Design of a hearing loop system in CMOS18 technology

Application description

The presentation A hearing loop system in CMOS18 technology shows the application and the high level requirements of the complete hearing aid system.

Hearing loop receive coil amplifier A1

The hearing loop sub system converts the magnetic field of a current-driven loop antenna into an audio signal. Hearing aid devices automatically switch from microphone reception to the hearing loop reception if correlation betweeen the signal at the output of the amplifier A1 and the output of the microphone is sufficently large.

Hence, the gain and the frequency characteristics of A1 should be such that its output signal corresponds with that of the microphone.

ADC driver A2

The ADC driver amplifier A2 drives the 10 pF input capacitance of the ADC to a level of 0.9 Vpp at 110 dB SPL audio input.

Loudspeaker driver A3

Amplifier A3 drives the loudspeaker that must produce 110 dBSPL when driven from a peak-peak voltage equal to the fully charged battery voltage.

Links

• Presentation Sonion

• Hearing aid DSP

• Patent Hearing Loop System

• Battery specification

Design exercise 1

Give the preferred amplifier types and idealized transfer functions of the three amplifiers.

Answer A1

The hearing transmit loop is a current driven loop. The current carries the audio signal and the conversion from the audio input to the current is frequency independent.

The receive coil converts this magetic field into a voltage. This converstion mechanism is described by Faraday's law: the induced voltage is proportional with the rate of change of the magnetic flux in the coil.

A voltage amplifier accurately converts the voltage at the output of the receive coil into a voltage equal to that produced by the microphone. If we take the short circuit current of the receive coil as information carrying quantity, the transfer depends on both the inductance and the resistance of the receive coil.

In order to determine the required voltage transfer \(Av_1\) of A1, we will write the output voltage generated by the receive coil and by the microphone both as a function of the sound pressure in Pa.

At 1 kHz the sensitivity of the receive coil equals -60.2 dBV/(A/m). Hence, at this frequency the sensitivity may also be written as:

\[V_{ind_{1kHz}}=10^{-\frac{60.2}{20}}=980\cdot 10^{-6}\, \mathrm{\frac{Vm}{A}}\]

The conversion gain of the receive coil can thus be writen as:

\[V_{ind}=980\cdot 10^{-6}\frac{s}{2000 \pi}=0.156\cdot 10^{-6} \cdot s \, \mathrm{\frac{Vm}{As}}\]

A field strength of 1 A/m corresponds with a sound pressure level of 90 dBSPL, which is equivalent to a sound pressure level of 633 mPa. Hence, we may write:

\[V_{ind}=\frac{0.156}{0.633}10^{-6}s = 0.246\cdot 10^{-6}\cdot s \, \mathrm{\frac{V}{s\,Pa}}\]

The output voltage of A1 must equal the output voltage of the microphone. According to the data sheet, at 1 kHz, the microphone produces a voltage \(V_{mic}\) of:

\[V_{mic} = -35.5 \, \mathrm{\frac{dBV}{Pa}}\]

This is equivalent to

\[V_{mic} = 0.0168 \, \mathrm{\frac{V}{Pa}}\]

The gain \(Av_1\) of A1 thus needs to be:

\[Av_1 = \frac{V_{mic}}{V_{ind}} = \frac{0.0168}{0.246 \cdot 10^{-6}} = \frac{68.2}{s} 10^3\]

Answer A2

The amplifier A2 brings the peak-peak voltage at the output of the microphone to 0.9 V. The maximum required sound level of 110 dBSPL corresponds with 6.33 Pa. At this sound pressure level, the microphone produces 106 mV. If the sound pressure level requirement is given in terms of an RMS value, an integration time interval should be given together with it. If this time interval maximally equals the occurence of a peak, we may assume a crest factor of unity. If the duration of a peak is much smaller than the RMS integration time interval, the crest factor is larger than unity. We will asume the latter case and work with the crest factor of 3. Hence, the peak-peak value at the output of the microphone can be as large as 636 mV.

From the above we conclude that A2 must be a voltage amplifier with a voltage gain \(Av_2\) of:

\[Av_2 = \frac{0.9}{0.636} = 1.41\]

Answer A3

When driven with an RMS voltage of 0.18 V, the loudspeaker produces a sound level of maximally 106 dBSPL at about 2.5 kHz. For a sound level of 110 dBSPL, it thus requires a voltage \(V_{LSP}\) of:

\[V_{LSP}=10^{\frac{110-106}{20}}0.18 = \mathrm{285} \, \mathrm{mV}\]

If we account for the crest factor of 3, it requires a peak-peak voltage of 1.71 V.

At 1 kHz a, the sensitivity is tipically 4.5 dB less. In that case a peak-peak voltage of 2.87 V is required. This is slightly more than two times the battery voltage which is the theoretical limit with a full-bridge voltage driver.

Hence, at this stage of the design we define A3 as a single-ended to full-bridge voltage amplifier with a voltage gain of two.