Three phase fully controlled bridge rectifier circuit, L 0 4H, R 5, requires Ud to vary between 0 22

6k harmonics. In single-phase rectifier bridge circuits.

Since the AC side current is full-wave or half-wave symmetrical. In the three-phase bridge rectifier circuit, the AC side current is full-wave symmetrical. However, the three-phase half-control rectifier circuit is different, and the current half-wave on the AC side is no longer symmetrical, so even harmonics are generated.

In the same way, the output voltage is equal to UBA, UCA, and UCB in that order.

At this time, the working condition and output voltage waveform are exactly the same as those of the three-phase bridge uncontrolled rectifier circuit, and the rectifier circuit is in a fully conductive state.

When >0, the conduction of the thyristor should be postponed, but the triggering and conduction sequence of the thyristor remain unchanged.

The base period of the output waveform is 1 6 of the input AC cycle, so the fundamental frequency of the output voltage is 6 times the input frequency, and for a 50Hz power supply, the fundamental is 300Hz. Since the waveform is not a sine waveform, according to the principle of Fourier series decomposition, it should also contain the harmonic components of the fundamental frequency, which is 6n times the power supply frequency, that is, it contains 300Hz, 600Hz, 900Hz, 1200Hz ......

After rectification, it is already direct current, so there is no talk about harmonics.

Summary. Rectifier output average voltage ud = = 156V, output DC current id = ud r = 156V 4 =39A;

2. Considering the voltage margin coefficient δu=, the rated voltage of the selected thyristor should be greater than =, and the rated voltage of the thyristor should be greater than or equal to 500V. Considering the current margin coefficient δi=, it is obtained that the rated current of the selected thyristor should be greater than , and the rated current of the thyristor should be greater than or equal to 50A. Due to the consideration of twice the safety margin, the above equation is multiplied by two to obtain a rated voltage of 1000V and a rated current of 100A

Single-phase bridge full-control rectifier circuit, U2=200V, the load is r=4, the L value is very large, when =30°, it is required: (1) Find UD, ID, IT, I2. 2) Considering the return of 2 times the safety margin, find the rated voltage and rated first grinding current of the thyristor.

The average voltage of the rectified output is ud = = 156V, and the DC current of the output vertical plex is id = ud r = 156V 4 =39A; 2. Considering the voltage margin coefficient δu=, the rated voltage of the selected thyristor should be greater than =, and the rated voltage of the thyristor should be greater than or equal to 500V. Considering the current margin coefficient δi=, it is obtained that the rated current of the selected thyristor should be greater than , and the rated current of the thyristor should be greater than or equal to 50A. Considering twice the full margin of Yu Sui Sakura by the slag, multiply the above formula by two to obtain a rated voltage of 1000V and a rated current of 100A

IT and i2 are what they are.

i2 equals id

IT equals id 2 equals.

A "rectifying circuit" is a circuit that converts alternating current energy into direct current energy. Most rectifier circuits consist of transformers, rectifier main circuits, and filters, among others. It is widely used in the speed regulation of DC motors, the excitation adjustment of generators, electrolysis, electroplating and other fields.

After the 70s of the 20th century, the main circuit was mostly composed of silicon rectifier diodes and thyristors.

Approaching 180 degrees, strictly defined as (180°-

the adjustment interval of the trigger angle or the formation interval of the conduction angle; When the freewheeling is special, it does not need to be triggered, but it is turned on.

1. Single-phase bridge full-lift pants control rectifier missing circuit, connected to resistive load, when required = 30, ud=80V, id=7A trial calculation of the rectifier transformer secondary noise voltage U2 and rectifier transformer secondary side current I2

Dear, I'm glad to answer for you! In a single-phase bridge fully controlled rectifier circuit, the secondary side current i2 and the secondary side voltage U2 of the rectifier transformer can be calculated by the following formula: I2 = id 2 + ud (2*r)) 2) u2 = u2 = ud 2 + 2*r*i2) 2) where id is the chain fluter vacant current of the blind circuit of the rectifier shed, ud is the output voltage of the rectifier circuit, and r is the load resistance.

According to the conditions given in the question, id = 7a, ud = 80v, and r = 30 is required. Bringing these values into the formula for calculation, we can get: i2 = 7 2 + 80 (2*30)) 2) au2 = 80 2 + 2*30* v Therefore, in the case of satisfying the requirement of 30, the secondary side current i2 of the rectifier transformer is about A, and the secondary side voltage U2 of the whole hole defect current transformer is about V. 