Ideal Switch Cell : First Building Block

Understanding the Basic Switching Cell

Our goal is to study the basic switching cell and understand how it works and how its output voltage changes.

The switching cell has one pole and two throws. The switch repeatedly connects the pole to the input voltage for a certain amount of time and then to ground for the remaining time. This switching action repeats periodically.

The time for one complete switching cycle is called the switching period, \(T_s\). Therefore, the switching frequency is

fs=1Tsf_s=\frac{1}{T_s}

Switch Voltage

The voltage at the pole with respect to ground is \(v_s(t)\).

When the switch is in position 1, the pole is connected to the DC input voltage \(V_g\):

vs(t)=Vgv_s(t)=V_g

When the switch is in position 2, the pole is connected to ground:

vs(t)=0v_s(t)=0

In practice, the ideal switch is realized using power semiconductor devices, such as transistors and diodes, which are controlled to turn ON and OFF as required.

The switching frequency \(f_s\) is the inverse of the switching period \(T_s\):

fs=1Tsf_s=\frac{1}{T_s}

Duty Ratio

The duty ratio, \(D\), is the fraction of one switching period for which the switch remains in position 1.D=TonTsD=\frac{T_{\mathrm{on}}}{T_s}

where \(T_{\mathrm{on}}\) is the time for which the switch is in position 1.

Therefore,0≤D≤10\leq D\leq1

Switched Volatge

The switching action produces a rectangular output voltage waveform that repeats periodically at a very high frequency, typically from hundreds of kHz to MHz. Because the waveform is periodic, it can be represented using Fourier series.

From Fourier analysis, any periodic waveform can be represented as a combination of a DC component and an infinite number of sine and cosine harmonics.

The general Fourier series is

f(t)=F0+∑n=1∞(ancos⁡(nωt)+bnsin⁡(nωt))f(t)=F_0+\sum_{n=1}^{\infty}\left(a_n\cos(n\omega t)+b_n\sin(n\omega t)\right)

where \(F_0\) is the DC component, and \(a_n\) and \(b_n\) are the coefficients of the harmonic components.

DC Component

The DC component can be found from the average value of the waveform over one complete period \(T\):

F0=1T∫0Tf(t)dtF_0=\frac{1}{T}\int_0^T f(t) dt

Therefore, \(F_0\) represents the average, or DC, value of the periodic waveform.

For the switched voltage \(v_s(t)\), the DC component is

⟨vs⟩=1Ts∫0Tsvs(t)dt\langle v_s\rangle=\frac{1}{T_s}\int_0^{T_s}v_s(t)dt
⟨vs⟩=1Ts∫0Tsvs(t)dt\langle v_s\rangle=\frac{1}{T_s}\int_0^{T_s}v_s(t)dt
⟨vs⟩=1Ts[∫0DTsVindt+∫DTsTs0dt]\langle v_s\rangle= \frac{1}{T_s} \left[ \int_0^{DT_s}V_{in}dt + \int_{DT_s}^{T_s}0dt \right]

Evaluating the first integral:

⟨vs⟩=1Ts[Vint]0DTs\langle v_s\rangle= \frac{1}{T_s} \left[ V_{in}t \right]_0^{DT_s}
⟨vs⟩=1Ts[Vin(DTs)−Vin(0)]\langle v_s\rangle= \frac{1}{T_s} \left[ V_{in}(DT_s)-V_{in}(0) \right]

Since the second integral is zero:

⟨vs⟩=DTsVinTs\langle v_s\rangle= \frac{DT_sV_{in}}{T_s}

Canceling \(T_s\) gives:

⟨vs⟩=DVin\langle v_s\rangle=DV_{in}

Therefore, the DC component of the switched voltage is equal to the duty ratio multiplied by the input voltage.

Harmonic Components

The cosine and sine coefficients are calculated from

an=2T∫0Tf(t)cos⁡(nωt),dta_n=\frac{2}{T}\int_0^T f(t)\cos(n\omega t),dt

and

bn=2T∫0Tf(t)sin⁡(nωt),dtb_n=\frac{2}{T}\int_0^T f(t)\sin(n\omega t),dt

where \(n=1,2,3,\ldots\) represents the harmonic number.

The first harmonic corresponds to the fundamental frequency, while the higher values of \(n\) represent the higher-order harmonics.

Therefore, the switched voltage can be viewed as

DC component + fundamental component + higher-order harmonic components.

This Fourier representation helps us understand the frequency components produced by the switching action and how the switching waveform is built from its DC and harmonic components.

Harmonic Components for D = 0.5

For the first case, let \(D=0.5\), which corresponds to a 50% duty cycle. The switch is connected to \(V_{in}\) for half of the switching period and to ground for the other half.

The switching waveform is therefore

vs(t)={Vin,0≤t<Ts2 0,Ts2≤t<Tsv_s(t)= \begin{cases} V_{in}, & 0\leq t<\frac{T_s}{2}\ 0, & \frac{T_s}{2}\leq t<T_s \end{cases}

The switching angular frequency is

ωs=2πTs\omega_s=\frac{2\pi}{T_s}

The general trigonometric Fourier series is

vs(t)=F0+∑n=1∞[ancos⁡(nωst)+bnsin⁡(nωst)]v_s(t)=F_0+\sum_{n=1}^{\infty}\left[a_n\cos(n\omega_s t)+b_n\sin(n\omega_s t)\right]

1. DC Component

The DC component is the average value over one switching period:

F0=1Ts∫0Tsvs(t),dtF_0=\frac{1}{T_s}\int_0^{T_s}v_s(t),dt

Since the waveform has two different values during the two intervals, the integral can be divided into two parts:

F0=1Ts[∫0Ts/2Vin,dt+∫Ts/2Ts0,dt]F_0=\frac{1}{T_s} \left[ \int_0^{T_s/2}V_{in},dt+ \int_{T_s/2}^{T_s}0,dt \right]

Evaluating the first integral:

F0=1Ts[Vint]0Ts/2;F0=1Ts[VinTs2−Vin(0)]F_0=\frac{1}{T_s} \left[ V_{in}t \right]_0^{T_s/2} ; F_0=\frac{1}{T_s} \left[ V_{in}\frac{T_s}{2}-V_{in}(0) \right]

Therefore,

F0=Vin2F_0=\frac{V_{in}}{2}

So the DC component is

F0=Vin2F_0=\frac{V_{in}}{2}

2. Cosine Coefficients

The cosine coefficient is

an=2Ts∫0Tsvs(t)cos⁡(nωst),dta_n=\frac{2}{T_s}\int_0^{T_s}v_s(t)\cos(n\omega_s t),dt

Dividing the integral into the two switching intervals gives

an=2Ts[∫0Ts/2Vincos⁡(nωst),dt+∫Ts/2Ts0cos⁡(nωst),dt]a_n=\frac{2}{T_s} \left[ \int_0^{T_s/2}V_{in}\cos(n\omega_s t),dt+ \int_{T_s/2}^{T_s}0\cos(n\omega_s t),dt \right]

Therefore,

an=2VinTs∫0Ts/2cos⁡(nωst),dta_n=\frac{2V_{in}}{T_s} \int_0^{T_s/2}\cos(n\omega_s t),dt

Integrating,

an=2VinTs[sin⁡(nωst)nωs]0Ts/2a_n=\frac{2V_{in}}{T_s} \left[ \frac{\sin(n\omega_s t)}{n\omega_s} \right]_0^{T_s/2}

Since

ωsTs2=π\omega_s\frac{T_s}{2}=\pi

we obtain

an=2VinTssin⁡(nπ)nωsa_n=\frac{2V_{in}}{T_s} \frac{\sin(n\pi)}{n\omega_s}

For every integer \(n\),

sin⁡(nπ)=0\sin(n\pi)=0

Therefore,

an=0a_n=0

Thus, there are no cosine components.

3. Sine Coefficients

The sine coefficient is

bn=2Ts∫0Tsvs(t)sin⁡(nωst),dtb_n=\frac{2}{T_s}\int_0^{T_s}v_s(t)\sin(n\omega_s t),dt

Dividing the integral into the two switching intervals gives

bn=2Ts[∫0Ts/2Vinsin⁡(nωst),dt+∫Ts/2Ts0sin⁡(nωst),dt]b_n=\frac{2}{T_s} \left[ \int_0^{T_s/2}V_{in}\sin(n\omega_s t),dt+ \int_{T_s/2}^{T_s}0\sin(n\omega_s t),dt \right]

Therefore,

bn=2VinTs∫0Ts/2sin⁡(nωst),dtb_n=\frac{2V_{in}}{T_s} \int_0^{T_s/2}\sin(n\omega_s t),dt

Integrating,

bn=2VinTs[−cos⁡(nωst)nωs]0Ts/2b_n=\frac{2V_{in}}{T_s} \left[ -\frac{\cos(n\omega_s t)}{n\omega_s} \right]_0^{T_s/2}

Substituting the limits,

bn=2VinTs1−cos⁡(nωsTs/2)nωsb_n=\frac{2V_{in}}{T_s} \frac{1-\cos(n\omega_s T_s/2)}{n\omega_s}

Since

ωsTs2=π\omega_s\frac{T_s}{2}=\pi

we get

bn=2VinTs1−cos⁡(nπ)nωsb_n=\frac{2V_{in}}{T_s} \frac{1-\cos(n\pi)}{n\omega_s}

Using

cos⁡(nπ)=(−1)n\cos(n\pi)=(-1)^n

we obtain

bn=2VinTs1−(−1)nnωsb_n=\frac{2V_{in}}{T_s} \frac{1-(-1)^n}{n\omega_s}

For even harmonics, \(n\) is even and

(−1)n=1(-1)^n=1

Therefore,

bn=0b_n=0

For odd harmonics, \(n\) is odd and

(−1)n=−1(-1)^n=-1

Therefore,

bn=4VinnωsTsb_n=\frac{4V_{in}}{n\omega_sT_s}

Since

ωs=2πTs\omega_s=\frac{2\pi}{T_s}

we get

bn=4Vinn(2π)b_n=\frac{4V_{in}}{n(2\pi)}

Therefore,

bn=2Vinnπb_n=\frac{2V_{in}}{n\pi}

Hence,

bn={2Vinnπ,n odd 0,n evenb_n= \begin{cases} \dfrac{2V_{in}}{n\pi}, & n\text{ odd}\ 0, & n\text{ even} \end{cases}

4. Complete Trigonometric Fourier Series

We have

F0=Vin2F_0=\frac{V_{in}}{2}
an=0a_n=0

and

bn={2Vinnπ,n odd 0,n evenb_n= \begin{cases} \dfrac{2V_{in}}{n\pi}, & n\text{ odd}\ 0, & n\text{ even} \end{cases}

Substituting these into the Fourier series gives

vs(t)=Vin2+2Vinπ[sin⁡(ωst)+13sin⁡(3ωst)+15sin⁡(5ωst)+17sin⁡(7ωst)+⋯]v_s(t)=\frac{V_{in}}{2} +\frac{2V_{in}}{\pi} \left[ \sin(\omega_s t) +\frac{1}{3}\sin(3\omega_s t) +\frac{1}{5}\sin(5\omega_s t) +\frac{1}{7}\sin(7\omega_s t) +\cdots \right]

Therefore, the 50% duty-cycle switching waveform contains a DC component and only odd harmonics.

5. Amplitude-Phase Form

The amplitude-phase form of the Fourier series is

vs(t)=F0+∑n=1∞Fnsin⁡(nωst+θn)v_s(t)=F_0+\sum_{n=1}^{\infty}F_n\sin(n\omega_s t+\theta_n)

where

Fn=an2+bn2F_n=\sqrt{a_n^2+b_n^2}

and

θn=tan−1⁡(anbn)\theta_n=\tan^{-1}\left(\frac{a_n}{b_n}\right)

Since \(a_n=0\),

Fn=|bn|F_n=|b_n|

For odd harmonics,

Fn=2VinnπF_n=\frac{2V_{in}}{n\pi}

For even harmonics,

Fn=0F_n=0

Therefore,

Fn={2Vinnπ,n odd0,n evenF_n= \begin{cases} \dfrac{2V_{in}}{n\pi}, & n\text{ odd} \\ 0, & n\text{ even} \end{cases}

Since \(a_n=0\), the phase of the nonzero harmonics is

θn=0\theta_n=0

Therefore, the amplitude-phase Fourier series becomes

vs(t)=Vin2+2Vinπsin⁡(ωst)+2Vin3πsin⁡(3ωst)+2Vin5πsin⁡(5ωst)+2Vin7πsin⁡(7ωst)+⋯v_s(t)=\frac{V_{in}}{2} +\frac{2V_{in}}{\pi}\sin(\omega_s t) +\frac{2V_{in}}{3\pi}\sin(3\omega_s t) +\frac{2V_{in}}{5\pi}\sin(5\omega_s t) +\frac{2V_{in}}{7\pi}\sin(7\omega_s t) +\cdots

The general form can also be written as

vs(t)=Vin2+2Vinπ∑k=0∞12k+1sin⁡[(2k+1)ωst]v_s(t)=\frac{V_{in}}{2} +\frac{2V_{in}}{\pi} \sum_{k=0}^{\infty} \frac{1}{2k+1} \sin\left[(2k+1)\omega_s t\right]

Thus, for a 50% duty-cycle switching waveform, the DC component is \(V_{in}/2\), the even harmonics are zero, and only the odd harmonics are present. The amplitude of the odd harmonics decreases with \(1/n\).

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