how do you know which one to use when calculating power in watts?
P=I*E
P=E2/R
P=I2*R
i just converted a 225 hcca to 34.5 volt rails....so if i followed the second way of figuring power thatd be:
(imagine full rail voltage at outputs)
P=34.5*34.5/4ohms = 297 watts per channel
if it was in mono it would be:
P = 69 * 69 /4 ohms = 1190 watts mono
what am i missing? i dont think im getting 1190 watts from this amp.
im guessing the only truly accurate way is the first way?
P=I*E
P=E2/R
P=I2*R
i just converted a 225 hcca to 34.5 volt rails....so if i followed the second way of figuring power thatd be:
(imagine full rail voltage at outputs)
P=34.5*34.5/4ohms = 297 watts per channel
if it was in mono it would be:
P = 69 * 69 /4 ohms = 1190 watts mono
what am i missing? i dont think im getting 1190 watts from this amp.
im guessing the only truly accurate way is the first way?
I've seen a lot of strange and outrageous claims before on this forum, but doubting Ohm's law is a new one
Those equations you posted apply to RMS voltages, not peak, so you need to figure out what is the maximum capable peak-to-peak voltage swing across the speaker, the use Vrms = Vpp/(2*sqrt(2)), then use ohm's law.
Those equations you posted apply to RMS voltages, not peak, so you need to figure out what is the maximum capable peak-to-peak voltage swing across the speaker, the use Vrms = Vpp/(2*sqrt(2)), then use ohm's law.
not doubting ohms law...i am a law abiding citizen.
just figured i was missing something, bad subject title
everytime i look for the way to calculate power i find like 2-3 different ways and dont know which one to trust...
your formula is too much for this mathematic defficient brain to hande...care to simplfy what each variable stands for?
....i thought output voltage cannot exceed rail voltage???
just figured i was missing something, bad subject title
everytime i look for the way to calculate power i find like 2-3 different ways and dont know which one to trust...
your formula is too much for this mathematic defficient brain to hande...care to simplfy what each variable stands for?
....i thought output voltage cannot exceed rail voltage???
Hi Clipped,
if your rail voltages are +/- 34.5V and the amplifier has no voltage drop under load you will have a sinus on the load of 34.5Vpeak.
1st method: P=U^2/R
34.5Vpeak are 34.5Vpeak/1.41=24.5Vrms
Your RMS power will be (24.5Vrms)^2/4ohm=149Wrms
2nd method: P=R*I^2
With 34.5Vpeak on a 4ohm load your current will be Ipeak=34.5Vpeak/4ohm=8.62Apeak.
Your RMS current will be 8.62Apeak/1.41=6.11Arms,
thus the RMS power is (6.11Arms)^2*4ohm=149Wrms
3rd method: P=U*I
The RMS power will be Prms=Urms*Irms=24.5Vrms*6.11Arms=149Wrms
As you can see the three methods have the same solution, it is the ohm's law...
Now, if you put the amp in mono (brigded mode) your peak voltage on the load is doubled, so that you will have 69Vpeak.
69Vpeak is 69/1.41=48.9Vrms and then
Prms=(48.9Vrms)^2/4=598W
Putting an amp in brigde will double the voltage on the load and thus multiplies the power by 4. This is truth only in ideal case, normally the output voltage is not doubled because of the increased current demand increase the losses in the output stage and also the power supply may loose some volts. A good amp, designed to work in brigde mode normally multiplies the power be 3 or something like that
ciao
if your rail voltages are +/- 34.5V and the amplifier has no voltage drop under load you will have a sinus on the load of 34.5Vpeak.
1st method: P=U^2/R
34.5Vpeak are 34.5Vpeak/1.41=24.5Vrms
Your RMS power will be (24.5Vrms)^2/4ohm=149Wrms
2nd method: P=R*I^2
With 34.5Vpeak on a 4ohm load your current will be Ipeak=34.5Vpeak/4ohm=8.62Apeak.
Your RMS current will be 8.62Apeak/1.41=6.11Arms,
thus the RMS power is (6.11Arms)^2*4ohm=149Wrms
3rd method: P=U*I
The RMS power will be Prms=Urms*Irms=24.5Vrms*6.11Arms=149Wrms
As you can see the three methods have the same solution, it is the ohm's law...
Now, if you put the amp in mono (brigded mode) your peak voltage on the load is doubled, so that you will have 69Vpeak.
69Vpeak is 69/1.41=48.9Vrms and then
Prms=(48.9Vrms)^2/4=598W
Putting an amp in brigde will double the voltage on the load and thus multiplies the power by 4. This is truth only in ideal case, normally the output voltage is not doubled because of the increased current demand increase the losses in the output stage and also the power supply may loose some volts. A good amp, designed to work in brigde mode normally multiplies the power be 3 or something like that
ciao
34.5V rail, in THEORY, produces 69V peak to peak. You need to work in RMS quantities, so for a sine wave, Vrms = Vp-p / 2.828 = 24.4Vrms.
Into a 4 ohm load (theory again, speakers are not resistors), you get 24.4/4 = 6.1A.
Do the watt calculations:
P = V^2/R = 24.4^2/4 = 149
P = I*V = 6.1 * 24.4 = 149
P = I^2*R = 6.1^2*4 = 149
Ohms law is one essentially V=IR. P=IV is an offshoot of that. You can derive all the above equations for power based on these two equations.
Into a 4 ohm load (theory again, speakers are not resistors), you get 24.4/4 = 6.1A.
Do the watt calculations:
P = V^2/R = 24.4^2/4 = 149
P = I*V = 6.1 * 24.4 = 149
P = I^2*R = 6.1^2*4 = 149
Ohms law is one essentially V=IR. P=IV is an offshoot of that. You can derive all the above equations for power based on these two equations.
so basicly multiply the rail voltage by 0.707 , square it by 2 then divide by R.
so, 34.5 * 0.707= 24.3915 (using calculator in mobile phone)
24.3915 * 24.3915 =594.94527 / 4 = 148.73632
ok thats what i thought, but i read somewhere that wasnt the right thing to do....i guess that site was wrong.
thanks for all the input, this goes to back it up.
so, 34.5 * 0.707= 24.3915 (using calculator in mobile phone)
24.3915 * 24.3915 =594.94527 / 4 = 148.73632
ok thats what i thought, but i read somewhere that wasnt the right thing to do....i guess that site was wrong.
thanks for all the input, this goes to back it up.
You must not forget to subtract the transistor voltage loss from your rail voltage.
From all the calculations I have seen, most folks subtract 1V from the peak rail voltage or 2V for the peak to peak rail voltage.
Your calculations will be the same, just take into account the loss from the transistors as they cannot swing "rail to rail."
So you are using the 2150sx rail voltage on a 225 board? Should make close to full power in 4 ohm stereo. I'm sure it will get HOT too. Did you change the power supply secondary capacitors to 50V types?
From all the calculations I have seen, most folks subtract 1V from the peak rail voltage or 2V for the peak to peak rail voltage.
Your calculations will be the same, just take into account the loss from the transistors as they cannot swing "rail to rail."
So you are using the 2150sx rail voltage on a 225 board? Should make close to full power in 4 ohm stereo. I'm sure it will get HOT too. Did you change the power supply secondary capacitors to 50V types?
You need to factor in the voltage drop across the transistors and also the current sharing resistors that most amps run, so with this in mind it is better to not use the rail voltage and measure actual output voltage with an oscilloscope to get an accurate measurement.
As for the formulas, they are only as accurate as the data you input into them....
As for the formulas, they are only as accurate as the data you input into them....
im getting 34.5 at the emittor resistors, with a 33 volt zener at the regulator transistor....
i had to rewind the primary twice, the first time i came up short with only 5 turns opposed to 6 since i used thicker wire and forgot to take that into account....with the 5 turns i would only get 29~ volts at 12 volt input and it would swing to 33~ at 14.4 volt input.
rewound it again with original size primary wire, with a full 6 turns and get a fairly constant 34.5 volts between 12-15 volt input.
toroid starts buzzing at 13.5 volts in both cases.
i havent measured it with a sinewave yet....tested it lastnight for 4 hours @4 ohms mono full range on a 6x9....we will see if it dies when hooked up to a sub.
it gets a little hotter with this rail voltage when playing full range, but nothing drastic...with a sub hooked up i imagine it will get much hotter though....i dropped it into the car today to test on my mid/hi's and no fire ensued.
i'll try it on the subs tomorrow....4 ohms mono
i changed the rail caps to 2200uf 50volt since this is what i have on hand, but there are 2 more smaller 1000uf caps in the same array where i could only fit a 35 volt.
i took out 3 bass amps today with one left installed, subs are history, spiders have stretched and softened after that competition 2 weeks ago...time to recone
i had to rewind the primary twice, the first time i came up short with only 5 turns opposed to 6 since i used thicker wire and forgot to take that into account....with the 5 turns i would only get 29~ volts at 12 volt input and it would swing to 33~ at 14.4 volt input.
rewound it again with original size primary wire, with a full 6 turns and get a fairly constant 34.5 volts between 12-15 volt input.
toroid starts buzzing at 13.5 volts in both cases.
i havent measured it with a sinewave yet....tested it lastnight for 4 hours @4 ohms mono full range on a 6x9....we will see if it dies when hooked up to a sub.
it gets a little hotter with this rail voltage when playing full range, but nothing drastic...with a sub hooked up i imagine it will get much hotter though....i dropped it into the car today to test on my mid/hi's and no fire ensued.
i'll try it on the subs tomorrow....4 ohms mono
i changed the rail caps to 2200uf 50volt since this is what i have on hand, but there are 2 more smaller 1000uf caps in the same array where i could only fit a 35 volt.
i took out 3 bass amps today with one left installed, subs are history, spiders have stretched and softened after that competition 2 weeks ago...time to recone
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