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Power Factor Correction

Many loads are highly inductive, such a lightly loaded motors and illumination transformers and ballasts. You may want to correct the power factor by adding parallel capacitors. You can also add series capacitors to "remove" the effect of leakage inductance that limits the output current.

Why correct the power factor?

The current flow through the circuit is increased by the reactive component. Normally, loads are represented by a series combination of a resistance and a purely imaginary reactance. For this explanation, it is easier to contemplate it as an equivalent parallel combination. The diagram below illustrates a partially reactive load being fed from a real system with some finite resistance in the conductors, etc.

The current through the reactive component itself dissipates no power, and neither does it register on the watt hour meter. However, the reactive current does dissipate power when flowing through other resistive components in the system, like the wires, the switches, and the lossy part of a transformer. Switches have to interrupt the total current, not just the active component. Wires have to be big enough to carry the entire current, etc. Correcting the power factor reduces the amount of oversizing necessary.

Correcting power factor

Given the reactive load component (Xload), you can calculate the capacitance to exactly match it using the equation:

Xc = 2*pi * 60 / C = 377/C

or, rearranging: C = 377/Xc

Power factor correction capacitors are often rated in kVar, instead of uF, because that is how the power company works. Say a factory has several thousand horsepower worth of motors at .85 power factor. They might have a reactive component of several hundred kVar. At a distribution voltage of 14,400 volts, this would require a capacitor with an impedance of about 1037 ohms, or about 2.5 microfarads, a reasonable sized and priced package. However, if you were crazy enough to try to compensate this at 230 volts, you would need about .01 Farads (i.e. 10,000 uF), a sizeable package.

For very large systems, even capacitors get unwieldy. One approach is to use large over excited synchronous motors which look like capacitors, electrically. Another approach is clever systems of thyristors and inductors which simulate the capactive reactance by drawing "displacement current".

Loads that draw non-sinusoidal current

Classic reactive loads, like transformers, lighting ballasts, and AC motors still have a sinusoidal current flow. The phase of the current is just shifted from that of the supply voltage. However, there are some loads which draw distinctly non-sinusoidal currents. The most recently notorious is the switching power supply in a PC. These power supplies start with a bridge rectifier feeding a capacitor, and so, particularly at part load, draw their current in little peaks, when the instantaneous line voltage is above the capacitor voltage, forward biasing the rectifier. Another notorious non-sinusoidal current draw is the popular phase controlled light dimmer, which uses a TRIAC or SCR to reduce the RMS voltage to the load by turning on partway through the half cycle. Not only is the current waveform highly non-sinusoidal, but it is also out of phase with the voltage supply. Hence, these loads have a non-unity power factor, and draw reactive power.

However, to compensate these loads, you have to come up with a means to supply the reactive current at the appropriate times. A simple capacitor doesn't do this. A capacitor only compensates nice sinusoidal power factor lags, like those from linear (non-saturating) inductors.

Example of Power Factor Correction

Let's take an example. A 3/4 HP electric motor has a power factor of .85. The nameplate current is 10 Amps at 115 Volts, or 1150 Volt Amps.

  • Apparent power = 1150 Volt Amps
  • Active power = .85 * 1150 = 977.5 Watts
  • Reactive Power = sqrt(1150^2 - 977.5^2) = 605 VAR

So, we need 605 var of power factor correction. Calculating the required impedance from Q = E^2/X

  • 605 = 115^2/X => X = 115^2/605 = 21 ohms
  • C = 1 /( 2 * pi * f *X) = 1/ (377 * 21) = 126 uF

which is a fairly large capacitor. Furthermore, it will have a current of about 5.2 amps flowing through it, so its series resistance should be pretty low, or it will dissipate a fair amount of energy.

If the line voltage were higher, the correction impedance would be increased as the square of the line voltage. The capacitance would be reduced as the square of the line voltage. That is, if the same motor were run off 230 Volts, the capacitor would only need to be 31.5 uF. And if we were to do power factor compensation at the distribution voltage of 4160 volts (for example), you would only need about .1 uF. This is why power factor correction is usually done in the distribution network at MV or HV, and not at the end voltage.

Copyright 1997, Jim Lux /revised 4 Dec 1997/ pfc.htm / HV Handbook / Home Page / Email to Jim

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