I have been given the above equation for the average voltage of a fully controlled bridge rectifier. Apparently this is only valid for continuous load current conditions. Why is this the case?
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\$\begingroup\$ As assemblies of this type (FULL BRIDGES or called PDn) were generally used for the supply of (inductive) motors, the hypothesis was therefore made that the load was "strongly" inductive... or that the current of load was "constant" \$\endgroup\$– Antonio51Dec 1, 2022 at 14:11
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\$\begingroup\$ The best is to run a simulation with different load conditions and see which operating mode (continuous or discontinuous) is described by the equations you have been given. \$\endgroup\$– Verbal KintDec 1, 2022 at 14:15
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2 Answers
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Here is the example for PD2 (classical bridge single phase).
All specifications (voltages, currents, real Power, Apparent power, efficiency ...) ...
are all calculated with a current load = "constant" (simpler de facto).
NB: the diode currents waveforms are quasi "rectangular", which is simpler for "calculations" of various variables like Iavg, Irms, Ipeak ...
For the fully controlled bridge ... just multiply (for average voltage) by cos(alpha).
Alpha is the controlled angle.
The factor just before your cos (alpha) is the average voltage for the "diode bridge".
NB: load current = "constant" is "equivalent" to "infinite inductive" load.
Alpha can go from 0 to Pi.
0 to Pi/2: active converter, V load is positive & I is positive ...
Pi/2 to Pi: non-autonomous converter = in fact, "braking" power, V load is negative & I is positive.
NB: Alpha generally is always < 150-160° (guard angle of 20-30°).
Here is the case for PD3, full bridge 3 phases (star system).
NB: when 3 phases are "delta", the full bridge is then called S3.
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The equation integrates a sine wave between alpha and pi+alpha, x2 times in cycle. During this range, the rectifier output will be equal to AC voltage, and because of this, SCR must be switched on, and current must flow (continuos current).
This condition is true in high inductive load, but not is satisfied in case of resistive load (discontinous current). In dicontinous current, during the time that current does not flow, voltage will be zero, and equation is not valid.
See rectified waves with an angle of 40 degrees, first one inductive, and second one resistive.
