Microphones and Speakers

Warren Koontz

May 9, 2015

In ”Introduction to Acoustics” we began with an audio system including a source,a medium and a receiver. This chapter deals a class of receivers, namely microphones,and a class of sources, namely loudspeakers. Both microphones and loudspeakers areelectromechanical systems that either derive electrical signals from acoustic waves (mi-crophones) or derive acoustic waves from electrical signals (loudspeakers). We will seethat microphones and loudspeakers are very similar and some devices can perform bothfunctions. We will consider microphones first

We will begin by introducing and analyzing a basic model of the mechanical sub-system of a microphone. The mechanical subsystem interacts directly with the mediumand is set into motion by acoustic pressure. We will then review several mechanisms toconvert this motion to an electrical signal.

1 Mechanical Analysis

All of the devices that we will discuss in this chapter contain some kind of diaphragm, aflexible surface that captures or creates acoustic waves. A simple second order systemmodel of a diaphragm is shown in Figure 1. In the case of a microphone, the acousticwave exerts a force F (t) on the diaphragm given by

F (t) = Ap(t) (1)

where A is the area of the diaphragm and p(t) is the acoustic pressure. There arealso two forces that act in opposition to F (t): an elastic restoration force Kx(t) and aviscous damping force Ddx(t)/dt where x(t) is the displacement of the diaphragm fromits equilibrium position. Thus the equation of motion for the diaphragm is

Md2x(t)

dt2= Ap(t) −D

dx(t)

dt−Kx(t) (2)

where M is the mass of the diaphragm. Equation 2 can be re-written as

d2x(t)

dt2+ 2δω0

dx(t)

dt+ ω2

0x(t) = ω20

A

Kp(t) (3)

where

ω0 =

√K

M(4)

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Figure 1: Model of a Diaphragm

is the natural frequency of the system and

δ =

√D2

4KM(5)

is the damping ratio. This is a standard equation for a damped resonant system. Asystem is said to be over-damped if δ > 1, under-damped if δ < 1, and critically dampedwhen δ = 1.

We can use (3) to determine the transfer function for the mechanical subsystem andthis can be used to find the subsystem frequency response. We begin by applying theLaplace transform to (3).

s2X(s) + 2δω0sX(s) + ω20X(s) = ω2

0

A

KP (s) (6)

Thus the transfer function H(s) is given by

H(s) =X(s)

P (s)=A

K

ω20

s2 + 2δω0s+ ω20

(7)

Figure 2 is a plot of the magnitude of the frequency response H(s) (with A/K = 1 fornow) for f0 = ω0/2π = 10 KHz and values of δ corresponding to under, over, and criticaldamping. The frequency response is obtained from the transfer function by replacing swith j2πf . For frequencies sufficiently below the natural frequency, the magnitude ofthe frequency response is unity (0 dB). Thus for displacement-sensitive microphones itis desirable to have the natural frequency of the diaphragm above the audio range andthe damping at or somewhat below the critical value.

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Figure 2: Frequency Response of Diaphragm Displacement

Some microphones respond to the velocity of the diaphragm rather than the dis-placement. In the frequency domain the velocity U(s) is given by

U(s) = sX(s) (8)

and the transfer function is therefore

U(s)

P (s)=

ω20s

s2 + ω20

D

Ks+ ω2

0

A

K

=2δω0s

s2 + 2δω0s+ ω20

A

D(9)

Figure 3 shows the frequency response of this transfer function (ignoring the A/Dterm) for a few values of δ (starting at critical dampling and increasing the dampingratio). The frequency response peaks at the natural (or in this case resonant) frequencyand drops off above and below this frequency. As the damping increases, the responsecurve broadens and flattens over more of the audio range. The 3 dB bandwidth of thisfrequency response curve is given by

∆ω = 2δω0 =D

M(10)

Thus for a velocity sensitive microphone we need to strike a balance between band-width, which increases with increasing damping, and mid-band sensitivity (A/D), whichdecreases with increasing damping.

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Figure 3: Frequency Response of Diaphragm Velocity

This completes our introduction to the basic mechanical subsystem of a microphone.We now turn our attention to the electrical subsystem and the electro-mechanical inter-action.

2 The Carbon Microphone

The carbon microphone was the result of the search for a suitable ”transmitter” for thetelephone, which had been invented by A. G. Bell in 1875. It was developed indepen-dently by D. E. Hughes in England and E. Berliner and T. A. Edison in the U.S. Edisoneventually received a patent for the invention and his version became the standard forBell’s telephone in 1881. Carbon microphones were used for POTS (plain old telephoneservice) until around the 1980s, long after they were displaced from other applicationsby newer microphone designs.

The carbon microphone uses a variable resistor to produce an electrical audio signal.The resistor is formed by enclosing carbon granules between two conducting surfaces,one of which is mechanically connected to a diaphragm. The resistance R(t) of thecarbon device is related to the diaphragm displacement according to

R(t) = R0 + αx(t)

where α is a property of the device. For frequencies sufficiently below the naturalfrequency of the diaphragm, the resistance varies with the acoustic pressure as

R(t) = R0 + αA

Kp(t) (11)

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Figure 4: Carbon Microphone Circuit

Figure 4 is a schematic diagram of a basic carbon microphone circuit consisting ofa DC voltage source in series with a fixed resistance R1 and the acoustically variableresistance R(t). The voltage v1(t) across R(t) is given by

v1(t) = VR0 + α

A

Kp(t)

R1 +R0 + αA

Kp(t)

≈ VR0 + α

A

Kp(t)

R1 +R0

= V0 +αV

R1 +R0

A

Kp(t) (12)

where

V0 =R0

R1 +R0V (13)

The approximation is valid as long as R1 is large enough. The capacitor blocks the DCcomponent V0 so that the output voltage vo(t) is

vo(t) = Sp(t) (14)

where

S =αV

R1 +R0

A

K(15)

is the sensitivity of the carbon microphone in V/Pa.The primary advantage of the carbon microphone is that is produces a signal voltage

of sufficient magnitude without the need for electronic amplification (which did not existin 1881). This property made it the ideal choice for landline voice communication, whichplaces relatively modest requirements on audio quality. It was less suitable, however, forother applications such as broadcast radio and acoustical measurement.

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3 The Condenser Microphone

The condenser microphone, invented in 1916 by E. C. Wente, offered higher audio qualityand lower noise than the carbon microphone. It was initially used for acoustic measure-ments and subsequently used for broadcasting, until it was supplanted by the dynamicmicrophone. Condenser microphones are still popular for acoustic measurement and asinstrument microphones. And a recent version of the condenser microphone, known asthe electret microphone, has become the standard for telephones and similar consumerelectronic devices.

The condenser microphone uses an acoustically variable capacitor (capacitors wereoriginally called condensers) to produce an audio signal. The capacitor is a parallel platecapacitor where one of the plates also serves as a diaphragm that is displaced by acousticpressure. The capacitance of a parallel plate capacitor with an air gap is given by

C =4πε0A

d(16)

where C is the capacitance in farads (F), A is the area of the plates in m2, d is the distancebetween the plates in m, and ε0 is an electromagnetic constant known as permittivity(ε0 = 8.85× 10−12 F/m). In the presence of acoustic pressure p(t) the distance betweenthe plates varies according to

d(t) = d0 − x(t)

= d0 −A

Kp(t) (17)

Thus in the presence of acoustic pressure the capacitance is time varying and is givenby

C(t) =4πε0A

d0 −A

Kp(t)

(18)

Figure 5 shows a basic condenser microphone circuit. In the absense of acoustic input,the capacitor charges to some level Q through a resistance R. If the time constant of thecircuit is sufficiently long, the charge will remain approximately constant at this leveleven in the presense of acoustic input. The voltage across the capacitor is therefore givenby

VC(t) =Q

C(t)

= Qd0 −

A

Kp(t)

4πε0A

=Q

C0

(1 − 1

d0

A

Kp(t)

)(19)

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Figure 5: Condenser Microphone Circuit

where

C0 =4πε0A

d0(20)

Since Q = C0V , (19) becomes

VC(t) = V − V A

d0Kp(t) (21)

and, by simple application of Kirchhof’s voltage law, the output voltage vL(t) is givenby

vL(t) =V A

d0Kp(t)

= Sp(t) (22)

where S is the microphone sensitivity in V/Pa.

4 The Dynamic Microphone

In a dynamic microphone, the diaphragm is mechanically connected to a coil of wire thatis situated within a magnetic field created by a permanent magnet as shown in Figure6. If the coil is set in motion, the change in magnetic flux induces a voltage across thecoil given by

v(t) = Blu(t) (23)

where B is the magnetic field strength in Tesla, l is the length of the wire in the coil inm, and u(t) is the velocity of the coil in m/s. This result follows from Faraday’s Law

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Figure 6: Dynamic Microphone Mechanism

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Figure 7: Dynamic Microphone Equivalent Circuit

and some assumptions about the geometry of the coil and the magnet. If a current i(t)is flowing through the coil, it creates a force F (t) on the diaphragm given by

F (t) = Bli(t) (24)

We have to include this force in the equation of motion for the diaphragm of a dynamicmicrophone. The resulting equation is (from (2))

Md2x(t)

dt2= Ap(t) −D

dx(t)

dt−Kx(t) −Bli(t) (25)

Applying the Laplace transform and rearranging, we have

AP (s) = s2MX(s) + sDX(s) +KX(s) +BlI(s)

= (sM +D +K/s)U(s) +BlI(s) (26)

The output voltage of the microphone V (s) is also related to the coil velocity and current

V (s) = BlU(s) − (sLc +Rc)I(s) (27)

where Lc and Rc are the inductance and resistance, respectively, of the coil. Of course,no current flows unless the microphone is connected to some kind of electrical load.Figure 7 illustrates (26) and (27) as an equivalent circuit with electrical analogs for themechanical components.

We will now develop a Thevenin equivalent of the dynamic microphone model shownin Figure 7. The open circuit voltage Voc(s) can be determined by setting I(s) to 0.

Voc(s) = BlU(s) =BlAP (s)

sM +D +K/s

=2δω0s

s2 + 2δω0s+ ω20

SpeakP (s) (28)

where δ and ω0 are defined in (4) and (5) and

Speak = BlA

D(29)

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Figure 8: Dynamic Microphone Thevenin Equivalent Circuit

is the peak, open circuit sensitivity of the microphone in V/Pa.The Thevenin impedance ZTh(s) can be found by setting P (s) to zero and calculating

the ratio −V (s)/I(s). The result is

ZTh(s) = Rc + sLc +(Bl)2

sM +D +K/s

= Rc + sLc + Zm(s) (30)

Note that the mechanical component of the Thevenin impedance Zm(s) has the form ofa parallel combination of capacitance Cm, resistance Rm and inductance Lm given by

Cm =M

(Bl)2

Rm =(Bl)2

D

Lm =(Bl)2

K(31)

Figure 8 shows the Thevenin equivalent circuit for our dynamic microphone model.We can simplify this model considerably if we assume that the diaphragm system is

sufficiently overdamped and the frequency is sufficiently close to the resonant frequencyω0. We can then approximate Voc(s) and ZTh(s) as

Voc(s) ≈ SpeakP (s) (32)

ZTh(s) ≈ sLc +Rc +Rm (33)

For lower frequencies we can simplify the equation for ZTh(s) further by retaining onlythe resistive terms.

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Example 4.1. For a certain dynamic microphone, the magnetic field is B = 1.5 T andthe coil length is l = 10 m. The diaphragm has a mass of 0.6 g and a radius of 1 cm.The resonant frequency and damping ratio of the diaphragm system are f0 = 800 Hzand δ = 5, respectively. Calculate the peak sensitivity of the microphone in mV/Pa.

Solution The formula for the peak sensitivity is

Speak = BlA/D

B and l are given and the diaphragm area in m2 is simply

A = πr2 = 10−4π

= 0.000314

We can use (5) to determine D.

D = 2δω0M

= 4πδf0M

= 4π(5)(800)(0.6)10−3

= 30.16

The peak sensitivity in mV/Pa is

S = 10001.5(10)(0.000314)

30.16= 0.156

Example 4.2. Plot the magnitude of the sensitivity of the dynamic microphone ofExample 4.1 as a function of frequency for Rc = 50 Ω and Lc = 500 µH. Assume thatthe microphone is driving a resistive 50 Ω load.

Solution The voltage across the resistive load RL is given by

VL(s) =Voc(s)RL

RL + ZTh(s)

so that the overall sensitivity is given by

S(s) =VL(s)

P (s)=

RLRL + ZTh(s)

2δω0s

s2 + 2δω0s+ ω20

Speak

where ZTh(s) is given by (30). This only missing parameter is K, which can easily bedetermined as

K = (2πf0)2M

The following MATLAB script puts everything together:

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B=1.5; l=10; r=0.01; M=0.6e-3; f0=800; d=5;

Rc=50; Lc=500e-6; Rl=50; N=100;

A=pi*r^2;

w0=2*pi*f0;

D=2*d*w0*M;

Speak=1000*B*l*A/D;

K=M*w0^2;

f=logspace(1,5,N);

w=2*pi*f;

S=zeros(1,N);

for n=1:N

s=1j*w(n);

Zth=Rc+s*Lc+(B*l)^2/(s*M+D+K/s);

H1=Rl/(Rl+Zth);

H2=2*d*w0*s/(s^2+2*d*w0*s+w0^2);

S(n)=H1*H2*Speak;

end

semilogx(f,20*log10(abs(S)));

grid on;

title(’Microhone Sensitivity’);

xlabel(’Frequency in Hz’);

ylabel(’Sensitivity in dBmV’);

Figure 9 shows the resulting plot.

The frequency response of our sample microphone is not very impressive. We canimprove the response by increasing the damping, but this comes at the expense of reducedsensitivity. The better dynamic microphone designs address this trade off by introducingresonant spaces in the microphone housing that boost both the low and high frequencyresponse.

5 The Loudspeaker

A loudspeaker includes the same elements as a dynamic microphone with the speakercone replacing the diaphragm. In fact, a loudspeaker can function as a microphone. Inthe case of application as a loudspeaker, however, the driving force is an audio signalvoltage v(t) and the output is an acoustic pressure p(t).

Figure 10, which is mostly a reversal of Figure 7, shows our equivalent circuit for aloudspeaker. The speaker drives an acoustic load impedance modeled by damping DL

and mass ML. The acoustic output PL(s) is given by

PL(s) = (DL + sML)U(s) (34)

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Figure 9: Dynamic Microphone Frequency Response

Figure 10: Loudspeaker Equivalent Circuit

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The circuit in Figure 10 yields the following pair of equations:

V (s) = (Rc + sLc)I(s) +BlU(s)

0 = −BlI(s) + (sMT +DT +K/s)U(s) (35)

where

MT = M +ML

DT = D +DL (36)

The solution to (35) is

I(s) =sMT +DT +K/s

(Rc + sLc)(sMT +DT +K/s) + (Bl)2V (s)

U(s) =Bl

(Rc + sLc)(sMT +DT +K/s) + (Bl)2V (s) (37)

We can use (37) to determine the input impedance Zin(s) of the speaker.

Zin(s) =V (s)

I(s)

= Rc + sLc +(Bl)2

sMT +DT +K/s(38)

Example 5.1. A loudspeaker has the following characteristics: B = 1.5 T, l = 10 m,Rc = 2 Ω and Lc = 800 µH. The mechanical subsystem has a resonant frequency off0 = 800 Hz and a damping ratio of δ = 5. The total mass of the speaker cone, includingthe effect of acoustic impedance, is MT = 5 g. Determine the speakers impedance at theresonant frequency.

Solution At the resonant frequency the impedance is given by

Zin = Rc + j2πf0Lc +(Bl)2

DT

The total damping DT (in nt-s/m) can be determined from the given mass, resonantfrequency, and damping ratio:

DT = 4πδf0MT

= 4π(5)(800)(0.005) = 251

Thus the resonant frequency impedance is given by

Zin = 2 + j2π(800)(0.0008) +(1.5 · 10)2

251= 2.90 + j4.02Ω

Note that the mechanical component contributes about 0.9 Ω to the total.

The responsiveness of the speaker PL(s)/V (s) can be determined from (34) and (37).

PL(s)

V (s)=

(DL + sML)Bl

(Rc + sLc)(sMT +DT +K/s) + (Bl)2(39)

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6 Microphone Directionality

In our analysis of microphones up to now we have assumed that the acoustic pressureexerts a force on only one surface of the diaphragm. A microphone where this is thecase is known as a pressure microphone. In the case of a gradient microphone, however,the acoustic pressure exerts a force on both surfaces of the diaphragm. The net forceon the diaphragm depends on the difference between the paths to the front and backof the diaphragm. Let us assume that the acoustic pressure on the front surface of thediaphragm is Pejωt. Then the pressure on the rear surface will be Pej(ωt−βd) where βis the wave number and d is the extra distance that the wave must travel to reach therear surface. Then the magnitude of the net acoustic pressure is given by

∆P = P (1 − ejβd) (40)

Recall that β = ω/c where c is the speed of sound. If d is small compared to thewavelength of the sound wave, then the net acoustic pressure can be approximated as

∆P ≈ jPβd (41)

This approximate equation points out a property of gradient microphones that maynot be immediately obvious from the exact version. Since β = ω/c, it seems that thepressure gradient increases linearly with frequency. That this is the case is illustrated inFigure 11, which is a plot of the magnitude of ∆P/P as a function of frequency usingboth the exact and approximate equations. The exact equation includes null pointsfor wavelengths equal to D, D/2, etc. This frequency dependence can be compensatedby electrical equalization but also by clever mechanical design. For example, if themechanical resonant frequency is low, then the roll-off of the resonant curve tends toflatten the response at higher frequencies. Even the design of the microphone’s screencan affect the frequency response.

Figure 12 is a (very) simple model of how an plane acoustic wave interacts with agradient microphone. The two circles represent access to the front and rear surfaces ofthe diaphragm. The internal distances from the access points to the respective surfacesare assumed to be equal. For a wave approaching the microphone at angle θ, the pathdifference is d = D cos θ where D is the distance between the two access points. The netacoustic pressure is then

∆P = P (1 − ejβD cos θ)

≈ jPβD cos θ (42)

Figure 13 is a polar plot of the magnitude of ∆P/P from (42) (exact version) for f = 1000Hz and D = 1 cm. The ”figure eight” shape of the plot indicates that the sensitivityof the microphone is greatest for sounds directed at the front or rear of the microphoneand least for sounds coming from the sides.

Figure 14 is a simple model of a gradient microphone and a point source, whichgenerates spherical waves. In this case, (42) becomes

∆P = P

(e−jβr1

r1− e−jβr2

r2

)(43)

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Figure 11: Frequency Behavior of Pressure Gradient (D = 1 cm)

Figure 12: Simple Model of a Gradient Microphone

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Figure 13: Gradient Directional Response (f = 1000 Hz, D = 1 cm)

Figure 14: Spherical Wave and Gradient Microphone

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Figure 15: Gradient Microphone Response to Spherical Wave (f = 1000 Hz, D = 1 cm)

It can be easily shown that r1 and r2 are related as

r22 = r21 +D2 + 2rD cos θ1 (44)

Figure 15 shows the response pattern for a gradient microphone to a spherical wavefor two values of r1. The pattern is still a figure eight, but it is no longer symetrical,especially when r1 is on the order of D. This result is an example of how the behaviorof a microphone can vary with the proximity of the source.

We can obtain a different sensitivity pattern by designing the microphone so thatsound entering the rear access point travels a greater distance to reach the diaphragm.Let this additional distance be αD. Then the path difference is d = D(cos θ + α) andthe net acoustic pressure is

∆P = P (1 − ejβD(cos θ+α))

≈ jPβD(cos θ + α) (45)

Figure 16 is a polar plot of the magnitude of ∆P/P from (45) (exact version) for f = 1000Hz, D = 1 cm, and various values of α. For α = 0 we of course get the figure eight

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pattern again, but for other values of α we get various cardiod patterns which indicatethat the microphone is more unidirectional - an often desireable feature.

6.1 Further Reading

For more detailed analysis of microphones and practical information about their use, seewww.viser.edu.rs/download.php?id=16489.

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Figure 16: Cardiod Directional Responses (f = 1000 Hz, D = 1 cm)

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