Sizing a hearing loop driver: current, voltage and impedance

A loop driver has two jobs. It must push enough current around the loop to reach the target field strength, and it must have enough voltage to hold that current as the frequency rises and the loop's impedance climbs. Most sizing mistakes come from checking the first and forgetting the second.

Key points

• Current sets the field. Bigger loops and greater listening heights need more ampere-turns.
• Impedance rises with frequency, because a loop is a resistance in series with an inductance.
• Voltage is current times impedance. The highest demand is toward the top of the speech range, not at 1 kHz.
• Metal loss hits twice. It raises the current you need, most of all at high frequencies, where the impedance is already highest.
• Count the feeder, and leave headroom. Compare current and voltage with the driver's continuous ratings, channel by channel.

Current: what sets the field strength

IEC 60118-4 uses 400 mA/m at listening height as its reference field strength (0 dB), with targets for how evenly the field is spread and how flat the response is from 100 Hz to 5 kHz (the targets explained). The driver's first job is to supply enough current to reach that level.

Three things set the field at a point:

• Ampere-turns. Double the current, or double the turns at the same current, and the field doubles.
• Loop size. In a larger loop the conductor is farther from the middle of the room. Once a loop is much wider than the listening height, doubling both its length and its width roughly doubles the current needed for the same field in the middle. The shorter side counts for more than the longer one.
• Listening height. The field is planned at head height, usually about 1.2 m for seated and 1.7 m for standing listeners (listening height). The farther the heads are from the plane of the loop, the more current the same field in the middle needs.

The current worked out this way is the 1 kHz figure in a room with no metal. Whether the field is even enough across the seating is a layout question, covered in the design method.

Impedance: resistance plus inductance

Electrically, a loop and its feeder behave like a resistor in series with an inductor.

Resistance comes from the conductor's length and cross-section. A metre of copper with a 1 mm² cross-section has a resistance of about 0.017 Ω at 20 °C, so 50 m of 2.5 mm² wire is about 0.34 Ω. Across the speech range, resistance barely changes with frequency.

Inductance comes from the loop's geometry. For a single turn of round wire it is roughly 1.5 to 2 µH per metre of perimeter, rising slowly with room size. It rises with the square of the turns: two turns on the same path have about four times the inductance of one. A wide flat conductor has slightly less inductance than round wire; see loop wire vs copper tape.

Reactance, X = 2π × f × L, grows in step with frequency: it is five times larger at 5 kHz than at 1 kHz. Impedance combines the two: Z = √(R² + X²). In a single-turn loop of 2.5 mm² wire, the loop's reactance overtakes its resistance below 1 kHz, in small rooms and large ones alike, so the impedance climbs steeply toward 5 kHz. Thinner wire, or a long feeder on a small loop, adds resistance and flattens the climb. Extra turns make it steeper: the loop's resistance rises in step with the turns, its inductance with their square.

Voltage: where drivers run out

Most loop drivers are built to control current. They deliver the current you set into whatever impedance they see, up to the limit of their output voltage. The voltage needed at any frequency is V = I × Z.

As the impedance rises with frequency, so does the voltage demand. A driver that runs out of voltage can no longer hold the current, and the high frequencies fall away or clip.

Speech carries much less energy at 5 kHz than around 1 kHz, so a speech system is not normally sized to deliver full current at 5 kHz. A common practical approach is to size for full current, after metal loss, up to about 1.6 kHz, and some data sheets state their rated load at 1.6 kHz. Keep the 5 kHz figure in view as a warning: allow more voltage when it is far beyond the driver, when the metal-loss boost is large, or when the loop will carry music.

The feeder adds resistance and inductance but no useful field, so include it in the calculation.

Metal loss and headroom

Metal in floors, ceilings and walls, mostly steel, weakens the field, and weakens the high frequencies more than the lows. You make up for it with more current overall and with the driver's metal-loss correction, a high-frequency boost. Both add voltage demand where it is already highest. Making up 3 dB of loss takes about 1.4 times the current; 6 dB takes twice the current and twice the voltage. See metal loss for spotting it on site.

Headroom is the margin between what the design asks for and what the driver can deliver. Check:

• Current and voltage against the driver's continuous ratings, per channel, after metal loss.
• Load resistance against any minimum and maximum the maker states.
• Channels. A two-channel phased array has two loads, often different, and coupling between the two channels' overlapping elements can change the load each one sees. Check each channel on its own figures.

A margin of about 2 to 3 dB on both current and voltage (roughly 25 to 40 percent) is a sensible starting point, with more when the metal loss is an estimate rather than a measurement.

Read the conditions behind each rating. IEC 62489-1 sets out how makers measure and state loop driver performance. It covers the rated load, given as a resistance in series with an inductance, the current a driver can deliver indefinitely, and the higher current it can deliver for a limited time. A short-term or peak figure is not the same as a continuous one.

A worked example: Community Hall

Example only: a synthetic room with rounded figures. Your loop will differ.

• Room: Community Hall, 15 m × 10 m, flat floor, seated audience.
• Listening height: 1.2 m above the loop.
• Loop: one turn of 2.5 mm² round wire around the perimeter at floor level, 50 m.
• Feeder: 15 m of four-core 1.5 mm² star-quad cable, opposite cores paralleled for each leg.

Current. At 1.2 m above the centre of this loop, each amp produces about 72 mA/m. Reaching 400 mA/m there takes about 5.5 A at 1 kHz with no metal.

Resistance. The loop is about 0.34 Ω. The feeder is 30 m of conductor path with an effective 3 mm² per leg, about 0.17 Ω. Total: about 0.52 Ω.

Inductance. About 90 µH for the loop (1.8 µH per metre) plus about 2 µH for the feeder: roughly 92 µH.

By 5 kHz the impedance is nearly four times its 1 kHz value.

Metal loss. Suppose the site survey suggests about 3 dB of loss at 1 kHz, rising to about 4.5 dB at 1.6 kHz. The current at 1.6 kHz becomes 5.5 A × 1.68 ≈ 9.2 A, and the voltage 9.2 A × 1.06 Ω ≈ 9.8 V.

Headroom. With a 2 to 3 dB margin, you would look for a driver channel rated for roughly 12 to 13 A and 12 to 14 V into this kind of load. Full current at 5 kHz would take 16.2 V even before metal loss, more than that rating. For speech that is normally acceptable, as explained above; for music or a large metal-loss boost, allow more voltage. Commissioning confirms the frequency response.

Two turns instead of one would halve the 1 kHz current to about 2.8 A. But the inductance would rise to about 360 µH, and the 1.6 kHz voltage from 5.8 V to about 10 V before any metal loss.

Checking the numbers as you design

Hearing Loop Designer does this arithmetic for each channel as you draw the loop. It works out resistance and inductance from the conductor, turns and feeder you choose, applies your metal-loss allowance, and shows the required current and the voltage at 1 kHz, 1.6 kHz and 5 kHz. Driver suggestions from its multi-brand catalog are filtered by the current, 1.6 kHz voltage, channels and load resistance the design needs, with the design's demands shown next to each driver's ratings.

The figures are still a plan. Measure the loop resistance before it is covered, compare it with the design, and confirm installed performance by commissioning.

Common questions

Can I just fit a bigger driver to be safe?

Extra current and voltage give you headroom, and the field is set by the current you adjust the driver to, not by its maximum. Check that the loop resistance suits the driver. A bigger driver will not fix an uneven layout.

Should I add turns to reduce the current?

Sometimes. N turns need 1/N of the current, but the inductance rises N² times, so the high-frequency voltage rises about N times. Extra turns suit small loops and drivers with spare voltage. In large rooms they often run the driver out of voltage.

Why does the loop sound dull or distorted when the 1 kHz reading looks fine?

There are two usual causes: metal loss that the correction has not made up, or a driver running out of voltage at higher frequencies, often because of a large metal-loss boost. Check the frequency response and the driver's overload indicators together. If the driver is already at its voltage limit, turning up the current or the boost makes it worse. The fix then is a lower-impedance load, such as fewer turns or a shorter feeder, or, after checking the load, a driver with more voltage.

Does the feeder cable really matter?

Yes. It carries the full loop current but adds no useful field, so its resistance and inductance use up driver voltage, and a feeder with separated conductors can put field where you do not want it. Use a twisted pair, or star-quad cable with opposite cores paralleled, and keep it short.