What is the relationship between watts & volts?

[NormanB]

I know that 2.83 volts is one watt, but is 1.415 volts 0.5 watts or 0.283 v 0.1 watt? Is the relationship linear or logrithmic? Just wondering how to interperate my voltage reading at the speakers.

Thanks

[artto]

Watts (power)=volts (pressure) x amps (current)

[WMcD]

In speaker testing it is often assumed that the load is 8 ohms. In reality the electrical load presented by the speaker, i.e., its impedance, varies with frequency.

The formula used is: Electrical power (watts) = V*V/R where R is 8 ohms.

( We know that per Ohms law, Amps =V/R, so my equation is just like Art's, with that substitution.)

You can see that V must be square root of 8 to make things come out at one watt.

If V = square root of 2 (i.e. 1.414) then power is 2/8 or 0.25 watts.

Other voltages generating power into 8 ohms are:

2.828 volts gives 1.0 watt

0.89 volts gives 0.1 watt

0.282 volts gives 0.01 watt

0.089 volts gives 0.001 watt.

You can see the pattern. These all are based on that equation of electrical power and the 8 ohm assumption.

This can give us some handy ways of testing speakers.

We expect the K-Horn to give 104 dB of sound pressure level when fed 2.82 volts and the sound meter is one meter (39 inches) away. This is a actually a "voltage sensitivity", relating volts input to SPL output.

But, again, people will assume we have and 8 ohm speaker, and then we call it a power efficency of SPL out with one watt electrical power input, with sound pressure measured at one meter (39 inches).

If the 104 dB/watt/meter speaker is fed with 0.89 volts, we expect the SPL of 94 dB.

If fed with 0.282 volts, we expect the SPL of 84 dB.

If fed with 0.089 volts, we expect the SPL of 74 dB.

It is a demonstration that various voltages result in 10 dB lower sound pressure.

We often see the equation relating dB and a voltage ratio. It is

dB = 20 log (V1/V2)

Let us try that with two voltages used above

dB = 20 log (2.828/ 0.089) = 20 log (31.77) = 20 * 1.5 = 30

If we look back at the example, the SPL output does indeed rise or fall 30 dB

Best,

Gil

Note: I edited the post for the sake of clarity and one error.

[NormanB]

Art & William,

Thanks so much! That is exactly what I needed. I knew I could count on you Klipscheads.

[WMcD]

I don't want to edit the above after it has been up a while. It is sort of like rewriting history.

I should point out that the power ratio is implied above.

That is

dB = 10 log (P1/P2)

Note we've changed the 20 to a 10 because here we're talking about power ratio rather than voltage. Voltage is easy to measure, but we're often interested power. That may have been the original question posed.

It takes either some fuzzy logic or a knowledge of math. However, multiplying a log by 2 is the same as squaring the operand, i.e. the thing in the parenthesis. So in the relation with the 20 rather than 10, we're squaring the voltage by some deeper, mathematical function of the log.

I had written it as V*V. That is the same as V^2 which is also written as V with a flying small 2.

This is a bit of Math 101 but perhaps some people will be entertained.

The square root is just saying, if we have Y, what is X, if Y=X*X. You can put a Y number into a calculator and ask for a square root, which will give you Y. Square root of 8 is 2.828. And 2.828 * 2.828 = 8.0. So in the P=V*V/8, we get 8/8=1 watt.

The log thing may be a little strange to non-math majors. But with some numbers it is easy to see.

Essentially we're saying, what number is it that 10 can be raised to the power of to get our result. English doesn't say it well. Consider.

10^ -3 = 0.001

10^ -2 = 0.01

10^ -1 = 0.1

10^ 0 = 1.0

10 ^ 1 = 10.0

10 ^ 2 = 100.0

10 ^ 3 = 1000.0

To a large extent, we're counting how the decimal point moves assuming there is a decimal point just to the right of the 1.

Looking back at our dB stuff and logs we see if the power ratio is 1000, then log (1000) = 3 and we mulitiply that by 10 to get +30 dB.

And very remarkably going the other way, if the ratio is 1/1000 or 0.001, the log is -3. We multiply that by 10 to get -30 dB. You can see the symetry. Also, because of logs, we can fool around with very big ratios in a different way.

The only other ratio which comes up so often is when power is reduced to one-half of what we started with as a reference.

dB = 10 log (0.5/1.0). The power ratio in the ( ) could be anything which equals 1/2 = 0.5.

So, what is the value of log(0.50)? It is -0.3. And going the other way, the log of 2 is +0.3. Try it on the calculator. (We used to do it with log tables and slide rules.)

Looking back at the power ratio and dB, we find that we have to multiply the + or - 0.3 by 10 to give us + or - 3 dB. This expresses doubling or halving of power. (Yeah sticklers will point out there are some numbers making it not exactly 3, but close enough.)

So, you say, where does the 3 dB stuff come up that I have read about?

First: It comes up in typical speaker frequency response. It is just about some frequency range where the speaker is working reasonably well. And we are told the frequency response range is, say 35 to 17,000 Hz, "plus or minus 3 dB."

This means that once an average output is determined, the acoustic power output, over the range, doesn't get more than twice as much, or half as much. So it defines the high end and low end of the response if we demand it be in a (total) 6 dB range or window. Essentially, the plus or minus 3 dB acknowledges the response is not totally "flat" but the variations are not too bad, either.

Second: We're interested in giving some frequency number in Hz where our woofers start to not doing a good job. That is called F3. That nomenclature gives us a hint of what it means. The frequency response in SPL is 3 dB down, -3 dB, from the average higher up in frequency.

Dwelling on it, at 200 Hz we might have 104 dB SPL (with one watt input) but at 35 Hz (still with one watt input, which is an electrical input of 2.828 volts, remember?), it is 101 dB SPL. So that is our F3. Or F3=35 Hz.

It has been a long trip. However, I did want to point out that all these math items are related, we read about them, and it is not too difficult to understand if everything is set down in one spot.

There is much more to the use of logs, exponentials, and dB in hour hobby. (We haven't spoken on perceived loudness. We haven't spoken about exponential horns.) But I've tried to make some inroads here.

Best,

Gil

[D-MAN]

Like Gil said... he's right.

I have one more thing to add:

technically wattage is a rating for heat dissipation arising from electrical current encountering resistance.

DM