Single Phase Full Bridge Inverter: Circuit, operation and waveforms

In this article, we will discuss the basics of a Single Phase Full Bridge Inverter such as its working using diagram, waveforms for various loads (R, RL, and RLC ) and in the last the mathematical analysis using the Fourier series. The diagram of a typical single phase full bridge inverter is given below:

Diode $D_1, D_2, D_3$ and $D_4$ are called Feedback Diodes and they functions only when the load is other than Resistive Load.

These Diodes are called Feedback diodes since they conduct only when the Power flow is negative means Power is being fed back to the DC source, when these diodes conduct.

For a Full Bridge Inverter, when $T_1, T_2$ conduct, Load voltage is $V_s$ and when $T_3, T_4$ conduct load voltage is $-V_s$ . Frequency of the output voltage can be controlled by varying the periodic time $T$ . The waveform of the output voltage for a single phase full bridge inverter is given below:

Steady State Analysis

Resistive Load

$v_o$ and $i_o \longrightarrow$ Square Wave

Each Switching device conducts for $180^{\circ}$ or $\left( \dfrac {T}{2} \right)$

Each feedback diode conducts for $0^{\circ}$

Pure Inductive Load

$v_o \longrightarrow$ Square Wave

$i_o \longrightarrow$ Triangular Wave

$\displaystyle i_o = \frac {1}{L} \; \int_{}^{} \; V_s \; \text {dt} = \left( \frac {V_s}{L} \right) \; t$

Each Switching device conducts for $90^{\circ}$ or $\biggr( \frac {T}{4} \biggr)$

Each feedback diode conducts for $90^{\circ}$ or $\left( \dfrac {T}{4} \right)$

$\displaystyle \boxed {\text {Peak value of Inductor Current} = \left( \dfrac {V_s}{L} \right) \; \left( \dfrac {T}{4} \right)}$

RL load

$v_o \longrightarrow$ Square Wave

$i_o \longrightarrow$ Exponential

RLC Load

When $\displaystyle (X_{L} > X_{C})$

Overdamped Response $(\zeta > 1)$

$v_o \longrightarrow$ Square Wave

$i_o \longrightarrow$ Sinusoidal

Characteristcs equation of an 2nd order series RLC System

$\displaystyle \Rightarrow s^2 + \biggr( \frac {R}{L} \biggr) s + \frac {1}{LC} = 0$

$\displaystyle \Rightarrow \omega_n = \frac {1}{\sqrt{LC}}$

$\displaystyle \Longrightarrow \zeta = \left( \dfrac {R}{2} \right) \; \sqrt{\dfrac {C}{L}}$

$\displaystyle \Longrightarrow \boxed { R > 2 \sqrt {\dfrac {L}{C}}}$

Forced Commutation is required because to turn OFF the switches $S_1, S_2$ at $\dfrac {T}{2}$ , anode current should comes to zero – Due to load, it is not possible because current becomes zero after the voltage becomes zero.

When $\displaystyle (X_{L} > X_{C})$

Underdamped Response $(\zeta < 1)$

$\displaystyle \Longrightarrow \zeta = \left( \dfrac {R}{2} \right) \; \sqrt{\frac {C}{L}}$

$\displaystyle \boxed { \Longrightarrow R < 2 \sqrt {\dfrac {L}{C}}}$

Forced Commutation is not required because to turn OFF the switches $S_1, S_2$ at $\left( \dfrac {T}{2} \right)$ , anode current should comes to zero – Due to load it is possible because current becomes zero before the voltage becomes zero.

Load Commutation Occurs due to Underdamped Response

$\displaystyle V_{0(\text {rms})} = \left[ \dfrac {1}{T}\; \left( \int_{0}^{\frac {T}{2}} \; V^2_s \; \text {dt} + \int_{\dfrac {T}{2}}^{T} \; V^2_s \; \text {dt} \right) \right]^{\dfrac {1}{2}}$

$\displaystyle V_{0(\text {rms})} = V_s$

$\displaystyle V_{0(\text {rms})} = \left[ \dfrac {1}{T} \; \left( V^2_{s} \; \times \dfrac {T}{2} + V^2_{s} \; \times \dfrac {T}{2} \right) \right]^{\dfrac {1}{2}} = V_s$

Output Voltage waveform is Half Wave Symmetric hence all even harmonics are absent

$\displaystyle V_{0(\text {avg})} = 0$

⇒ $\displaystyle \boxed { V_{0} = \sum_{n = 1,3,5}^{\infty} \; \left( \dfrac {4V_{s}}{n \pi} \right) \; \sin(n \omega t)}$

⇒ $\displaystyle i_{0} = \sum_{n = 1,3,5}^{\infty} \; \left( \dfrac {4V_{s}}{n \pi Z_{n}} \right) \; \sin(n \omega t - \phi_{n})$

$\displaystyle Z_{n} = \left[ R^2 + \left( n \omega L - \dfrac {1}{n \omega C} \right)^2 \right]^{\dfrac {1}{2}}$

6 Comments

hi there,
the article is beautiful. i want to know these statements:
1. Due to load it is not possible because current becomes zero after the voltage becomes zero.
2. Due to load it is possible because current becomes zero before the voltage becomes zero.
my question is how we can arrive at this conclusion from damping ratio. i want to understand the concept. for your kind information, i have studied rlc series ckt.

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Ashok Saini says:

19 Apr 2020 at 8:14 PM

Hello Gourab

Thanks for the compliment, Nice to hear from you.

For your answer please check the below link –

https://electricalbaba.com/load-commutation/

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1. GOURAB DUTTA ROY says:
  
  19 Apr 2020 at 10:57 PM
  
  tanks for your reply
  but my question is different. i want to know:
  how can you say for overdamped system, “current becomes zero after the voltage becomes zero.”
  i specificall want to know this………….
  
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  Reply