Power Electronics - Philadelphia University...Power Electronics Inverters 1 Dr. Firas Obeidat 2...

Power Electronics Inverters

1

Dr. Firas Obeidat

2

Table of contents

1 • Single Phase Half Bridge Inverter – Resistive Load

2 • Single Phase Half Bridge Inverter – RL Load

5 • Single Phase Full Bridge Inverter

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

Single Phase Half Bridge Inverter – Resistive Load

Basic Operation

Consists of 2 choppers, 3-wire DC source.

Transistors switched ON and OFF alternately.

Each provides opposite polarity of Vs/2 across

the load.

When T1 is ON through the period 0<t<T/2,

the output voltage equal to Vs/2.

When T2 is ON through the period T/2<t<T,

the output voltage equal to -Vs/2.

2

sV

SVR

Vo

C1

C2

io i1

i2

T1D1

T2

D22

sV

ao

2

sV

SVR

Vo

C1

C2

io i1

i2

T1D1

T2

D22

sV

ao

T1 ON, T2 OFF, Vo = Vs/2 T1 OFF, T2 ON, Vo = -Vs/2

2

sV

SVR

Vo

C1

C2

io i1

i2

T1D1

T2

D22

sV

ao


Single Phase Half Bridge Inverter – Resistive Load

The frequency can be changed by controlling the conduction time of the

transistors.

The rms value for the output

voltage can be found as

𝑉𝑜,𝑟𝑚𝑠 =2

𝑇

𝑉𝑠2

2

𝑑𝑡𝑇 2

0

=𝑉𝑠2

When T1 is ON through the

period 0<t<T/2, the output

current equal to Vs/2R.

When T2 is ON through the

period T/2<t<T, the output

current equal to -Vs/2R.

The output voltage frequency is

𝑓𝑜 =1

𝑇

2

sV

2

sV

R

V s

2

R

V s

2

t

t

t

Vo

i2

i1

R

Vs

2

t

io

R

Vs

2

2

T T

2

T T

2

T T

2

T T


Single Phase Half Bridge Inverter – RL Load

At t=0, the control signal is removed from T2

and a control signal is applied to T1. At this

moment the current is negative maximum.

The current can’t change the direction

directly to positive value due to the inductive

load. In this case, the current will flow from

the load through D1 to the source and T1 stay

disconnected in spite of existence the control

signal on it due to reverse biased. At t=t1 the

current become zero and T1 start to be

forward biased.

The operation of single phase half bridge

inverter with RL load can be divided into

four periods

2

sV

SVR

Vo

C1

C2

io i1

i2

T1D1

T2

D22

sV

ao L

0<t<t1

2

sV

R

Vo

C1

io i1

T1D1

ao L



At t=t1, the current change its direction to be

positive and T1 start to conduct. The positive

current is increased until reach its positive

maximum value at t=T/2. At t=T/2 the control

signal is removed from T1 and applied to T2.

t1<t<T/2

2

sV

R

C1

io i1

T1D1

ao L

T/2<t<t2

Vo

C2

i2

T2

D22

sV

ao

At this moment the current is positive

maximum. The current can’t change the

direction directly to negative value due to the

inductive load. In this case, the current will

flow from the load through D2 to the source

and T2 stay disconnected in spite of existence

the control signal on it due to reverse biased.

At t=t2 the current become zero and T2 start

to be forward biased. At this period the

voltage become negative and the current

positive.



At t=t2, the current change its direction to be

negative and T2 start to conduct. The negative

current is increased until reach its negative

maximum value at t=T. At t=T the control

signal is removed from T2 and applied to T1.

t2<t<T

Vo

C2

i2

T2

D22

sV

ao

The rms value for the output

voltage can be found as

𝑉𝑜,𝑟𝑚𝑠 =2

𝑇

𝑉𝑠2

2

𝑑𝑡𝑇 2

0

=𝑉𝑠2

2

sV

2

sV 2

T Tt

t

Vo

io

t1 t2

D1

ON

T1

ON

D2

ON

T2

ON

D1

ON



0<t<T/2

The output current can be found as

𝑉𝑠2= 𝑅𝑖𝑜 𝑡 + 𝐿

𝑑𝑖𝑜 𝑡

𝑑𝑡

𝑖𝑜 𝑡 =𝑉𝑠2𝑅

1 − 𝑒−𝑅𝐿𝑡 + 𝐼𝑜𝑒

−𝑅𝐿𝑡

At t=T/2, io(t)=Io

𝐼𝑜 = 𝑖𝑜𝑇

2=𝑉𝑠2𝑅

1 − 𝑒−𝑅2𝐿𝑇 − 𝐼𝑜𝑒

−𝑅2𝐿𝑇 =

𝑉𝑠2𝑅

1 − 𝑒−𝑅2𝐿𝑇

1 + 𝑒−𝑅2𝐿𝑇

Substitute the value of Io in io(t) equation

𝑖𝑜 𝑡 =𝑉𝑠2𝑅

1 − 𝑒−𝑅𝐿𝑡 −

𝑉𝑠2𝑅



𝑒−𝑅𝐿𝑡 =

𝑉𝑠2𝑅

1 −2𝑒−

𝑅𝐿𝑡




T/2<t<T

−𝑉𝑠2= 𝑅𝑖𝑜 𝑡

′ + 𝐿𝑑𝑖𝑜 𝑡

′

𝑑𝑡

𝑖𝑜 𝑡′ = −

𝑉𝑠2𝑅

1 − 𝑒−𝑅𝐿 𝑡

′

+ 𝐼𝑜𝑒−𝑅𝐿(𝑡

′)

At t=T/2, io(t)=Io

𝐼𝑜 = 𝑖𝑜𝑇

2= −

𝑉𝑠2𝑅



Substitute the value of Io in io(t) equation

𝑖𝑜 𝑡′ = −

𝑉𝑠2𝑅

1 −2𝑒−

𝑅𝐿(𝑡

′)


𝑡′ = 𝑡 −𝑇

2


Single Phase Full Bridge Inverter

The output voltage Vo in single phase full

bridge inverter can be Vdc, -Vdc, or zero,

depending on which switches are closed. SV

Load

Vo

ioT3

D3

T2D2

baT1

T4

D1

D4

i3

i2

i1

i4

is

Switched Closed Output Voltage Vo

T1 and T2 +Vdc

T3 and T4 -Vdc

T1 and T3 0

T2 and T4 0

SVLoad

Vo=Vs

io ba

is

T1

T2

SVLoad

Vo=0

ba

is

T1 T3

SVLoad ba

is

T2T4

Vo=0SV

Load io ba

is

Vo=-VsT4

T3

T1 and T2 Closed T3 and T4 Closed T1 and T3 Closed T2 and T4 Closed



SVLoad

Vo

ioT3

D3

T2D2

baT1

T4

D1

D4

i3

i2

i1

i4

is

The switches connect the load to +Vdc when

T1 and T2 are closed or to -Vdc when T3 and

T4 are closed. The periodic switching of the

load voltage between +Vdc and -Vdc

produces a square wave voltage across the

load. Although this alternating output is

nonsinusoidal, it may be an adequate ac

waveform for some applications.

T1 and T4 should not be closed at the same

time, nor should T2 and T3. Otherwise, a

short circuit would exist across the dc source

The current waveform in the load

depends on the load components.

For the resistive load, the current

waveform matches the shape of the

output voltage.

The waveforms when resistive load

2

sV

2

sV

R

V s

2

R

V s

2

t

t

t

Vo

i2

i1

R

Vs

2

t

io

R

Vs

2

2

T T

2

T T

2

T T

2

T T



An inductive load will have a

current that has more of a

sinusoidal quality than the voltage

because of the filtering property of

the inductance.

The waveforms when RL load

Switches T1 and T2 close at t=0. The

voltage across the load is +Vs, and

current begins to increase in the load

and in T1 and T2. The current is

expressed as the sum of the forced and

natural responses.

where A is a constant evaluated from the initial condition and τ=L/R. at t=0,

i(0)=Imin.

→



In steady state, the current waveforms for RL load can be described by

At t=T/2, T1 and T2 open, and T3 and T4 close. The voltage across the RL load

becomes -Vs, and the current has the form

where B is a constant evaluated from the initial condition and τ=L/R. at

t=T/2, i(T/2)=Imax.

→

𝑖𝑜 𝑡 =

𝑉𝑑𝑐𝑅+ 𝐼𝑚𝑖𝑛 −

𝑉𝑑𝑐𝑅

𝑒−𝑡 𝜏 0 < 𝑡 <𝑇

2−𝑉𝑑𝑐𝑅

+ 𝐼𝑚𝑎𝑥 +𝑉𝑑𝑐𝑅

𝑒−(𝑡−𝑇 2 ) 𝜏 𝑇

2< 𝑡 < 𝑇



An expression is obtained for Imax by evaluating the first part of io(t)

equation at t=T/2.

And by symmetry

Substituting –Imax for Imin in i(T/2) equation yields

The power absorbed by the load can be determined from (Pac=Irms2R)

The power supplied by the source must be the same as absorbed by the load.

Power from a dc source is determined from (Pdc=VdcIs)

The rms load current is determined by



Example: The full-bridge inverter has a switching sequence that produces a

square wave voltage across a series RL load. The switching frequency is 60

Hz, Vs=100 V, R=10 Ω, and L=25 mH. Determine (a) an expression for load

current, (b) the power absorbed by the load, and (c) the average current in

the dc source.

(a)



Example: The full-bridge inverter has a switching sequence that produces a

square wave voltage across a series RL load. The switching frequency is 60

Hz, Vs=100 V, R=10 Ω, and L=25 mH. Determine (a) an expression for load

current, (b) the power absorbed by the load, and (c) the average current in

the dc source.

(b)

(c)

The power absorbed by the load is

Average source current can also be computed by equating source and load power

Power Electronics - Philadelphia University...Power Electronics Inverters 1 Dr. Firas Obeidat 2...

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