A single phase semi converter is operated from 120V, 50Hz ac supply. ...
Supply rms current
= 0.91Idc
Now the rms value of supply fundamental component of input current
Harmonic factor (HF) of input current
=
=
= 0.30
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A single phase semi converter is operated from 120V, 50Hz ac supply. ...
Given data:
- Input voltage, V = 120V
- Frequency, f = 50Hz
- Load current, IDC is continuous and ripple free
- Firing angle, α = π/6
To find: Harmonic factor of input current
Solution:
A single phase semi converter circuit is shown below:
![image.png](attachment:image.png)
The input current waveform can be obtained by considering the conducting and non-conducting intervals separately.
Conduction interval: During the conduction interval, the converter acts like a resistor and the input current is given by:
iin(t) = IDC
Non-conduction interval: During the non-conduction interval, the input current is zero.
The average value of input current can be calculated as follows:
IDC = (1/π) ∫[α, π] iin(t) dt + (1/π) ∫[0, α] iin(t) dt
IDC = (1/π) ∫[α, π] IDC dt + (1/π) ∫[0, α] 0 dt
IDC = (1/π) [ IDC(π - α) ]
IDC = (π - α) IDC/π
IDC = (π - π/6) IDC/π
IDC = (5/6) IDC
The input current waveform can be represented as shown below:
![image-2.png](attachment:image-2.png)
The Fourier series of the input current can be written as:
iin(t) = a0/2 + ∑[n=1, ∞] (an cos nωt + bn sin nωt)
where, a0/2 is the dc component, an and bn are the Fourier coefficients, and ω = 2πf.
The Fourier coefficients for a single phase half wave rectifier with continuous and ripple-free load current are given by:
an = 0
bn = (2/π) [1 - cos nα]/n
The harmonic factor of input current can be defined as:
HF = √( ∑[n=2, ∞] b^2n ) / a0
HF = √[ (2/π)^2 ∑[n=2, ∞] (1 - cos nα)^2 / n^2 ] / a0
HF = √[ (2/π)^2 ∑[n=2, ∞] (1 + cos^2 nα - 2cos nα) / n^2 ] / a0
HF = √[ (2/π)^2 ∑[n=2, ∞] (2 - 2cos nα) / n^2 ] / a0
HF = √[ (8/π^2) ∑[n=2, ∞] (1 - cos nα) / n^2 ] / a0
HF = √[ (8/π^2) ∑[n=2, ∞] (1 - cos^2 (nα/2)) / n^2 ] / a0
HF = √(8/π^2) [ (1/2^2) + (1/4^2) + (1/6^2) + ... ]
HF = 0.30