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8th September 2006, 07:02 AM  #1 
diyAudio Member
Join Date: Jul 2004
Location: Sweden

Output power for a power amplifier
Output power for a power amplifier
If I have +/ 15V sinus top – top (se picture) and capacity for 2 Ampere and a 4 Ohm load. What is den the amplifier output power effective and top value. Thanks 
8th September 2006, 07:23 AM  #2 
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Join Date: Sep 2006

hmm I can see no picture attached
cheers 
8th September 2006, 07:47 AM  #3 
diyAudio Member
Join Date: Feb 2004
Location: Recife  Brasil Northeast

Having plus 15 and negative 15 you will have 30 volts maximum theorical output
And will be 30 volts peak to peak....
30 X 30 = 900 900 divided by the load..... 4 ohms the load...result is 225 watts peak to peak. but normally amplifiers have not that efficiency, they do not swing the total supply voltage...there are losses with voltages that will be from colector to emitter and other problems related saturations, impedances mismatches, and EMF. It is normal that the audio amplifier's efficiency goes around 65 percent maximum...and the more common to happen is 50 percent...this means that 50 percent is transformed in audio, and the other 50 percent is transformed in heat....as circuit has internal resistances. So...maximum will be around 110 watts or more if 65 percent of efficiency was reached. But really....real world, you will have less power....i think 80 watts will be the maximum real power using this voltage and this output impedance.....but beeing more realistic, the real produced power will be around 20 watts RMS. I think our forum friends can be more precise than i can...as i am a "changeneer"..... not an engineer...and i know things because experience and i am not good with calculations. As a rule to me, to have a fast idea of the output power, the output voltage will be around half or the positive or half of the negative rail voltage.... around 8 volts...that result in 16 watts without distortion. Also i use to divide the consumption by two, to obtain the maximum possible power, that will be divided by two channels and will the power with the lowest impedance allowed by the industrial equipment that i would be inspecting to discover the real power. regards, Carlos 
8th September 2006, 07:57 AM  #4 
diyAudio Member

15v peak / sqroot of 2 = about 10V RMS over 4 ohms is 10 squared / 4 is 25 W RMS, That is if you drive the amp continuously at full power.
For 10V RMS over 4 ohms you need 10 / 4 = 2.5A RMS or 2.5 times sqroot of 2 = about 3.5 amps peak. Jan Didden
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8th September 2006, 08:29 AM  #5 
diyAudio Member
Join Date: Apr 2004
Location: Gütersloh

With a capacity of 2 amperes into 4 ohms you would get max ~8v peak, giving 8*8/(4*2) = 8watts.
With the 15v > 15*15/(4*2) = 28watts. Mike 
8th September 2006, 11:09 AM  #6 
diyAudio Member
Join Date: Feb 2006
Location: Yangon

I have Bridge Amplifier for woofer.
Bridge With IRF 540 & IRF 9540 I have Two woofer 8 ohm SPK.( parallel ) Design GIGAAMP SubWoofer. You can saw in picture. I connected with two speaker parallel with my bridge AMP. AMP output (low frequency) is Max 50V p to p ( saw in oscilloscope ). (Two speaker parallel)Test with woofer tester is 3.9 So my MAX output power will be 50v peak / sqroot of 2 = about 35V RMS over 4 ohms is 35 squared / 4 is 306.25 W RMS. Is that true? 
8th September 2006, 12:31 PM  #7  
diyAudio Member
Join Date: Feb 2005
Location: Zagreb

Re: Having plus 15 and negative 15 you will have 30 volts maximum theorical output
Quote:
Progg stated +15V, i.e. + and  15V peak SINE WAVE at the output  so all saturation voltages and rail losses have been accounted for already. This means 30V peak to peak. The actual maximum voltage  remember, this is AC we are talkinga bout  is half of the peak to peak voltage, i.e 15V peak. For a sine wave, effective voltage is peak / sqrt(2), this is the equivalent DC voltage that would produce the same power on a given load. Power P=Veff^2/Rload = Vpeak^2/(2*Rload) = 28.125W into 4 ohms, or 14.0625W into 8 ohms. Also, because a current limit is given, we have to check that it is enough. P = Ipeak^2 * Rload/2, for a maximum of 2A, this comes out as only 8W into 4 ohms. A full 14W is available into 8 ohms as the required current at peak voltage is less than 2A. Other considerations may apply, we are assuming an 'ideal' resistive load for all of these calculations. 

8th September 2006, 12:35 PM  #8  
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Hi My,
Quote:
Ignore Destroyer's calculation, he has little prowess with numbers. MikeB describes the solution well. 50Vpp=25Vpk=25/(root2)Vrms or Vac. Power delivered into a 4r load is P=Vpk*Vpk/2/Load=25*25/2/4=78.1W Maximum current when delivering this power is Ipk=Vpk/Load=25/4=6.2Apk However you quote 3.9ohms for a measurement of the two drivers. This implies a combined impedance higher than 4ohms. I would estimate that the combined speaker impedance will be in the range 5ohms to 7ohms. The average power delivered into the higher impedance will be less than the example quoted above. The peak current will be variable as frequency changes but at a few frequencies the peak current could be determined by the worst case resistance of 3r9 giving Ipk=6.4Apk edit. Ilimzn can also be relied on for accurate information.
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8th September 2006, 12:38 PM  #9 
diyAudio Moderator

Hi ilimzm, Andrew,
Don't forget that we lose a couple volts across the output stage as well. Add some loss to keep things linear. Power output is looking a little more grim, but that is the reality. Chris 
8th September 2006, 12:57 PM  #10 
diyAudio Member
Join Date: Feb 2004
Location: Recife  Brasil Northeast

Yes guys...thank you...i had forgot the 2 amps that will limit the consumption into
60 watts..... and with 50 percent efficiency the limit will be around 30 watts.
Sorry my fault. Ilyzm.... sometimes i use to produce errors.... but as i use to enter thousand times.... they are very small related the rigth things i use to say. Take a look at my personal info and you will see that i am not an enginneer but i have some practice and many thousand amplifier constructed, i am sorry that i am bothering you with my failures...i am happy that you are not a moderator, that you are only trying to help, even not beeing your obligation. I hope you can open your mail to receive direct messages, as you close the messages entrance that is turning difficult to us to find a solution to our empathy problems. This remembers me the guy that was always asking a kiss for the girls he met..... and he received a lot of agressions from the girls...and them someone told him:  " Tell me the reason you do those foolishes all time long?" The guy answered:  "Well, i use to invite 1000 girls to kiss me..... 1 percent use to accept.....so, at least i succeed related 10 girls!" regards, Carlos 
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