3rd part a
u i 0
tv
Im 0.5T
tv
U1
S1 0
T
tv
b u10
t2 i 0
–U1
0.5T
0.5T
t u20
Im
S3
0 S2 t
t2
u i
0.5T – tv
t
U1/2
tv
0.5T
T
S4
Im
S2
t
U1 I1
+
S1 1
U1
S4
–
i, u L
R
S3
0
S2
t
t1
2 –I1
–Im
Fig. 3.37. Voltage control by means of dead-time (a) and switch leg operation shift (b)
Analysing the load voltage harmonious composition, the amplitude of the first harmonic is determined 2U U (1)m = 1 2(1− cos ωt v ) , (3-142) π but the amplitude of the current components generated by the voltage of the first harmonic, which is very close to the actual load current amplitude, is I(1)m ≈ I m =
2U1 π
2(1− cos ωt v ) R2 + X L2
. (3-143)
Here ω = 2πf , X L = 2πfL. Applying a similar method as in Section 3.2.2 as well as assuming that over the shor tening load current is i = I m , one can find the approximate expression to determine the time interval t1 and the load current amplitude Im: t1* =
−(2τ * + t v* ) + 4τ*2 + 8τ *t v* + t v*2
I m* =
2 −(2τ* + t v* ) + 4τ*2 + 8τ*t v* + t v*2 4τ* + t v* − 4τ*2 + 8τ*t v* + t v*2
, . (3-144) 221
EnergoelektronikaEN BOOK.indb 221
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