Power electronics digital notes b. Tech III year

 
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load 
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and 
conventions. 
1) The current entering node a is considered to be positive. 
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction. 
3) The current through S1 is denoted as i1 and the current through S2 is i2. 
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time 
duration
0 ≤ t ≤ T1 and the switch S2 is on for the time duration T1 ≤ t ≤ T2. When switch S1 is turned 
on, the instantaneous voltage across the load is
ν o
= Vin/ 2 
When the switch S2 is only turned on, the voltage across the load is 
ν o =
Vin/ 2. 

---

146 | 
P a g e
 
Figure: 5.2 Single phase Half Bridge DC-AC inverter output waveforms 
The r.m.s value of output voltage ν o is given by,
 
The instantaneous output voltage ν o is rectangular in shape. The instantaneous value of ν o can be 
expressed in Fourier series as, 
 
Due to the quarter wave symmetry along the time axis , the values of a0 and an are zero. The value of bn 
is given by, 
 
Substituting the value of bn from above equation , we get 

---

147 | 
P a g e
The current through the resistor ( iL ) is given by, 
 
Half Bridge DC-AC Inverter with L Load and R-L Load
The DC-AC converter with inductive load is shown in Figure below. For an inductive load, the load 
current cannot change immediately with the output voltage. 
 
Figure: 5.3 Single phase Half Bridge DC-AC inverter with RL load 
The 
working 
of 
the 
DC-AC 
inverter 
with 
inductive 
load 
is 
as 
follow 
is: 
Case 1: In the time interval 0<=t<= T1 the switch S1 is on and the current flows through the inductor 
from points a to b. When the switch S1 is turned off (case 1) at t-T1, the load current would continue to 
flow through the capacitor C2 and diode D2 until the current falls to zero, as shown in Figure below.