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A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is 15.53Ω and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230V, 50Hz single phase source. For a firing delay angle of 60°, the average value of diode current is
- a)10 A
- b)8.165 A
- c)5.774 A
- d)3.33 A
Correct answer is option 'D'. Can you explain this answer?
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Given Information:
Single-phase full-bridge converter with a free-wheeling diode feeds an inductive load.
Load resistance is 15.53 and it has a large inductance providing constant and ripple-free DC current.
Input to the converter is from an ideal 230V, 50Hz single-phase source.
Firing delay angle is 60.
To Find:
Average value of diode current
Solution:
The given circuit diagram for a single-phase full-bridge converter with an inductive load is shown below.
The voltage across the load can be given as
V = Vm sin(ωt)
where Vm = peak value of input voltage
ω = angular frequency = 2πf
f = frequency of input voltage = 50Hz
t = time
The load current is given by
iL = V/R
where R = Load resistance = 15.53
The voltage across the inductor is given by
VL = L(diL/dt)
where L = Inductance of the load
diL/dt = rate of change of load current
The input current is given by
iS = iL + iD
where iD = current through the diode
Average value of diode current can be calculated as follows:
The firing delay angle is 60 degrees, which means the converter will be turned ON at 120 degrees of the input voltage waveform.
The diode will be conducting when the converter is turned OFF, i.e., from 180 degrees to 360 degrees of the input waveform.
During this period, the voltage across the load will be negative, and the diode will conduct.
The average value of diode current can be calculated using the following formula:
Id_avg = (1/T) ∫[Td,T] iD dt
where T = time period of the input voltage waveform = 1/50s
Td = delay angle for the converter to turn ON = 60 degrees
∫[Td,T] iD dt = area of the negative half cycle of the load current waveform
The load current waveform can be obtained by solving the differential equation
VL = L(diL/dt)
The solution is
iL = (Vm/R) [1 - exp(-t/(L/R))]
The waveform of load current is shown below.
The area of the negative half cycle of the load current waveform can be calculated as
∫[180,360] iL dt = (Vm/R) [0.5T + (L/R)(1 - exp(-2π/ω)) - (L/R)(1 - exp(-π/ω))]
Substituting the values, we get
∫[180,360] iL dt = (230/15.53) [0.5(1/50) + (L/R)(1 - exp(-2π/50)) - (L/R)(1 - exp(-π/50))]
∫[180,360] iL dt = 3.33A
Therefore, the average value of diode current is 3.33A (Option D).
Single-phase full-bridge converter with a free-wheeling diode feeds an inductive load.
Load resistance is 15.53 and it has a large inductance providing constant and ripple-free DC current.
Input to the converter is from an ideal 230V, 50Hz single-phase source.
Firing delay angle is 60.
To Find:
Average value of diode current
Solution:
The given circuit diagram for a single-phase full-bridge converter with an inductive load is shown below.
The voltage across the load can be given as
V = Vm sin(ωt)
where Vm = peak value of input voltage
ω = angular frequency = 2πf
f = frequency of input voltage = 50Hz
t = time
The load current is given by
iL = V/R
where R = Load resistance = 15.53
The voltage across the inductor is given by
VL = L(diL/dt)
where L = Inductance of the load
diL/dt = rate of change of load current
The input current is given by
iS = iL + iD
where iD = current through the diode
Average value of diode current can be calculated as follows:
The firing delay angle is 60 degrees, which means the converter will be turned ON at 120 degrees of the input voltage waveform.
The diode will be conducting when the converter is turned OFF, i.e., from 180 degrees to 360 degrees of the input waveform.
During this period, the voltage across the load will be negative, and the diode will conduct.
The average value of diode current can be calculated using the following formula:
Id_avg = (1/T) ∫[Td,T] iD dt
where T = time period of the input voltage waveform = 1/50s
Td = delay angle for the converter to turn ON = 60 degrees
∫[Td,T] iD dt = area of the negative half cycle of the load current waveform
The load current waveform can be obtained by solving the differential equation
VL = L(diL/dt)
The solution is
iL = (Vm/R) [1 - exp(-t/(L/R))]
The waveform of load current is shown below.
The area of the negative half cycle of the load current waveform can be calculated as
∫[180,360] iL dt = (Vm/R) [0.5T + (L/R)(1 - exp(-2π/ω)) - (L/R)(1 - exp(-π/ω))]
Substituting the values, we get
∫[180,360] iL dt = (230/15.53) [0.5(1/50) + (L/R)(1 - exp(-2π/50)) - (L/R)(1 - exp(-π/50))]
∫[180,360] iL dt = 3.33A
Therefore, the average value of diode current is 3.33A (Option D).
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A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is 15.53Ω and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230V, 50Hz single phase source. For a firing delay angle of 60°, the average value of diode current is a)10 Ab)8.165 Ac)5.774 Ad)3.33 ACorrect answer is option 'D'. Can you explain this answer?
Question Description
A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is 15.53Ω and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230V, 50Hz single phase source. For a firing delay angle of 60°, the average value of diode current is a)10 Ab)8.165 Ac)5.774 Ad)3.33 ACorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is 15.53Ω and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230V, 50Hz single phase source. For a firing delay angle of 60°, the average value of diode current is a)10 Ab)8.165 Ac)5.774 Ad)3.33 ACorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is 15.53Ω and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230V, 50Hz single phase source. For a firing delay angle of 60°, the average value of diode current is a)10 Ab)8.165 Ac)5.774 Ad)3.33 ACorrect answer is option 'D'. Can you explain this answer?.
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