### General Information

The input and no-load output voltages of a rectifier transformer normally have sine wave form. Due to the fact that a rectifier is a non-linear load, the transformer currents have harmonics. The type and the amount of harmonics depend on the rectifier circuit, the smoothing choke Xd and the transformer leaking reactance Xt.

In the following text there are the most commonly rectifier circuits with the voltage and current diagrams with 5 harmonics of the transformer secondary currents which can be used in the Large Transformers Program. Also, there are screen captures of the input mask for Ud=100Vdc, Udiode=0 and Id=100Adc at Xt=0 and Xd=o/o.

Note that the program calculates the transformer AC output voltage on load. Due to this fact the no-load rectifier will be higher than 100Vdc but on-load rectifier voltage lower than 100Vdc because the program does not calculatr the commutation voltage drop of the rectifier. Using the calculated the inductive short-circuit voltage you can easy recalculate the turns of the secondary winding.

### Single Phase, 2 Pulses Center Tap Rectifier with EI and UI core  No-load voltage per winding U1=U2=1.11*(Ed+Udiode) Harmonic k Isec.k.rms/Id Angle I22k.rms/Id Angle 1 0.505 0 0.505 0 0=dc 0.5 0 0.5 180 2 0.236 0 0.236 180 4 0.047 0 0.047 180 6 0.002 180 0.002 0
 No-load voltage per winding U1=U2=1.11*(Ed+Udiode) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.45 0 0.45 0 0=dc 0.5 0 0.5 180 3 0.15 180 0.15 180 5 0.09 0 0.09 0 7 0.065 180 0.065 180

In order to get symmetrical operation mode of both secondary windings both secondary windings have to have equal short-circuit voltages. This condition can be realized when the secondary windings are wound bifilar (not supported by program) or if the secondary windings are made in 2 sectors which are cross connected : the sector 1 of the secondary 1 in series with the sector 2 of the secondary 2, ...) The following construction is not optimal but it is a good comprise. Due to the fact that the input for design a transformer with an UI core is per leg all voltages in the input mask above have to be divided by 2.

For optimal, symmetrical operation mode for both secondary windings the inside secondary winding on the leg 1 has to be connected with outside secondary winding on the leg 2, ... . Note that the parallel connection of the primary windings is allowed only with this order of the secondary windings. ### Single Phase, 2 Pulses Bridge Rectifier  No-load voltage per winding U2=1.11*(Ed+2*Udiode) Smoothing choke Xd=0 Xd=o/o Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 1.11 0 0.9 0 3 0.3 0 5 0.18 0 7 0.13 0 9 0.1 0 ### Three Phase DY Transformer with 3 Pulses Rectifier  No-load voltage per winding U1=U2=U3=0.855*(Ed+Udiode) Smoothing choke Xd=0 Xd=o/o Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 u.39 0 0.39 0 0=dc 0.333 0 0.333 0 2 0.236 0 0.195 0 3 0.07 4 0.047 180 097 180 5 0.059 180 0.077 180

With the primary connection in delta the ampere-turns of the harmonics 2, 3, 4, ... are balanced. These harmonics do not exist in the lines of the net.

Note that the ampere-turns of the secondary dc current are not balanced. For this reason there is a strong dc magnet field outside around the transformer. ### Three Phase YZZ Transformer with 3 Pulses Rectifier  No-load voltage per winding Ea=Eb=Ec?=0.5*(Ed+Udiode/2) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.39 0 0.39 0 0=dc 0.333 0 0.333 180 2 0.19 0 0.19 0 4 0.09 0 0.09 0 5 0.07 180 0.07 0
 No-load voltage per winding Ea=Eb=Ec==0.5*(Ed+Udiode/2) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.39 0 0.39 0 0=dc 0.333 0 0.333 180 2 0.236 0 0.236 0 3 0.07 0 0.07 180 4 0.047 180 0.047 0

Note that the input is per leg. All voltages are per leg The phase angle of the primary voltage on the leg is the reference angle. Normally it is set to "0".

Phase difference between the currents of 2 secondary windings on the same leg is 60° (set with angles 30° and 330°, +- 30° to the peumary voltage)

The harmonics 0, 2, 3, 4, ... are balanced within the secondary windings and do not exist in the primary ### Three Phase Transformer with 6 Pulses Bridge Rectifier  No-load voltage per winding Ua=Ub=Uc=0.427*(Ed+2*Udiode) Smoothing choke Xd=0 Xd=o/o Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.808 0 0.779 0 5 0.177 0.155 180 7 0.088 180 0.111 0 11 0.071 0 0.07 180 13 0.05 180 0.06 0 ### Three Phase DYY Transformer with 6 Pulses 2x3 Diodes Rectifier  No-load voltage per winding U1=U4 = 0.74*(Ed+Udiode) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.224 0 0.224 0 0=dc 0.166 0 0.166 180 2 0.196 0 0.196 180 3 0.153 0 0.153 0 4 0.096 180 0.096 0
 No-load voltage per winding U1=U4 = 0.74*(Ed+Udiode) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.224 0 0.224 0 0=dc 0.166 0 0.166 180 2 0.19 0 0.19 180 3 0.15 0 0.15 0 4 0.097 180 0.097 0

There are 3 windings per leg: primary in delta connection and 2 secondary windings in star connection with phase difference of 180°.

The harmonics 0, 2, 4 ,8... are balanced within the secondary windings. 3. harmonic is balanced within the delta of the primary winding. ### Three Phase yYY Transformer with 6 Pulses 2x3 Diodes Rectifier, in Series connected  No-load voltage per winding U1=U4?=0.427*(Ed+Udiode/2) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.39 0 0.39 0 0=dc 0.333 0 0.333 180 2 0.196 0 0.196 180 4 0.102 0 0.102 180 5 0.088 0 0.088 0
 No-load voltage per winding U1=U4=0.427*(Ed+Udiode/2) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.39 0 0.39 0 0=dc 0.333 0 0.333 2 0.196 0 0.196 180 4 0.108 180 0.108 0 5 0.076 0 0.076 0

There are 3 windings per leg: primary in star connection and 2 secondary windings in star connection with phase difference of 180°. Each secondary drives a 3 pulses rectifier. Both rectifiers are connected in series.

The harmonics 0, 2, 4 ,8... are balanced within the secondary windings. ### Three Phase yYY Transformer with 6 Pulses 2x3 Diodes Rectifier, parallel connected via Drainage Reactor  No-load voltage per winding U1=U4 =0.855*(Ed+Udiode) Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.195 0 0.195 0 0 0.166 0 0.166 180 2 0.098 0 0.098 180 4 0.051 0 0.051 180 5 0.044 0 0.044 0
 No-load voltage per winding U1=U4?=(Ed+Udiode)/1.17 Harmonic k I21k.rms/Id Angle I22k.rms/Id Angle 1 0.195 0 0.195 0 0 0.166 0 0.166 2 0.098 0 0.098 180 4 0.059 180 0.059 0 5 0.038 0 0.038 0

This circuit has two 3 pulse rectifiers, connected parallel via the drainage reactor. The ampere-turnsof the harmonics 0, 2, 4, 8, ... are balanced within the secondary windings. Normally the inductance of the drainage reactor should be set so high that the 3. harmonic of the transformer current is not higher than 0.5-1% of the DC output current. If so, then the primary must not be connected in delta. For an optimal current distribution between the parallel connected rectifiers the secondary windings have to have equal short-circuit voltages. For this reason the order of the windings per leg should be: secondary 1- primary - secondary 2 (NOT USED IN THE FOLLOWING INPUT). In order to create the input for the design of the drainage reactor we need to know the inductance Lab for:

• max. value of the 3. current harmonic I3ab = 0.01*Id = 1Arms
• f = 150Hz
• nominal rms value of the 3. voltage harmonic U3oa = Ed/4/1.41=17.7Vrms
• L0a=U3oa/I3ab/2/pi/f = 17.7/1/6.28/150 =18.7mH
• Lab=4*187 =75mH
• If the asymmetry of the rectifier currents is q% of the DC output current then 2.41 instead 1.41

Note that +50 means that this current does NOT produce any magnetizing of the core ### Three Phase Y/YZZ Transformer with 6 Pulses 3x2 Diodes Rectifier  No-load voltage per winding Ea=Ea1=Ea2=0.427*(Ed+Udiode/2) Harmonic k I21k.rms/Id Ia1=Ia2 Angle a1/a2 I22k.rms/Id Ia Angle a 1 0.224 30/330 0.39 0 0=dc 0.166 0 0.333 180 2 0.196 60/300 0.196 180 3 0.15 0 180 4 0.102 120/240 0.102 0
 No-load voltage per winding Ea=Ea1=Ea2=0.427*(Ed+Udiode/2) Harmonic k I21k.rms/Id Ia1=Ia2 Angle a1/a2 I22k.rms/IdIa Angle a 1 0.224 30/330 0.39 0 0=dc 0.166 0/0 0.333 180 2 0.196 60/300 0.196 180 3 .15 90/270 0 0 4 0.1081 120/240 0.108 0

Note that there are 4 windings per leg; for example A, a, a1 and a2.

3. harmonic is balanced within the windings a1 and a2. The harmonics 0,2,4,8,.. are balanced within the secondary windings a, a1 and a2 ### Three Phase D/3x Center Tap Windings with 3x2 Diodes gee Rectifier, connected parallel via 3 Phase Drainage Reactor  No-load voltage per winding U1=U4?=1.11*(Ed+Udiode) Harmonic k I1k.rms/Id Angle I4k.rms/Id Angle 1 0.15 0 0.15 0 0=dc 0.166 0 0.166 180 3 0.05 180 0.05 180 5 0.03 0 0.03 0 7 0.023 180 0.023 180

Due to the fact that there are not balanced 3. current harmonics in the secondary currents (line 9) the primary is connected in delta.

The DC output voltage of both secondary windings per leg (line 3) has 0, 2, 4, 6, 8, 10, ..., .. voltage harmonics. The voltage harmonics 6, 12, ... are limited by the smoothing choke Xd. The 3 phase drainage reactor can limit only the current harmonics 2, 4, 8, 10, .. . The DC current is limited by the load Rd.

Note that all current harmonics 2, 4, 8, 10, ... run in the 3 phase symmetrical mode. The current harmonics 0, 6, 12, ... have the same phase angle. They flow throw the 3 phase drainage reactor but they do not magnetize the core. In order to create the input for the design of the 3 phase drainage reactor we need to know the inductance Loa=Lob=Loc=Lr for:

• max. value of the 2. current harmonic I2oa = 0.01*Id = 1Arms
• f =100Hz
• nominal rms value of the 2. voltage harmonic U2oa = 0.47*Ed1=147Vrms
• Lr=U2oa/I2oa/2/pi/f =47/1/6.28/100 =75mH

Note that +33.3 means that this current does NOT produce any magnetizing of the core  