Back EMF in Average Loudspeaker Coil: Compare to 18V

In summary, the coil in a loudspeaker has an inductance of L = 56uH and a sound frequency of 20 kHz. The average back emf induced in the coil during the current variation is 9.856 V, which is higher than the applied emf of 18V. This can be calculated using the formula emf = L(ΔI/Δt) and taking into consideration the duration of one half period. However, the question is not clearly stated and could potentially have multiple interpretations.
  • #1
David Truong
9
0

Homework Statement


The coil in a loudspeaker has an inductance of L = 56uH (or 5.6 x 10^-5 H). To produce a sound frequency of 20 kHz, the current must oscillate between peak values of +2.2 A and -2.2 A in one half of a period. What average back emf is induced in the coil during this variation? How does this compare to the applied emf of 18V?

Homework Equations


T = 1/f

emf = L(ΔI/Δt)

L = N(ΔΦB/Δl)

The Attempt at a Solution



back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/Δt]

My problem here is I do not know where I can obtain Δt. I am assuming that it has something to do with the frequency of 20kHz in one half period (1/2T).

T = 1/f --> 1/2T = 1/f --> 2/f -- > T = 2/20000 = 1.0 x 10^-4 secs

If this is the case then, by subbing, Δt = 1.0 x 10^-4 secs, into the above equation, I get:

back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs]
= (5.6 x 10^-5)[4.4/1.0 x 10^-4 secs]
= (5.6 x 10^-5)(44000)
= 2.464 V

Is this correct?
 
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  • #2
You are nearly there. But remember what happens in one period. It says it goes from +2.2A to -2.2 in a half period. How many half periods in a whole period? How many whole periods per second to get a frequency of 20 KHz? So how many half periods per second?

If I give you 100 glasses of water per second, then it takes me 1/100th of a second to give you a glass of water. But how long does it take me to give you a half glass of water?
 
  • #3
DEvens said:
You are nearly there. But remember what happens in one period. It says it goes from +2.2A to -2.2 in a half period. How many half periods in a whole period? How many whole periods per second to get a frequency of 20 KHz? So how many half periods per second?

If I give you 100 glasses of water per second, then it takes me 1/100th of a second to give you a glass of water. But how long does it take me to give you a half glass of water?

Okay, so if I am considering a whole period than my currents are actually +4.4 and - 4.4. Additionally, my period is actually just T = 1/f = 1/20000 = 5.0 x 10^-5.

back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5 secs]
= (5.6 x 10^-5)[8.8/5.0 x 10^-5 secs]
= (5.6 x 10^-5)(176000)
= 9.856 V
 
  • #4
What is the maximum current? (2.2A) ... but how long does it take for the +2.2A to become -2.2A ?
 
  • #5
David Truong said:
back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs]
That would be correct if the current slides uniformly from one extreme to the other. But it's a sinewave (well, I presume it is) so di/dt is changing all the time.

You could try using L.di/dt at the moment the sinewave crosses the time axis, because that's where di/dt is a maximum so it's where the voltage will peak, and it's simply the peak voltage that you're after. You can do calculus? See what you get for an answer using that approach.

Once you've solved it this way, we'll look at another method. ;)

EDIT: Hmmm, does the question really ask for the "average back emf"??
 
  • #6
Yes, the question does explicitly ask for the average emf during this half-cycle. (centered on the maximum emf)
... we know how V_RMS relates to V_average and V_max .
 
  • #7
This question is sufficiently ill-explained that, were it an exam question, I'd accept either of two analyses: either with the current ramping linearly from +2 to -2 (i.e., a triangular wave), or with the current following a sinusoid. It is just not clear to me which the examiner intends.

Is there any detail you have left out that could decide what is intended? Have you examined a problem like this in class?
 
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Likes DEvens
  • #8
lightgrav said:
Yes, the question does explicitly ask for the average emf during this half-cycle. (centered on the maximum emf)
... we know how V_RMS relates to V_average and V_max .
Can you attach a sketch of the waveforms you are expecting?
 
  • #9
NascentOxygen said:
This question is sufficiently ill-explained that, were it an exam question, I'd accept either of two analyses: either with the current ramping linearly from +2 to -2 (i.e., a triangular wave), or with the current following a sinusoid. It is just not clear to me which the examiner intends.

Is there any detail you have left out that could decide what is intended? Have you examined a problem like this in class?

No I didn't leave anything out, the question is copied verbatim from the question sheet I was given.
 
  • #10
Was this answered correctly? 9.8 V?
 

Related to Back EMF in Average Loudspeaker Coil: Compare to 18V

1. What is back EMF in average loudspeaker coil?

Back EMF (electromotive force) is a phenomenon that occurs in an average loudspeaker coil when the current flowing through it is suddenly interrupted. This interruption creates a reverse voltage that opposes the flow of current, which is referred to as back EMF.

2. How does back EMF affect the performance of a loudspeaker?

Back EMF can have a significant impact on the performance of a loudspeaker. It can cause distortion in the sound produced, particularly at higher volumes. It can also affect the efficiency of the loudspeaker, as the back EMF can require more power to overcome and maintain the desired sound output.

3. What is the relationship between back EMF and the voltage applied to the loudspeaker?

The back EMF in an average loudspeaker coil is directly proportional to the voltage applied to it. This means that as the voltage is increased, the back EMF will also increase. This is why it is important to carefully consider the voltage used when designing or using a loudspeaker.

4. How does back EMF compare in a loudspeaker with 18V versus one with a lower voltage?

Generally, a loudspeaker with a lower voltage will have a lower back EMF compared to one with a higher voltage. This is because the higher voltage will result in a stronger current flow, which in turn creates a higher back EMF. However, other factors such as the design and components of the loudspeaker can also play a role in the back EMF.

5. Can back EMF be controlled or minimized in a loudspeaker?

Yes, there are various techniques that can be used to control or minimize back EMF in a loudspeaker. These include using a crossover network, which divides the audio signal into different frequency ranges and sends them to different drivers, and using a feedback circuit to reduce the effects of back EMF. Additionally, careful design and selection of components can also help to reduce back EMF.

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