Welcome to EC&M's monthly forum where power quality experts Mark McGranaghan, vice president of Electrotek Concepts, and Mike Lowenstein, president of Harmonics Limited, address your PQ problems and concerns. This month, editorial director John DeDad is pinchhitting for Lowenstein.
Q. Can shielded isolation transformers provide power conditioning? Designers of electronic instrumentation and data processing systems have made recommendations for their installation, but I'm not sure about their effectiveness.
DeDad's answer. Shielded isolation transformers aren't true powerconditioning devices because they can't adjust their output voltage to compensate for sags or surges on the input line. Nor can they protect your system against voltage spikes of subcycle duration. To accomplish the latter, you would need to add surge suppressors to the system.
Instead, isolation transformers are “noise rejectors” — they provide good electrical noise isolation. Most of the noise currents that couple onto victim communications cabling are commonmode type. This noise can corrupt the communications signals sent by the equipment connected to the cable ends. In extreme cases, these noise currents can be disruptive enough to cause electrical damage, such as component burnout to circuit elements used at both ends of communications cables. The industry expresses achieved commonmode noise attenuation either as a ratio (V_{ti}/V_{to}) or in decibels (dB) (Figure above and Table at right). Typical applications of isolation transformers include protecting critical loads, such as computers, electronic instrumentation, and data process peripherals, from electrical noise.
Isolation transformers can be designed with a 1:1 ratio, for isolation only, or to provide voltage transformation. While unshielded transformers use capacitance coupling between the primary and secondary windings, a shielded isolation transformer uses primarytoshield capacitance. It also uses the shieldtosecondary capacitance resulting from the small electrostatic field around the shield. The primarytoshield capacitance conducts most of the transient energy to ground so the effective capacitance between primary and secondary windings of a shielded isolation transformer is much less than that of an unshielded transformer.
Q. Our machine shop is equipped with a number of computercontrolled machines. A high harmonic content (3rd and 5th harmonics) present on the building's incoming power supply is causing our machines to alarm out. The outside power supply, not the AC drives, is causing the problem. Can you suggest what we can do to address the problem on the 3phase, 240V, open delta power supply?McGranaghan's answer. It's unlikely that the harmonic problems you're experiencing are related to the fact that the supply is open delta. I've heard of some problems with open delta systems, but they're usually associated with very unbalanced load conditions. So let's discuss other possibilities for your harmonic problem.
You say the building's incoming supply has a high content of 3rd and 5th harmonics. I don't know if you're referring to voltage or current distortion, but you should consider a few things regardless.
Assuming you have an open delta supply, third harmonic components in your facility would have to be caused by unbalanced conditions. Delta connections on the transformer secondary (closed or open delta) prevent the flow of zero sequence harmonics from the primary system into your facility or from your facility onto the primary system. Balanced third harmonic components are zero sequence, so they shouldn't be showing up in the supply current. However, unbalanced third harmonic components aren't zero sequence and they can exist. As a matter of fact, if the voltage is unbalanced, adjustable speed drives (ASDs) can draw a significant amount of unbalanced third harmonic current from the system. If you measure it, you'll see that it's not in all the phases. It's possible that the open delta connection is contributing to unbalances in the voltage, which in turn leads to the third harmonic components in your load current.
The fifth harmonic is more common and could be coming from your own AC drives or from the supply system because it goes right through the open delta transformer connection. The fifth harmonic component of the voltage is usually the highest component for most supply systems, but it should be less than 3%, according to IEEE Standard 5191992. If the fifth harmonic voltage on the supply to your facility (primary of the transformer) is greater than 3%, then it's an issue you should bring up with your utility.
Your facility is more likely interacting with the supply system to cause the high harmonic components. For instance, if you have any power factor correction equipment on the lowvoltage system, it could easily magnify harmonic components from your own equipment and from the supply system. This would result in highvoltage distortion on your lowvoltage system and thus cause the harmonic problems you're experiencing.
Consider installing a tuned capacitor bank (tuned to about 4.7X the fundamental) that can supply both power factor correction and prevent resonance problems. This would help solve the problem, regardless of whether you currently have power factor correction equipment installed on your system.
Q. I've seen references to the term “decibel” in electrical measurements. In this context, what is a decibel and how is it calculated?DeDad's Answer: The decibel (dB) is used several ways in electrical measurements. For example, it describes the magnitude of signal lost over a transport path, such as a coaxial cable or similar metallic path, or the gain of an amplifier.
A dB is expressed in relation to the electrical units to which it's being used. For example, dBV and dBmV represent decibels expressed in terms of volts and millivolts, respectively. Similarly, dBA and dBmA represent decibels expressed in terms of amperes and milliamperes. Prefixing dB with a minus sign () means loss while no sign or a positive sign (+) means gain.
The equation for calculating dB gain (+dB) or loss (dB) for any set of two voltages on a path of equal impedance is expressed as:
dB=20_{log} (E_{1}÷E_{2}).
Solving this equation isn't difficult if you have a calculator with a LOG key. Compute the ratio of the input to output voltage (E_{1}÷E_{2}), press the log key, and then multiply the result by 20. This will give you the result in terms of dB or +dB, depending upon whether the signal was lost or added to in the circuit.
Power calculations, on the other hand require you to use a factor of 10 instead of 20. The power equation for computing dB in a circuit of the same impedance is expressed as:
dB=10_{log}(P_{1}÷P_{2})
If you need a shortcut, you can use the Table at right as a quick reference. These dB values were calculated using the equations based on varying ratios of power and voltage/current.
There are also some important dB relationships worth memorizing to quickly estimate a value:

Doubling or halving of voltage or current is a ±6 dB change.

A numerical ratio of voltage or current of: 10=20 dB; 100=40 dB; 1,000=60 dB; and so on.

Doubling or halving of power is a ±3 dB change.

A numerical ratio of power of: 2=3 dB; 4=6 dB; 10=10 dB; and so on.
One benefit of dB notation is that you only need to add or subtract arithmetically the gains and losses in any given circuit to find the final value of gain or loss. Then with a little algebra or using the ANTILOG key (10X) on your calculator, you can find the voltage, current, or power ratio from the resulting answer.