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Hello,

I need help with selecting the correct line reactor for my application. I'm dealing with up to 270A/100V DC that is derived from AC current produced with an automotive alternator and fed through a 300A/1000V bridge rectifier. If I have correctly comprehened what I have read then I need a

I have been reading as much as I can find on the subject and the following is as close as I have found to offering a formula. The problem is that my lack of formal education in electrons prevents me from being able to interpret the characters in the equation that are not found on a standard keyboard. I tried to research

It is widely publicized that 3% is a constant used for smoothing and 5% for dampening harmonics. Using that value I have calculated the amount impedance to be entered into the Rockwell Automation equation below.

100V / 270A = 0.370 Ohms

3% * .370 = .011 impedance

This is the equation that I used to calculate the frequency of the voltage ripple:

Alternator revolving at 6000 RPM / 60 = 100 Revolutions Per Second

100 RPS * (6 North and 6 South poles) = 1200

1200 alternating currents * 3 phases = AC @3600Hz

I would appreciate assistance with interpreting the equation in the following:

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Line Reactors and AC Drives

Rockwell Automation

Mequon Wisconsin

Sizing a reactor:

The first rule is make sure you have a high enough amp rating. In terms of the impedance value, you will usually find that 3% to 5% is the norm with most falling closer to 3%. A 3% reactor is enough to provide line buffering and a 5% reactor would be a better choice for harmonic mitigation if no link choke is present. Output reactors, when used, are generally around 3%. This % rating is relative to the load or drive where the reactor impedance is a % of the drive impedance at full load. Thus a 3% reactor will drop 3% of the applied voltage at full rated current. To calculate the actual inductance value we would use the following formula.

Your drive distributor should be able to help you size a reactor for use with a drive. If you wish to calculate the value yourself, the following example may be helpful. If a 3% reactor was required for a 100 amp 480 volt drive, a 100 amp or larger current rating would be required. The drive impedance would be: Z=V/I or 480/100 = 4.8 ohms. 3% X 4.8 ohms = 0.114 ohms inserting this 0.114 impedance in the equation for inductance we get a value of about 300 Microhenrys.

Yep,

I'mClueless

I need help with selecting the correct line reactor for my application. I'm dealing with up to 270A/100V DC that is derived from AC current produced with an automotive alternator and fed through a 300A/1000V bridge rectifier. If I have correctly comprehened what I have read then I need a

**swinging**type of reactor to cope with the variable A/V.I have been reading as much as I can find on the subject and the following is as close as I have found to offering a formula. The problem is that my lack of formal education in electrons prevents me from being able to interpret the characters in the equation that are not found on a standard keyboard. I tried to research

**2¶FL**which is used in the Rockwell Automation equation below and all I came up with was ads featuring products that are packaged in 2 fluid ounce volumes.It is widely publicized that 3% is a constant used for smoothing and 5% for dampening harmonics. Using that value I have calculated the amount impedance to be entered into the Rockwell Automation equation below.

100V / 270A = 0.370 Ohms

3% * .370 = .011 impedance

This is the equation that I used to calculate the frequency of the voltage ripple:

Alternator revolving at 6000 RPM / 60 = 100 Revolutions Per Second

100 RPS * (6 North and 6 South poles) = 1200

1200 alternating currents * 3 phases = AC @3600Hz

I would appreciate assistance with interpreting the equation in the following:

Copied from:

Line Reactors and AC Drives

Rockwell Automation

Mequon Wisconsin

Sizing a reactor:

The first rule is make sure you have a high enough amp rating. In terms of the impedance value, you will usually find that 3% to 5% is the norm with most falling closer to 3%. A 3% reactor is enough to provide line buffering and a 5% reactor would be a better choice for harmonic mitigation if no link choke is present. Output reactors, when used, are generally around 3%. This % rating is relative to the load or drive where the reactor impedance is a % of the drive impedance at full load. Thus a 3% reactor will drop 3% of the applied voltage at full rated current. To calculate the actual inductance value we would use the following formula.

**L =XL/(2¶FL)**Where L is inductance in Henrys, XL is inductive reactance or impedance in Ohms (0.370 Ohms) and F is the frequency (3600Hz). In general Frequency will be the line frequency for both input and output reactors.Your drive distributor should be able to help you size a reactor for use with a drive. If you wish to calculate the value yourself, the following example may be helpful. If a 3% reactor was required for a 100 amp 480 volt drive, a 100 amp or larger current rating would be required. The drive impedance would be: Z=V/I or 480/100 = 4.8 ohms. 3% X 4.8 ohms = 0.114 ohms inserting this 0.114 impedance in the equation for inductance we get a value of about 300 Microhenrys.

Yep,

I'mClueless

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