Student Neighbour Problem

[kuiperbelt]

A low frequency amplifier has a power gain of 50 dB. The input of the amplifier is 800 ohms resistive. The output is arranged for a 8 ohm loudspeaker. What will be the rms current in the speaker if a 1 volt rms signal is applied to the input of the amplifier?

Given that the loudspeaker is a passive device rated at 50 watts with a sensitivity of 87 dB, find the following:

a.) The increase in Amplifier gain to meet the loudspeaker rating.

b.) With the new amplifier gain, what will be the sound pressure level (SPL) at a distance 10 metres from the speaker?

c.) How will this change in SPL be perceived by those stood 10 metres away?

d.) The sound pressure reference in air is 2 x 10-5 N/m^2, what is the sound pressure at 10 metres. Express your answer in pascals?

[Hymie]

A 50dB power gain is a (power) gain of 100,000 times – but power gain is normally based on input and output impedances being equal.

1V rms input will dissipate 1.25mW into an 800 ohm impedance; with a 50dB power gain the output power will be 125W.

If the 50dB is an amplitude (voltage) gain, the signal output will be 316V rms.

The question might make more sense if the amplifier input impedance is 8k ohm.

[Hymie]

or the amplifier power gain is 40dB (10,000 times).

[ZebuSanctuary]

Answer to the first part;

Input power to Amplifier = V²/R = 1²/800 = 1.25 mWatts.

Calculating output power ---> 50 dB = 10log (Po/Pi)

Where Po is the output power and Pi the input power.

Po = 1.25 x 10 x antilog (50/10) = 125 Watts

Since P = I²R, current in loudspeaker is given by √(P/R);

√(125/8) = 3.95 Amps

This question is flawed for the following reasons;

Loudspeaker rated at 50 Watts. Output power 125 Watts. That loudpeaker coil is going to start cooking pretty quickly.

Part a.) asks for the increase in power gain of the Amp. Surely this should be an attenuation?

Ask the student to clarify.

[kuiperbelt]

Sorry guys, my mistake. It's a 500 Watt loudspeaker!!

[ZebuSanctuary]

Continued from last post.

a.) Doubling the power equates to +3dB. Therefore the amplifier gain will have to increase by 6dB to 56dB. (125W - 500W)

b.)b.) Since SPL is inversely proportional to the square of distance;

SPL α 1/d² then the SPL will be reduced by a factor of 1/10² = 1/100

dB = 10 log 1/100 = -20db (attenuation due to distance at 10 metres from loudspeaker)

Sensitivity of loudspeaker = 87 dB i.e 1W/1m

With 56dB powergain, SPL = (87 + 56)dB = 114dB (500 W/1m)

At a distance 10 metres away, SPL = (114 - 20)dB

= 94dB

c.)To a 'normal' listener stood at 10 metres the perceived loudness will seem to have diminished by a factor of 4.

Although loud it is not 'deafening' as if stood 1 metre away.

d.)Sound Pressure at 10 metres away from loudspeaker;

94 dB = 20 log (SP/0.00002) ----> Make SP subject of the formula

SP = 0.00002 x Antilog 94/20 = 1.0023

Sound Pressure = 1 pascal

Albeit subjective, this is the equivalent to being a few feet from a car horn blast for a 'normal' listener.