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26th March 2008, 05:23 PM  #1 
diyAudio Member

accuracy of ohms law
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? 
26th March 2008, 05:34 PM  #2 
diyAudio Member
Join Date: Mar 2007

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 peaktopeak voltage swing across the speaker, the use Vrms = Vpp/(2*sqrt(2)), then use ohm's law. 
26th March 2008, 05:47 PM  #3 
diyAudio Member

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 23 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??? 
26th March 2008, 06:10 PM  #4 
diyAudio Member
Join Date: Jan 2008
Location: Chiasso

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 
26th March 2008, 06:16 PM  #5 
diyAudio Member
Join Date: Oct 2006
Location: Charlotte, NC

34.5V rail, in THEORY, produces 69V peak to peak. You need to work in RMS quantities, so for a sine wave, Vrms = Vpp / 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. 
26th March 2008, 06:57 PM  #6  
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

Re: accuracy of ohms law
Quote:
They are all the same thing. P=I*V, but I=V/R so P also = V*V/R, and V=I*R so P also = I*I/R. /sreten. 

26th March 2008, 07:28 PM  #7 
diyAudio Member
Join Date: Sep 2005

They are the same but differ in that they calculate the same thing useing 2 diffirent known values...
Current and voltage Voltage and resistance Current and resistance The method you use will depend on which of these two values are known or measurable... 
26th March 2008, 08:52 PM  #8 
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

typo :
P=V*I, but I=V/R so P also = V*V/R, and V=I*R so P also = I*I*R /sreten. 
26th March 2008, 10:21 PM  #9 
diyAudio Member
Join Date: Dec 2007

Why not use a clamp meter, volt meter, and an oscilloscope to find real output??

27th March 2008, 03:05 AM  #10 
diyAudio Member

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. 
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