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Example on a Half Wave Rectfier with RL load. Example 3-2 of "Power Electronics" by Daniel Hart.

p 0 0.0093 s * (0.00001 s * range :

f 60 Hz * :

V.m 100 V * :

source voltage

ω 2 π * f * :

ω 376.9911 Hz *

R 100 Ω * :

L 0.1 H * :

Load parameters

Z1 R 2^ω L * (2^(+ sqrt :

Z1 106.8701 Ω *

θ ω L R / * atan :

L R / 0.001 s *

θ deg 20.656

"Load angle"

θ 0.3605

p I V.m Z1 / ω s * p * θ - sin θ sin p R s * L / * - exp * + (* :

Current as a function of time

x I x 0 ≥ ( * x 0.0093 ≤ ( *

p I p 0 0.0093 s * int 0.0166 / 0.3093 A *

average current

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Same as before but now as a function of ωt.

a β 0.361 - sin 0.361 sin β 0.377 / - exp * + 0 ≡ β 0.001 2 π * solve :

a is the value at whcih the current is zero.

a rad 3.5026

solve is used to solve the non linear equation

ωt 0 4.0 rad * (0.001 rad * range :

ωt B V.m Z1 / (ωt θ - sin θ sin ωt R * ω L * / - exp * + (* :

x B x 0 ≥ ( x a ≤ ( * *

I.ave ωt B ωt 0 a int 2 π * / :

Average Current

I.ave 0.3081 A *

I.rms 12 π * (/ ωt B (2^(ωt 0 a int * sqrt :

I.rms 0.4736 A *

RMS current

Power 12 π * / 100 V * ωt sin * ωt B * ωt 0 a int (* :

Power 22.4281 W *

Load Power

pf Power 100 V * 2 sqrt / I.rms * / :

pf 0.6697

Power Factor of the source

a rad 3.5026

β f β 0.361 - sin 0.361 sin β 0.377 / - exp * + (:

x f

Graphical verification