AHF efficiency

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    How much power will an AHF draws when it generates 50A harmonic current? Is billing increases while removing harmonics from system?

    M R Srinivas


    The AHF (Active Harmonic Filter) will have a loss of approx. 40 to 50Watts per Amps.

    Fot 50A, AHF the loss will be approx. 2kW to 2.5kW.

    Yes, by adding AHF the kWH consumption will increase in the system, but the main reason for adding AHF if to improve the quality of power. The AHF is always promoted as Power Quality improving equipment and not as energy saving device.

    Having said that, there are lot more benefits of filtering harmonic. In many cases where Induction motors are used in the network having higher harmonic levels, the AHF will reduce the energy consumption in motors due to reduced negative sequence harmonic.

    Hope above answers your query.

    Best regards,



     2.5KW will be the loss in AHF,but to generate 50A @440V the energy drawn will be same?

    ie output power(KVA) + AHF loss(KW) = Input power (KVA) so that KVA demand will increase.Is that correct assumption?

    AHF draws fundamental current from system and produce harmonic current to remove harmonics that means for generating 50A harmonic current a 50A funda.current + (current for AHF loss) has to be drawn by AHF which will increase KVA consumption.

      My assumption is – to remove 50A harmonic current from the system AHF must draw more than 50A fundamental current.

    Is it correct?

    M R Srinivas

    The kVA of AHF is as you rightly pointed out = Losses + VxI.

    But, the point is, what ever the harmonic current injected by AHF is used to cancel out the harmonic current from Non-linear loads (VFDs, DC Drives, Furnace, UPS, Etc). Hence as seen from system side, the net current will be the vector difference of Load harmonics and Active filter harmonic current.

    It is like Capacitor connected to motor.

    The capacitor will draw leading current & the motor will draw lagging current. The Capacitor current & motor inductive current will cancel out vectorially, and as seen from the source side the effective current will reduce, and hence the kVA will be less.

    Active Filter topology is similar to above.

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