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Sine wave into wattage

  1. Aug 24, 2015 #1
    So my question is about calculating wattage with a sine wave.

    So with speakers I always just thought of it as the basic vi=w
    So a sine wave of 12v through 4 ohms will produce three amps and thus 36watts.

    But when you put a square wave and use pulse width modulation on something like an led or fan you can simulate a different voltage and then come out with a different wattage.
    I know that sine waves acts something like a 70% duty cycle square wave, so would 12v into 4 ohms actually give me 17.6 watts on a speaker instead of the 36 that I previously calculated? (or anything else for that matter)?
     
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  3. Aug 25, 2015 #2

    meBigGuy

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    Supplying a 0 to 12V 50% square wave into a resistive load of 4 ohms will give you E^2/R watts half the time. (144/4)/2 = 36 watts/2 = 18 watts.

    supplying a +- 12V squarewave will give you 36 watts.

    Supplying a +6 -6 voltage square wave will give you 36/4 watts continuous = 9 watts.

    Supplying a + - 6 sine wave will give you 4.49 watts

    +- 12V sine wave will be 18 watts.

    But, when you PWM you reduce power by the output filter, and by the duty cycle. If you end up with a +- 12V sine wave after PWM it is the same answer as a +- 12V sine wave ...., 18 watts
     
    Last edited: Aug 25, 2015
  4. Aug 25, 2015 #3

    anorlunda

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    There is a big caveat on all of this, is the 12v a peak-to-peak measurement or RMS?

    Most voltmeters will measure RMS voltage for most waveforms sine or other, not peak-to-peak. Ditto for AC amp meters, most will measure RMS current. For a pure sine wave, the RMS voltage is 0.707 times the peak-to-peak voltage.

    Regardless of the waveform, RMS power = RMS voltage * RMS current for purely resistive loads. That is what makes RMS so popular.
     
  5. Aug 25, 2015 #4
    By convention AC (sine) voltages and currents are measured in their DC equivalent value, the rms (for root mean squared) and not their peak-peak.

    Vrms=0.707Vpp. (That's 1/√2, BTW.)

    This value changes with pulse width modulation -- by a lot. You will need to integrate both the voltage and the current (squaring, taking the mean, then the root; considering phase differences, etc.) when finding the values like power. Everything is set up for sine waves. There are power factors and all kinds of conventions that are designed for them. But pulse width modulation isn't the convention, so it's more work.

    Music is also different. The power divides over the frequency spectrum. So it's not as simple as assuming a sine wave. I would tell you more, but here be dragons (as far as I know anyway).
     
  6. Aug 25, 2015 #5

    sophiecentaur

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    Do you really mean that or do you mean they 'assume' it's a sine wave? It's fairly hard to measure accurate RMS for a fast waveform unless you use a thermal based method. Not long ago, all that a bog standard RMS meter did was to rectify and measure peak volts - then multiply by 1/√2.
     
  7. Aug 25, 2015 #6

    anorlunda

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    n
    Yes, I do mean it. Actually I was thinking of analog voltmeters, or gavananometers as they used to be called. Becuase the needle has intertia, they are quite incapable of vibrating with the frequency of AC. Simply via the averaging effect of intertia, they inherently measure RMS.

    Actually, measuring peak voltage to do mathematical manipulations as you describe is much more difficult and advanced.
     
  8. Aug 25, 2015 #7

    sophiecentaur

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    A Galvanometer was a particular form of very sensitive current meter and should not be confused with a milli or even micro Ammeter. The return spring on a Galvo was very weak, which allowed the instrument move with tiny curents. In the case of a ballistic galvonometer, its deflection could depend on the Charge that flowed during a short pulse than on the actual current.
    If you feed AC directly to a meter movement, however sensitive, the average position of the needle is Zero. A half wave or full wave rectifier will keep the needle on one side and it will end up at an 'average' position on the scale. However, that average is not an RMS value it will be the mean of the magnitude of the current value (no 'squaring' of the waveform values). (The Maths are totally different) There is another problem with such meters and that is the voltage drop across the diodes will 'subtract' a small voltage from the voltage across the meter coil. You can correct for all this, to some extent, by using a suitably doctored scale on the meter. That scale will only work over a certain range of values and frequencies, because of the dynamics. A 'good' meter would give a very good measure of the RMS value of a good quality Sine wave (Mains frequency) but it's all a bit of a fudge and it all goes out of the window, once the waveform is non-sinusoidal.
    There is an excellent analogue / passive method in which you look at the temperature of a resistor which is actually dissipating a 'sample' of the power in the circuit being tested. It works for any waveform (within the limits of the frequency response of the system). Problem is that it is not very sensitive.
    Posh DVMs will work on samples of the waveform and give as good an accuracy as you're prepared to pay for.

    BTW, IRMS2R and VRMS2/R are also used for Power.
     
    Last edited: Aug 25, 2015
  9. Aug 25, 2015 #8

    anorlunda

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    In all those years before we had digital, we had lots of AC voltmeters. For exampled the one below labeled RMS volts.

    60079.jpg
    Or this cheap one.

    http://www.allmobilerepaircourse.com/wp-content/uploads/2014/12/10.jpg [Broken]
    I suppose they could be thermal. If the thermal element was small enough, response would be fast. But were they not more likely done using the circuit below and a suitably fudged scale?

    02225.png
     
    Last edited by a moderator: May 7, 2017
  10. Aug 25, 2015 #9

    anorlunda

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    There are many ways to skin this cat.

    Wikipedia says the following about analog voltmeters. https://en.wikipedia.org/wiki/Voltmeter


    Moving-coil instruments with a permanent-magnet field respond only to direct current. Measurement of AC voltage requires a rectifier in the circuit so that the coil deflects in only one direction. Moving-coil instruments are also made with the zero position in the middle of the scale instead of at one end; these are useful if the voltage reverses its polarity.

    Voltmeters operating on the electrostatic principle use the mutual repulsion between two charged plates to deflect a pointer attached to a spring. Meters of this type draw negligible current but are sensitive to voltages over about 100 volts and work with either alternating or direct current.

    VTVMs and FET-VMs[edit]
    The sensitivity and input resistance of a voltmeter can be increased if the current required to deflect the meter pointer is supplied by an amplifier and power supply instead of by the circuit under test. The electronic amplifier between input and meter gives two benefits; a rugged moving coil instrument can be used, since its sensitivity need not be high, and the input resistance can be made high, reducing the current drawn from the circuit under test. Amplified voltmeters often have an input resistance of 1, 10, or 20 megohms which is independent of the range selected. A once-popular form of this instrument used a vacuum tube in the amplifier circuit and so was called the vacuum tube voltmeter, or VTVM. These were almost always powered by the local AC line current and so were not particularly portable. Today these circuits use a solid-state amplifier using field-effect transistors, hence FET-VM, and appear in handheld digital multimeters as well as in bench and laboratory instruments. These are now so ubiquitous that they have largely replaced non-amplified multimeters except in the least expensive price ranges.

    Most VTVMs and FET-VMs handle DC voltage, AC voltage, and resistance measurements; modern FET-VMs add current measurements and often other functions as well. A specialized form of the VTVM or FET-VM is the AC voltmeter. These instruments are optimized for measuring AC voltage. They have much wider bandwidth and better sensitivity than a typical multifunction device.
     
  11. Aug 25, 2015 #10
    All i need to know is if i read (for example) 12vac across my speaker and lets say 3 amps of ac with a clamp meter.

    What calculations do i need to do in order to find my wattage?
     
  12. Aug 25, 2015 #11

    anorlunda

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    Your answer is in post #3
     
  13. Aug 25, 2015 #12

    sophiecentaur

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    But you stated that the system "inherently" measures RMScurrent. It doesn't. It measures the Mean modulus of the current. The scale on those meters would have been calibrated to indicate the probable value of RMS on the assumption of a sinusoidal waveform. If you look at the waveform of the Mains Voltage, it is seldom a very good sine wave so an 'old fashioned' analogue meter would give an inaccurate result even for a simple case like that.
    Reading what is written on the front of the meter is not enough. You need to read the spec sheet and the caveats about operating conditions if you want to be sure of the accuracy.
     
  14. Aug 25, 2015 #13

    anorlunda

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    You're right. Strike inherently.
     
  15. Aug 25, 2015 #14
    Get a scope, trace the waveforms and integrate.

    I've heard horror stories (not sure if they are true) of audio systems where each frequency component carried the full rated load. Then the power amps blew.

    I'm not sure how a broad band audio signal would show up on a random meter. It could seriously underestimate the power levels if it handled frequency poorly.
     
  16. Aug 25, 2015 #15
    I do have a digital scope, so should i put the scope on the speaker, find the amplitude value of my sine wave, multiply that by 0.7, then multiply that by the current i read on my digital clamp meter?
     
  17. Aug 26, 2015 #16

    sophiecentaur

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    Looking at Jeff Rosenbery's post (above your's) you seem to be saying that Integration is the equivalent of multiplying by 0.707.
    The 1/√2 factor applies only to sine waves. Is that stated clearly enough for you? RMS is the Root of the Mean of the Squares of all the values over the period of the waveform (or over a sufficiently long period that is doesn't matter where you start of stop). That is the average of the Powers of all of the elementary sections of the waveform.
     
  18. Aug 26, 2015 #17
    Ok that eleborates a bit better.
    My point of this thread is to test amplifiers wattage output at clipping.
    So i need to take my scope, put it across my speaker, play a 60hz pure sine wave (60 incase clamp meter is best at wall socket frequency), read my clipping voltage, multiply by .707, and multipply that by my current?
     
  19. Aug 27, 2015 #18

    meBigGuy

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    That would tell you the maximum rms power for a 60 Hz sine wave into that particular load.

    You also should put an AC meter across the speaker and verify you get the same voltage as you calculated from 0.707 * peak voltage (not pk-pk)
     
  20. Aug 27, 2015 #19

    sophiecentaur

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    That's fine as long as you don't exceed clipping. If you drive it hard enough to get a near square wave, the RMS would be the same as the peak.
    If you are short of a test waveform generator, there are websites that will give you waveforms at a range of frequencies. I remember long ago, finding something of the sort that was very handy (and free!)
     
  21. Aug 27, 2015 #20
    If it is a speaker system, that isn't enough unless you are only using it for that sine wave.

    Speaker impedance varies with frequency. Voltage and current relationships vary as well.

    But all of this depends on how you will use the data. If it's simple bragging rights, those are 50W speakers. Because I said so. If you are doing an EM noise analysis for an FCC mask and you are near the limit, you will need to be more careful.

    There is usually no reason to need that sort of precision.
     
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