Single Phase Inverter

Single Phase Inverter

Fig below shows a simple inverter circuit with two nos. of transistor acting as switches. And this type of inverter gives a square wave as ac output. If the desired frequency of ac output if ‘f’ Hz, then time period is T = 1/f Hz. The transistor Q1(S1) is switched ON when transistor Q2(S2) is switched off. From the figure below, we can observe that Q1 is in on position for initial T/2 second and Q2 is on position for next T/2 second.

The rms value of output voltage is given by:

$$Vo(rms) = \;\sqrt {\frac{1}{\pi }\int_0^\pi {\mathop {\left( {\frac{{Vs}}{2}} \right)}\nolimits^2 } dt} \; = \;\frac{{Vs}}{2}$$

The Fourier series of the output voltage Vo(t) can be obtained as follow:

From the waveform symmetry, it is clear that dc component is zero.

Ao = 0

$$\begin{array}{l} An = \frac{1}{\pi }\int\limits_0^\pi {\frac{{{\rm{Vs}}}}{2}} \;\;Cos(n\omega t)\;d(\omega t) - \frac{1}{\pi }\int\limits_\pi ^{2\pi } {\frac{{{\rm{Vs}}}}{2}} \;\;Cos(n\omega t)\;d(\omega t)\\ \;\;\;\;\; = \;0 \end{array}$$

$$\begin{array}{l} Bn\; = \;\frac{1}{\pi }\int\limits_0^\pi {\frac{{{\rm{Vs}}}}{2}} \;\;Sin(n\omega t)\;d(\omega t) - \frac{1}{\pi }\int\limits_\pi ^{2\pi } {\frac{{{\rm{Vs}}}}{2}} \;\;Sin(n\omega t)\;d(\omega t)\\ \;\;\;\;\; = \;\frac{{Vs}}{{2n\pi }}( - 2\cos (n\pi ) + Cos(0) + Cos(2n\pi ))\\ \;\;\;\;\; = \;\frac{4}{{n\pi }}\frac{{Vs}}{2}\;for\;all\;odd\;values\;of\;n\\ \;\;\;\;\; = \;0\;\;\;\;\;\;\;for\;all\;even\,values\;of\;n \end{array}$$