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• hamza abdo mohamoud

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• Rectificador
• Impedancia eléctrica
• Ingenieria Eléctrica
• Electromagnetismo
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Full-Wave Controlled Rectifier RL Load: Discontinuous Operation

Full-Wave Controlled Rectifier: RL Load (discontinuous mode) There are many ways we can develop a full-wave controlled rectifier for an inductive load. One of the possible configurations is shown below. Once again, the idea here is to control the dc output voltage and current through the inductive load when the maximum value of the input voltage cannot be changed. In this case, there are two possible operations:(a) discontinuous, and (b) continuous. In this section, we consider the discontinuous operation.

Given data: R := 100⋅ Ω

Vm := 170⋅ V

L := 100⋅ mH f := 60⋅ Hz ω = 376.991

The impedance of the RL circuit: Magnitude and phase of the impedance:

ω := 2⋅ π⋅ f Rad/s

z := R + j ⋅ ω⋅ L

z = 100 + 37.699j Ω

Z := z

Z = 106.87 Ω

φ := arg ( z)

φ = 0.361 rad

tanφ := tan ( φ )

tanφ = 0.377

The operation is discontinuous if the current in the inductor is negligibly small prior to the firing angle of the SCRs. In other words, the current is zero when the SCRs turn on. Let us sketch the waveform for one time period:

Guru/PE424/FCDRL

1

ωt := 0 , 0.01 .. 2⋅ π +

π 3

May 05, 2006

Full-Wave Controlled Rectifier RL Load: Discontinuous Operation

Let us also assume that the firing angle is

α :=

π 4

α = 45 deg

vs ( ωt) := Vm⋅ sin ( ωt)

The source voltage:

When the SCR is conducting, the differential equation for the RL circuit is

L⋅

dio ( t)

+ R⋅ io ( t) = Vm⋅ sin ( ωt)

dt

− ( ωt−α ) 



Vm  tanφ  io ( ωt) := ⋅ sin ( ωt − φ ) − sin ( α − φ ) ⋅ e  Z

The Solution:

Let us now verify if the operation is really discontinuous. To do so, find the extinction angle β when the current is zero again. For the discontinuous mode, the angle β ≤ π + α, where π + α = 225 deg. Let us guess the value of extinction angle as

β := π

We use the root function to obtain the extinction angle as β := root io ( β ) , β

(

)

β = 3.502 rad

or

β = 200.638 deg

Since β = 200.638 deg is less than π + α = 225 deg, the condition is satisfied. Hence, the operation is discontinuous. Then the expression for the output voltage is vo ( ωt) := if α ≤ ωt ≤ β , vs ( ωt) , if π + α ≤ ωt ≤ π + β , −vs ( ωt) , 0

(

Guru/PE424/FCDRL

(

2

)) May 05, 2006

Full-Wave Controlled Rectifier RL Load: Discontinuous Operation

Input (solid), Output (Dotted)

Input and Output Voltages 200 160 120 80 40 0 40 80 120 160 200

0

30

60

90

120 150 180 210 240 270 300 330 360 390 Angle in degrees

The Load current: i ( ωt) := if α ≤ ωt ≤ β , io ( ωt) , 0 + if α + π ≤ ωt ≤ β + π , io ( ωt − π) , 0

(

)

(

)

Output current ( A)

Load Current 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0

0

30

60

90 120 150 180 210 240 270 300 330 360 390 Angle (degrees)

Guru/PE424/FCDRL

3

May 05, 2006

Full-Wave Controlled Rectifier RL Load: Discontinuous Operation

Compute the rms values of the input voltage, output voltage, and output current as:

2⋅ π

Vsrms :=

1 ⌠ ⋅ 2⋅ π ⌡0

Vorms =

2 ⌠ 2 ⋅  vo ( a) da 2⋅ π ⌡

2

Vsrms = 120.208 V

vs ( a) da

β

α

Upon integration, we get

Vorms :=

Vm

β−α

⋅

π

2

−

sin ( 2⋅ β ) π⋅ 2

+

sin ( 2⋅ α ) 2⋅ π

Vorms = 115.226 V

The rms current through the load is

β

Iorms :=

2 ⌠ 2 ⋅  io ( a) da 2⋅ π ⌡α

Iorms = 1.051 A

The dc values of the output voltage is β

2 ⌠ Vodc = ⋅  vo ( a) da 2⋅ π ⌡α Upon integration, the above equation yields

Vodc :=

Guru/PE424/FCDRL

Vm π

⋅ ( cos ( α ) − cos ( β ) )

4

Vodc = 88.903 V

May 05, 2006

Full-Wave Controlled Rectifier RL Load: Discontinuous Operation

Since the average voltage across the inductor is zero, we can determine the average current in the load as Vodc

Iodc :=

Iodc = 0.889 A

R

DC Power Output:

Podc := Vodc⋅ Iodc

Total Power Output:

PoT := Iorms ⋅ R

PoT = 110.543 W

Apparent power output:

So := Vorms⋅ Iorms

So = 121.147 VA

Rectification ratio:

η :=

2

Podc

A ≡1

Hz ≡ 1

VAR ≡ 1

Guru/PE424/FCDRL

η = 0.652

So

V ≡1

Podc = 79.038 W

W≡1 VA ≡ 1

5

Ω ≡1 mH ≡ 0.001

May 05, 2006