chrz@ihlpf.ATT.COM (Chrzanowski) (07/08/88)
>>>> >>> ...12 volts driving a 4 ohm load yields 36 watts RMS... >>> ...The unit bench tested at 35 watts RMS which is about what I was >>> hoping for. >> >> wouldn't that be closer to 25W: 12**2/4 = 36W peak, 36 * 2**(-0.5) ~= 25W. >> i'd expect a sharp change in the price/power curve where the power hits >> (an honest) 25W, because of the need for a voltage-boosting gizmo. >> >Try dividing by 2 and then divide by the square root of 2 for RMS. [Correct] >Sine wave is + and - so that gives you + and - 6 volts peak to peak. [No. It can vary from zero to +12, that's 12 volts peak TO PEAK] >6 / 1.41 = 4.24 V RMS which yields ~4.5 Watts into 4 ohms. >Since cars are usually above 13 Volts you get slightly more than >5 watts per channel. If you want more than 5 watts you need to To convert volts peak to peak (VPP) to RMS (assuming a sinewave) >Try dividing by 2 and then divide by the square root of 2 Soo, If we've got 12 VPP into 4 ohms then RMS power is (12volts * 12volts) / ((2*(2**.5) * (4ohms)) = 12.7 watts. Of course, to actually deliver 12 volts to the load with a twelve volt supply the amp's output impedance would have to be zero. In fact, with a nominal 13.6 volt source this (~12 watts) is about the max you will get from such an amp WITH mucho distortion, as the output transistors will be at or near saturation (don't forget that a current limiting/sensing resistor is in series with the transistors). Actually, many modern car stereos use "bridged" amps. This means there are actually TWO amps driving each speaker: the amps are (180 degrees) out of phase; each is connected to one terminal on the speaker. Remember that warning not to connect ANY speaker wires to chassis ground? With a bridged amp, you can get (theoretically -- assuming the amp's output impedance is zero) FOUR times the power into the load: with 12VPP out of each amp, that's 50.9 watts into 4 ohms. To visualize how this happens, let's call the "+" speaker terminal V2 and the "-" speaker terminal V1. An oscilloscope connected from EITHER V2 or V1 to GROUND will see 12VPP -- but a 'scope connected BETWEEN V2 and V1 will see 24VPP! Try drawing it on paper. V2 is a sine wave starting at T=0, V1 is out of phase. At t=1/4 wave, V2 = +12, V1 = 0 and (V2-V1) = +12. At t=3/4 wave, V2 = 0, V1 = +12 and (V2-V1) = -12. BTW, with a bridged amp driving a 4 ohm load, EACH amp in the bridge will "see" a 2 ohm load. One problem with DC-to-DC converters (other than cost and size) is that they are power pigs: efficiency goes down in a hurry when not delivering max power, as is normally the case with a stereo. Remember when the U.S.A. had a consumer electronics industry? Oh, well ...
guido@twitch.UUCP ( G.Bertocci) (07/11/88)
In article <5244@ihlpf.ATT.COM>, chrz@ihlpf.ATT.COM (Chrzanowski) writes: | >I wrote: | >Sine wave is + and - so that gives you + and - 6 volts peak to peak. My apologies, peak to peak is not the right term. I was indicating that the sine wave would be 6*sin(x) volts. As opposed to 4.24*sin(x) volts rms. | | [No. It can vary from zero to +12, that's 12 volts peak TO PEAK] | | >6 / 1.41 = 4.24 V RMS which yields ~4.5 Watts into 4 ohms. | >Since cars are usually above 13 Volts you get slightly more than | >5 watts per channel. If you want more than 5 watts you need to | | To convert volts peak to peak (VPP) to RMS (assuming a sinewave) | >Try dividing by 2 and then divide by the square root of 2 | Soo, If we've got 12 VPP into 4 ohms then RMS power is | | (12volts * 12volts) / ((2*(2**.5) * (4ohms)) = 12.7 watts. Since your are squaring the voltage you need to square the part under the division sign too. (12volts * 12volts) / (((2*(2**.5)**2) * (4ohms)) = 4.5 watts. -- Guido Bertocci ...!ihnp4!twitch!guido AT&T Bell Labs Holmdel, NJ