=s. kp and ki are the proportional and the integral gain. 3: S3 is the gate drive signal. S3 ¼ S2GPWM þ Vswitch harmonics, where GPWM ¼ 1=Cpk and

4: S4 is the output voltage of the inverter Vinv, S4 ¼ S3Ginv, where Ginv ¼ VDC. The VDC can be either a time-varying or a constant signal; these two cases need to be treated

contain the voltage harmonics Vg harmonics:S4 � Vg is the voltage difference between the output filter. The transfer function of the filter is Gf ¼ 1=ð Þ Ls , where L is the filter's

The main causes of harmonic in PV inverter can be summarized into several categories: grid background voltage distortion, switch harmonics (high frequency), DC-link voltage variation due to MPPT, and some other causes (PLL blocks, etc.). Harmonic distortion for both cases, with or without voltage ripple on the DC link, can be analyzed by using this general-

Gf . The grid voltage Vg may

described as follows:

separately.

inductance.

ized model.

S1 ¼ Iref � Iout ¼ Iref � S5.

Cpk is the carrier signal's peak value.

5: S5 is the output current of the inverter Iout. S5 ¼ S4 � Vg

Figure 4. Model of current-controlled PWM inverter with harmonic information.

In this section, the current harmonics caused by DC-link voltage ripple has been analyzed. The model for considering the double-line frequency voltage ripple has been built. The closed-form solution for the current harmonics has been provided.

Figure 5 shows the model of the inverter based on Figure 4, and the DC-link voltage ripple has been taken into account. The inverter transfer function Ginv shown in Figure 4 is replaced by the section under the triangle shading, which is a sinusoidal signal Vrip at double-line frequency on top of the DC component VDC. Since the voltage ripple is time-varying, the transfer function for this section cannot be derived. In [23], the authors point out that closed-form solutions cannot be derived when the harmonic ripple components are not neglected. However, numeric solutions can be evaluated for any particular operating condition by using this model.

The harmonic characteristics of the output current shown in Figure 5 can be identified by qualitatively analyzing the simplified loop model. The section under the triangle shading is also known as the amplitude modulation; the feedback loop with unit delay is shown in Figure 6, where Z�<sup>1</sup> denotes the delay of a unit sample period. Compared with Figure 6, in this simplified model, several linear blocks are left out. Due to the system linearity, the signal frequency characteristics will remain the same. A similar analysis method, which has been used in sound processing research [24], is adopted in this chapter to analyze this time-varying system. Two discrete-time sinusoidal example signals Iref ½ �¼ n cos ð Þ ωon and Vrip½ �¼ n cos 2ð Þ ωon are used. The output signal y n½ � can be illustrated as the result of subtraction between the reference signal cos ð Þ ωon and the delayed output signal y n½ � � 1 then timed with the AM section, which is cos 2ð Þþ ωon VDC

$$\begin{split} y[n] &= [\cos(\omega\_o n) - y[n-1]][\cos(2\omega\_o n) + V\_{D\mathbb{C}}] \\ &= \cos(\omega\_o n)[\cos(2\omega\_o n) + V\_{D\mathbb{C}}] - y[n-1][\cos(2\omega\_o n) + V\_{D\mathbb{C}}] \end{split} \tag{2}$$

For n ≤ 0, ω<sup>o</sup> is the angular velocity of a signal in the fundamental frequency, VDCis constant, at the initial condition, y n½ �¼ 0,. The delay exists at any point in time n, and we need to store y n½ � � 1 so that it can be used in the computation of y n½ �. The y n½ � � 1 is

Figure 5. Model of inverter with the DC-link voltage ripple.

$$\begin{split} y[n-1] &= \left[ \cos \left( \omega\_o [n-1] \right) - y[n-1-1] \right] \left[ \cos \left( 2 \omega\_o [n-1] \right) + V\_{D\mathbb{C}} \right] \\ &= \cos \left( \omega\_o [n-1] \right) \left[ \cos \left( 2 \omega\_o [n-1] \right) + V\_{D\mathbb{C}} \right] - y[n-2] \left[ \cos \left( 2 \omega\_o [n-1] \right) + V\_{D\mathbb{C}} \right] \end{split} \tag{3}$$

Substitute Eq. (3) into Eq. (2)

$$\begin{aligned} y[n] &= \cos\left(\omega\_o n\right)[\cos\left(2\omega\_o n\right) + V\_{D\subset}] \\ &- \cos\left(\omega\_o[n-1]\right)[\cos\left(2\omega\_o[n-1]\right) + V\_{D\subset}][\cos\left(2\omega\_o n\right) + V\_{D\subset}] \\ &+ y[n-2][\cos\left(2\omega\_o[n-1]\right) + V\_{D\subset}][\cos\left(2\omega\_o n\right) + V\_{D\subset}] \end{aligned} \tag{4}$$

obtained by using this assumption, and a partial linearization for the control loop can be achieved. The amplitude of the feedback signals can be set as variables. After equating like terms in the equation of the output current and in the preset amplitude of feedback signal, a number of equations can be obtained. By solving these equations, the closed-form solution for the amplitude of a certain order of harmonic can be derived. Detailed analysis of the simplified model by

The output signal y tð Þ of the simplified model is the difference between the current reference

IrefðÞ¼ <sup>t</sup> Acosð Þ¼ <sup>ω</sup>ot ae<sup>j</sup>ωot <sup>þ</sup> ae�jωot

VripðÞ¼ <sup>t</sup> Bcosð Þ¼ <sup>2</sup>ωot be<sup>j</sup>2ωot <sup>þ</sup> be�j2ω<sup>o</sup> <sup>t</sup>

where A and B are the amplitudes of Irefð Þt and Vripð Þt . In order to simplify the analysis to a level that is suitable for manual calculation, the feedback signal Ifbð Þt is assumed to include

where C<sup>1</sup> and C<sup>3</sup> are the assumed variables for the amplitude of the fundamental component and third harmonics, and c<sup>1</sup> ¼ 0:5C1, c<sup>3</sup> ¼ 0:5C3. This can be easily extended to any number of

Eq. (11) can be obtained by substituting Eqs. (8)–(10) into Eq. (7). Since theIfbð Þt is the low-order part of y tð Þ, the harmonic amplitude equation can be found by equating like terms in Eqs. (10)

c<sup>1</sup> ¼ ð Þ VDC þ b ð Þ� a � c<sup>1</sup> bc<sup>3</sup>

All the parameters are fixed values ða, b and VDCÞ for a specific inverter; therefore, the harmonic amplitude can be obtained by substituting these values. By using the same method, the closedform solution for the averaged model in Figure 5 with PI controller can be derived. Two integral sections will be involved into the calculation due to the integrator in the controller and the filter. This calculation for the practical models becomes significantly complicated, and it is impossible to calculate manually. Matlab can be utilized as an effective tool to conduct these calculations.

c<sup>3</sup> ¼ b að Þ� � c<sup>1</sup> VDCc<sup>3</sup>

þ½ � b að Þ� � <sup>c</sup><sup>1</sup> VDCc<sup>3</sup> <sup>e</sup><sup>j</sup>3ω<sup>o</sup> <sup>t</sup> <sup>þ</sup> ½ � b að Þ� � <sup>c</sup><sup>1</sup> VDCc<sup>3</sup> <sup>e</sup>�j3ωot � bc3e<sup>j</sup>5ωot � bc3e�j5ωot (11)

y tðÞ¼ ½ � ð Þ VDC <sup>þ</sup> <sup>b</sup> ð Þ� <sup>a</sup> � <sup>c</sup><sup>1</sup> bc<sup>3</sup> <sup>e</sup><sup>j</sup>ωot <sup>þ</sup> ½ � ð Þ VDC <sup>þ</sup> <sup>b</sup> ð Þ� <sup>a</sup> � <sup>c</sup><sup>1</sup> bc<sup>3</sup> <sup>e</sup>�jωot

(

<sup>j</sup>ω<sup>o</sup> <sup>t</sup> <sup>þ</sup> <sup>c</sup>1<sup>e</sup>

y tðÞ¼ Irefð Þ� <sup>t</sup> Ifbð Þ<sup>t</sup> � � Vripð Þþ <sup>t</sup> VDC � � (7)

Harmonic Distortion Caused by Single-Phase Grid-Connected PV Inverter

http://dx.doi.org/10.5772/intechopen.73030

59

, a <sup>¼</sup> <sup>1</sup> 2

> , b <sup>¼</sup> <sup>1</sup> 2

�jωot <sup>þ</sup> <sup>c</sup>3<sup>e</sup>

<sup>j</sup>3ωot <sup>þ</sup> <sup>c</sup>3<sup>e</sup>

A (8)

B (9)

�j3ω<sup>o</sup> <sup>t</sup> (10)

(12)

using this method in a time domain is given as follows:

The signals in Figure 6 can be expressed in polar form

IfbðÞ¼ t C<sup>1</sup> cos ð Þþ ωot C<sup>3</sup> cos 3ð Þ¼ ωot c1e

harmonics with the help of computer-aided calculations.

only fundamental and third harmonics

and (11)

Irefð Þt and the feedback signal Ifbð Þt multiplied by the DC-link voltage

This feedback expression can be expanded into an infinite summation of products given by

$$\begin{aligned} \mathbf{y}[\mathbf{n}] &= \cos(\omega\_{\rm o}\mathbf{n})[\cos(2\omega\_{\rm o}\mathbf{n}) + \mathbf{V}\_{\rm DC}] \\ &- \cos(\omega\_{\rm o}[\mathbf{n}-1])[\cos(2\omega\_{\rm o}[\mathbf{n}-1]) + \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}\mathbf{n}) + \mathbf{V}\_{\rm DC}] \\ &+ \cos(\omega\_{\rm o}[\mathbf{n}-2])[\cos(2\omega\_{\rm o}[\mathbf{n}-2]) + \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}[\mathbf{n}-1]) + \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}\mathbf{n}) + \mathbf{V}\_{\rm DC}] \\ &- \cos(\omega\_{\rm o}[\mathbf{n}-3])[\cos(2\omega\_{\rm o}[\mathbf{n}-3]) + \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}[\mathbf{n}-2]) + \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}[\mathbf{n}-1]) \\ &+ \mathbf{V}\_{\rm DC}][\cos(2\omega\_{\rm o}\mathbf{n}) + \mathbf{V}\_{\rm DC}] \end{aligned} \tag{5}$$

A series of odd harmonics is caused by this amplitude modulation in a feedback loop. In Eq. (5), the first product term is illustrated as an example. According to Euler's formula, this term can be expressed as the sum of sinusoids with angular velocity ω<sup>o</sup> and 3ωo, which is the fundamental and third harmonics.

$$y[n] = \cos\left(\omega\_o n\right)[\cos\left(2\omega\_o n\right) + V\_{\rm DC}] = \cos\left(\omega\_o n\right)V\_{\rm DC} + \frac{1}{2}\cos\left(3\omega\_o n\right) + \frac{1}{2}\cos\left(\omega\_o n\right) \tag{6}$$

The closed-form solution is derived based on an idea which is similar to the harmonic balance method for radio frequency (RF) circuit [25]. The harmonic balance method is a frequency domain method for calculating steady states of a nonlinear circuit.

All the signals in the control loop can be expressed in polar forms by taking a Fourier transform. The high-order harmonics will be attenuated by the feedback loop, and only the low-order harmonics will be considered. The low-pass filter (LPF) can be assumed as an ideal filter, which can eliminate all the harmonics above a certain order. A finite number of equations can be

Figure 6. Amplitude modulation in unit delay feedback.

obtained by using this assumption, and a partial linearization for the control loop can be achieved. The amplitude of the feedback signals can be set as variables. After equating like terms in the equation of the output current and in the preset amplitude of feedback signal, a number of equations can be obtained. By solving these equations, the closed-form solution for the amplitude of a certain order of harmonic can be derived. Detailed analysis of the simplified model by using this method in a time domain is given as follows:

The output signal y tð Þ of the simplified model is the difference between the current reference Irefð Þt and the feedback signal Ifbð Þt multiplied by the DC-link voltage

$$y(t) = \left(I\_{r\notin}(t) - I\_{\emptyset}(t)\right) \left(V\_{r\not\mid p}(t) + V\_{D\subset}\right) \tag{7}$$

The signals in Figure 6 can be expressed in polar form

y n½ �¼ � 1 ½ � cos ð Þ� ωo½ � n � 1 y n½ � � 1 � 1 ½ � cos 2ð Þþ ωo½ � n � 1 VDC

58 Power System Harmonics - Analysis, Effects and Mitigation Solutions for Power Quality Improvement

y n½ �¼ cos ð Þ ωon ½ � cos 2ð Þþ ωon VDC

y n½ �¼ cosð Þ ωon ½ � cos 2ð Þþ ωon VDC

fundamental and third harmonics.

Figure 6. Amplitude modulation in unit delay feedback.

Substitute Eq. (3) into Eq. (2)

¼ cos ð Þ ωo½ � n � 1 ½ cos 2ð Þþ ωo½ � n � 1 VDC� � y n½ � � 2 ½ � cos 2ð Þþ ωo½ � n � 1 VDC

� cos ð Þ ωo½ � n � 1 ½ � cos 2ð Þþ ωo½ � n � 1 VDC ½ � cos 2ð Þþ ωon VDC þy n½ � � 2 ½ � cos 2ð Þþ ωo½ � n � 1 VDC ½ � cos 2ð Þþ ωon VDC

This feedback expression can be expanded into an infinite summation of products given by

þ cosð Þ ωo½ � n � 2 ½ � cos 2ð Þþ ωo½ � n � 2 VDC ½ � cos 2ð Þþ ωo½ � n � 1 VDC ½ � cos 2ð Þþ ωon VDC

þ VDC�½ � cos 2ð Þþ ωon VDC (5)

1

<sup>2</sup> cos 3ð Þþ <sup>ω</sup>on

1

<sup>2</sup> cos ð Þ <sup>ω</sup>on (6)

� cosð Þ ωo½ � n � 3 ½ � cos 2ð Þþ ωo½ � n � 3 VDC ½ � cos 2ð Þþ ωo½ � n � 2 VDC ½cos 2ð Þ ωo½ � n � 1

A series of odd harmonics is caused by this amplitude modulation in a feedback loop. In Eq. (5), the first product term is illustrated as an example. According to Euler's formula, this term can be expressed as the sum of sinusoids with angular velocity ω<sup>o</sup> and 3ωo, which is the

The closed-form solution is derived based on an idea which is similar to the harmonic balance method for radio frequency (RF) circuit [25]. The harmonic balance method is a frequency

All the signals in the control loop can be expressed in polar forms by taking a Fourier transform. The high-order harmonics will be attenuated by the feedback loop, and only the low-order harmonics will be considered. The low-pass filter (LPF) can be assumed as an ideal filter, which can eliminate all the harmonics above a certain order. A finite number of equations can be

� cosð Þ ωo½ � n � 1 ½ � cos 2ð Þþ ωo½ � n � 1 VDC ½ � cos 2ð Þþ ωon VDC

y n½ �¼ cos ð Þ ωon ½ cos 2ð Þþ ωon VDC� ¼ cos ð Þ ωon VDC þ

domain method for calculating steady states of a nonlinear circuit.

(3)

(4)

$$I\_{\rm ref}(t) = A\cos(\omega\_o t) = ae^{j\omega\_o t} + ae^{-j\omega\_o t}, a = \frac{1}{2}A \tag{8}$$

$$V\_{rip}(t) = B\cos(2\omega\_o t) = be^{j2\omega\_o t} + be^{-j2\omega\_o t}, b = \frac{1}{2}B \tag{9}$$

where A and B are the amplitudes of Irefð Þt and Vripð Þt . In order to simplify the analysis to a level that is suitable for manual calculation, the feedback signal Ifbð Þt is assumed to include only fundamental and third harmonics

$$I\_{\mathfrak{H}}(t) = \mathbb{C}\_{1}\cos\left(\omega\_{\mathfrak{o}}t\right) + \mathbb{C}\_{3}\cos\left(3\omega\_{\mathfrak{o}}t\right) = \varepsilon\_{1}e^{j\omega\_{\mathfrak{o}}t} + \varepsilon\_{1}e^{-j\omega\_{\mathfrak{o}}t} + \varepsilon\_{3}e^{j3\omega\_{\mathfrak{o}}t} + \varepsilon\_{3}e^{-j3\omega\_{\mathfrak{o}}t} \tag{10}$$

where C<sup>1</sup> and C<sup>3</sup> are the assumed variables for the amplitude of the fundamental component and third harmonics, and c<sup>1</sup> ¼ 0:5C1, c<sup>3</sup> ¼ 0:5C3. This can be easily extended to any number of harmonics with the help of computer-aided calculations.

Eq. (11) can be obtained by substituting Eqs. (8)–(10) into Eq. (7). Since theIfbð Þt is the low-order part of y tð Þ, the harmonic amplitude equation can be found by equating like terms in Eqs. (10) and (11)

$$\begin{split} y(t) &= [(V\_{D\mathbb{C}} + b)(a - c\_1) - bc\_3]e^{j\omega\_s t} + [(V\_{D\mathbb{C}} + b)(a - c\_1) - bc\_3]e^{-j\omega\_s t} \\ &+ [b(a - c\_1) - V\_{D\mathbb{C}}c\_3]e^{j3\omega\_b t} + [b(a - c\_1) - V\_{D\mathbb{C}}c\_3]e^{-j3\omega\_b t} - bc\_3e^{j5\omega\_b t} - bc\_3e^{-j5\omega\_b t} \end{split} \tag{11}$$

$$\begin{cases} \mathbf{c}\_1 = (V\_{\rm DC} + b)(a - \mathbf{c}\_1) - b\mathbf{c}\_3 \\\\ \mathbf{c}\_3 = b(a - \mathbf{c}\_1) - V\_{\rm DC}\mathbf{c}\_3 \end{cases} \tag{12}$$

All the parameters are fixed values ða, b and VDCÞ for a specific inverter; therefore, the harmonic amplitude can be obtained by substituting these values. By using the same method, the closedform solution for the averaged model in Figure 5 with PI controller can be derived. Two integral sections will be involved into the calculation due to the integrator in the controller and the filter. This calculation for the practical models becomes significantly complicated, and it is impossible to calculate manually. Matlab can be utilized as an effective tool to conduct these calculations.
