Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reducing the Power supplied to a Car audio speaker

  1. May 28, 2009 #1
    Hey guys, I'm not an Electrical Engineer, which is why I'm here asking for your help!

    I have a car audio system I am building and installing my self, and I have one specific question. I am planning on having a 60 watt (RMS), 2 ohm speaker powered by a 90 watt (RMS @ 2 ohms) amplifier, and I would like to eliminate the extra 30 watts of power, so that I don't blow the speaker. Through my college physics class, I know that P=E*I, P=I^2*R, and P=E^2/R. Through my math, operating under the assumption that E is approx 13.6V, I find that I need a 6.2ohm resistor.

    HOWEVER, there are several other assumptions I am operating under that I want to check if I am right or not.
    1. The change in power I am looking for can be solved using the P=E^2/R equation.
    2. The resistance of the speaker doesn't need to be taken into account, since it is all ready accounted for in the RMS Power rating of the speaker.

    Can anyone clarify this for me/tell me where I'm wrong/etc??? I would appreciate it very much if it were explained to me so that I can calculate this stuff on my own! Thanks in advance for the help!
  2. jcsd
  3. May 28, 2009 #2


    User Avatar
    Science Advisor

    You can't drop the power with a resistor. Well, you can, but you shouldn't.

    It is important to have the low impedance of the amplifier output across the speakers.
    This provides a damping effect and stops the speaker voice coil from ringing. This effect sounds awful.

    So, do not put any resistors in that line.

    The point here is that the amplifier will not produce 90 watts unless you drive it that hard.
    Even 10 watts in a car is really LOUD so you would have to know if you are anywhere near 60 watts.

    Just keep the volume to a reasonable level and your speakers will be fine.
    So will your ears. These sound levels are very destructive for your hearing.
  4. May 29, 2009 #3
    I agree with everything vk6kro said, however I would like to add a few things.

    Let's look at your math. If you get 90 W out with a 2 ohm speaker then the RMS voltage to the speaker is P = E^2/R; P*R = E^2; sqrt(P*R) = E; sqrt(90*2) = E; E = 13.4 V.

    You want to dissipate 30 W, so you calculated R = P/E^2; R = 1/(30/13.6^2) and got 6.2 ohms. This is not correct.

    If you then add the 6.2 ohms to the 2 ohm speaker you have 8.2 ohms altogether. The maximum power out you would get is 13.4^2/8.2 = 22 W and 2/8.2 of that would be in the speaker. You would have a maximum output of 5.4 watts.

    The simple answer is if you want 60 W through the speaker then the current through that speaker is sqrt(60 W/2 ohms) = 5.5 Amps. To get 5.5 Amps you need a total resistance of 13.4 Vrms/5.5 A = 2.43 ohms. Your resistor would have to be 0.43 ohms at a minimum rating of 15 W. Good luck.

    Other than do what vk6kro suggested, which is good advice, you might substitute 3.2 ohm speakers for the 2 ohm ones which would give you a max output of 56 W.
  5. May 29, 2009 #4


    User Avatar

    Staff: Mentor

    He could use a 3-resistor "T" pad to reduce the drive signal, while still matching the output impedance of the amp and the input impedance of the speaker....
  6. May 30, 2009 #5


    User Avatar
    Science Advisor

    If you did have 90 watts feeding into a resistor and a 60 watt 2 ohm speaker in series:

    Speaker gets 60 watts.

    Current in the speaker = sqrt ( 60 /2 ) = 5.477 Amps

    This current flows in the resistor and it dissipates 30 watts

    So R = 30 Watts / (5.477 * 5.477) = 1 ohm
  7. May 30, 2009 #6
    The difference is so small it's hardly worth quibbling over.

    If 90 watts can be dissipated in a 2 ohm speaker then the voltage required voltage is sqrt(90 * 2) or 13.416 V RMS.

    If you say you can add a 1 ohm resistor to the load and still get 90 watts out total means the voltage must have increased. sqrt(90 * 3) = 16.432 V

    This may in fact be true if the power supply is not well regulated but its not part of the problem as given. Assuming the available voltage is constant, adding 0.43 ohms to the speaker and keeping the voltage constant results in a total power available of (13.416^2)/2.43 = 74.074 watts. Of that total power, 2/2.43 * 74.074 = 60.996 watts is dissipated in the speaker and 0.43/2.43 * 74.074 = 13.108 is dissipated in the resistor.

    The amount of resistance that needs to be added is so small that using a little extra speaker wire may be enough especially if it is of a smaller gauge than normally used.
  8. May 30, 2009 #7


    User Avatar
    Science Advisor

    Yes, but if the amplifier could deliver the same power into a 3 ohm load, then using a smaller resistor would not give enough protection, would it?
    They don't specify, so you need to assume the worst.

    It would just depend on how close to the amplifier supply rails the output voltage was swinging, whether it had any spare voltage. As long as it could still deliver 5.5 amps, it could produce 90 watts into 3 ohms.

    As you say, though. Not worth worrying about.

    The proper solution here is to put speaker protection circuitry in line or to just keep the volume down.
  9. May 31, 2009 #8


    User Avatar
    Science Advisor
    Gold Member

    Inserting a resistor in series with a speaker is a common trick done as an attempt to get a tube amplifier sound.
  10. May 31, 2009 #9
    I've done this many times on shortwave receivers and ham radios. But not for the reason you stated. I did it to improve the signal to noise ratio. If you put your ear to the speaker of a small radio, with the volume turned all the way down, you will hear a hissing sound. This can be annoying when you are in a very quiet environment. By placing a resistor in series with the speaker you lower the overall output so that the internal noise produced by the amplifier is inaudible. It can result in a much more pleasing sound. Of course you will not be able to get the same power output, but if you're in a quiet environment, it doesn't matter. And you get the same effect regardless of whether the output is solid state or tube type.

    But this probably would not be practical or necessary for high quality, high power audio systems. If I had the same situation as the OP I would just use the speakers as is and not crank the volume all the way up.
  11. May 31, 2009 #10


    User Avatar
    Science Advisor

    I am amazed that 90 watt amplifiers are available for cars. I have heard 100 watts in a large lecture theatre and it was deafening. We had to try it though ! :)

    That would be 90 watts per channel too and also 90 watts into 4 or 8 ohms.

    Just like you can't legally buy cocaine because it is harmful for you, 90 watt amplifiers for cars should not be available either.
    The sound levels would be something like 130 dB at least.

    Still, if anyone is silly enough to buy them.....
  12. Jun 2, 2009 #11
    I apologize for being practical, but look on the amplifier for a small screw which adjusts the amplification. Every amplifier has one. Turn it about 66% of full amplification.
  13. Jun 2, 2009 #12


    User Avatar
    Science Advisor
    Gold Member

    Lot's of people buy them. 90 watts is nothing. Class D amps go WAY up from there. Mostly to power subwoofers, but still. Making a certain power car audio amp illegal is a slippery slope. Not only that, it isn't too difficult to series them up and quadruple the power. People have done it. The car audio community is nothing less than obsessed.
  14. Jun 2, 2009 #13


    User Avatar
    Gold Member

    I think he would need a full Z 'T' network (i.e. a transformer). A resistor T could be made to change the apparent source power and load, i.e make the speaker look like a specified wattage load to the amp and vice versa, but it would not match not the impedance, which is another way of stating vk6kro's comment that adding just resistors will degrade the sound. Still, the T would be better than just a series resistor.
  15. Jun 2, 2009 #14
    Perhaps you could do our requester a favor and find him a suitable transformer part number and also calculate the resistor values and power ratings for the network.

    Frankly it seems like a lot of time and expense when he really needs nothing more than a 1/2 ohm of speaker cable between the speaker and amp.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook