A question on v coil resistance (NOT impedance)

Fossil · 28th August 2002, 02:41 AM

For a voice coil with a DC resistance of 8 ohms, if I were to have a input of 1A (AC), is the 8W generated purely heat energy?

f4ier · 28th August 2002, 03:21 AM

No, when we're dealing with AC signals, we will always need to consider the impedance, not just the real (resistive) part alone. In this case, there will be real power and reactive power. The real power can be calculated from voltage*current*Cos x, where x is the phase difference between the voltage and current input.

hth

AndrewJ · 29th August 2002, 12:14 AM

f4ier, I think you missed the exact statement that Fossil made. He said for an input of 1A (ac). Therefore the reactive part, which is in series with the voice coil resistance, does not influnce the amount of heat generated in the voice coil. This is I^2*R, and so is 8W. Of course, there are other sources of dissipation, such as in the mechanical losses of the supension and cabinet, and in the air load resistance, but these are tiny compared to the voice coil.

Andrew

Ron E · 29th August 2002, 04:27 AM

efficiency of a speaker is Sound power out/electrical power in, which reduces to:

no=9.614*10^-10*Fs^3*Vas/Qes

100*(1-no) is the percentage of power that goes to heat, assuming the speaker to be a resistor. Since the speaker is not a resistor, the efficiency actually varies with frequency. The true efficiency curve roughly follows the impedance curve, but there is generally no need for you to know that - unless your barroom conversations involve large bets about arcane audio topics.

28th August 2002, 02:41 AM	#1
Fossil diyAudio Member Join Date: Sep 2001 Location: SG	A question on v coil resistance (NOT impedance) For a voice coil with a DC resistance of 8 ohms, if I were to have a input of 1A (AC), is the 8W generated purely heat energy?

28th August 2002, 03:21 AM	#2
f4ier diyAudio Member Join Date: Sep 2001 Location: Sydney	No, when we're dealing with AC signals, we will always need to consider the impedance, not just the real (resistive) part alone. In this case, there will be real power and reactive power. The real power can be calculated from voltagecurrentCos x, where x is the phase difference between the voltage and current input. hth __________________ Crossover/Subwoofer Simulator

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29th August 2002, 12:14 AM	#3
AndrewJ diyAudio Member Join Date: Sep 2001	f4ier, I think you missed the exact statement that Fossil made. He said for an input of 1A (ac). Therefore the reactive part, which is in series with the voice coil resistance, does not influnce the amount of heat generated in the voice coil. This is I^2*R, and so is 8W. Of course, there are other sources of dissipation, such as in the mechanical losses of the supension and cabinet, and in the air load resistance, but these are tiny compared to the voice coil. Andrew

29th August 2002, 04:27 AM	#4
Ron E diyAudio Member Join Date: Jun 2002 Location: USA, MN	efficiency of a speaker is Sound power out/electrical power in, which reduces to: no=9.61410^-10Fs^3Vas/Qes 100(1-no) is the percentage of power that goes to heat, assuming the speaker to be a resistor. Since the speaker is not a resistor, the efficiency actually varies with frequency. The true efficiency curve roughly follows the impedance curve, but there is generally no need for you to know that - unless your barroom conversations involve large bets about arcane audio topics.

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