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
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
I have Bridge Amplifier for woofer.
http://www.diyaudio.com/forums/showthread.php?s=&threadid=80972&perpage=10&pagenumber=6
I have Two woofer 8 ohm SPK.( parallel )
http://www.diyaudio.com/forums/showthread.php?s=&threadid=73456&perpage=10&pagenumber=3
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?
http://www.diyaudio.com/forums/showthread.php?s=&threadid=80972&perpage=10&pagenumber=6
I have Two woofer 8 ohm SPK.( parallel )
http://www.diyaudio.com/forums/showthread.php?s=&threadid=73456&perpage=10&pagenumber=3
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?
Re: Having plus 15 and negative 15 you will have 30 volts maximum theorical output
Carlos, I respect your amp building prowess, but with all due respect, I think you need some learning in the basics of electricity before jumping to answer this question.
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.
destroyer X said:
Carlos, I respect your amp building prowess, but with all due respect, I think you need some learning in the basics of electricity before jumping to answer this question.
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.
Hi My,
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.
no.
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.
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
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
Well Andrew..... no one told me that we need to be perfect and big experienced EE to
contribute in our forum.
And you can see that i use to help.
I will see what is prowess....as ignore i know the meaning.
Ahahahahha!....not a good meaning i suppose.
I hope you do not comdemned people to be crucified when produce errors in your school.
And to be objective...the metering instruments will be the one that are able to be precise...as we calculate and we have to measure to find the real result, that depends on the circuit.
regards,
Carlos
contribute in our forum.
And you can see that i use to help.
I will see what is prowess....as ignore i know the meaning.
Ahahahahha!....not a good meaning i suppose.
I hope you do not comdemned people to be crucified when produce errors in your school.
And to be objective...the metering instruments will be the one that are able to be precise...as we calculate and we have to measure to find the real result, that depends on the circuit.
regards,
Carlos
Re: Yes guys...thank you...i had forgot the 2 amps that will limit the consumption into
Please, Carlos, don't get this the wrong way - we really don't have an empathy problem, nor was it my intention to scald you in public or even worse, humiliate you.
The problem is simple:
1) electrical power is fixed by definition, it's not subject to subjective interpretation - and all the required data for a precise calculation was given. Your calculation was dead wrong - and enough information was provided for all of us, including you, to learn the right answer, by many contributors.
2) As a moderator of other forums, I know very well that one wrong answer is difficult to rectify even with ten right answers. This is an Audio forum, and you will agree there is already WAY too much voodoo in audio, and these things tend to propagate it - not that I am accusing you of doing this, especially not on purpose.
We all make mistakes, because we are all human. But it is really up to the more prominent board members to be careful what they write, because they enjoy the respect and trust of many board members, so in a way they have a responsibility towards them. So that's all - a mistake was made, correction was supplied, we all know (including me) you are a good guy, especially because of all the right answers and all the good info you give to this board
Please, Carlos, don't get this the wrong way - we really don't have an empathy problem, nor was it my intention to scald you in public or even worse, humiliate you.
The problem is simple:
1) electrical power is fixed by definition, it's not subject to subjective interpretation - and all the required data for a precise calculation was given. Your calculation was dead wrong - and enough information was provided for all of us, including you, to learn the right answer, by many contributors.
2) As a moderator of other forums, I know very well that one wrong answer is difficult to rectify even with ten right answers. This is an Audio forum, and you will agree there is already WAY too much voodoo in audio, and these things tend to propagate it - not that I am accusing you of doing this, especially not on purpose.
We all make mistakes, because we are all human. But it is really up to the more prominent board members to be careful what they write, because they enjoy the respect and trust of many board members, so in a way they have a responsibility towards them. So that's all - a mistake was made, correction was supplied, we all know (including me) you are a good guy, especially because of all the right answers and all the good info you give to this board
Oh..that's nice Ilyzm...i was thinking i could make another persecutor.
My God!...those last days i have beeing crashed constantly, and i discovered that one of the guys are crashing me...annoying me, was someone that i gave enormous support when the guy was suffering as a hell...because problems he had in this forum.
I am beeing a little Paranoid...sorry for that.
Let's forget that Ilyzm.
regards,
Carlos
My God!...those last days i have beeing crashed constantly, and i discovered that one of the guys are crashing me...annoying me, was someone that i gave enormous support when the guy was suffering as a hell...because problems he had in this forum.
I am beeing a little Paranoid...sorry for that.
Let's forget that Ilyzm.
regards,
Carlos
Getting back on topic : Loading a bridge amp with less than 4 ohms is probably inviting disaster. Because of the bridge action each amp actually sees half the load. So with 8 ohms speaker each amp sees 4 ohms, which probably is not too bad. But with 4 ohms load, each amp sees 2 ohms, probably NOT ok!
Jan Didden
: Loading a bridge amp with less than 4 ohms is probably inviting disaster. Because of the bridge action each amp actually sees half the load. So with 8 ohms speaker each amp sees 4 ohms, which probably is not too bad. But with 4 ohms load, each amp sees 2 ohms, probably NOT ok!Jan Didden
Hi,
Ilimzn already got it right
Take some time to understand the question, even ocassionally interpreting the needs, and then try to provide an accurate, or at least usefull answer.
My replies sometimes appear as gobble de gook, but maybe I will become more eloquent. My peers manage eloquence often. I am envious. Ocassionally complete untruths are espoused simply because the poster has not taken time to think. But that mistake does not warrant a defence when an apology or withdrawal would be more appropriate.
Ilimzn already got it right
Why do posters fail to comprehend the message contained therein?
Take some time to understand the question, even ocassionally interpreting the needs, and then try to provide an accurate, or at least usefull answer.
My replies sometimes appear as gobble de gook, but maybe I will become more eloquent. My peers manage eloquence often. I am envious. Ocassionally complete untruths are espoused simply because the poster has not taken time to think. But that mistake does not warrant a defence when an apology or withdrawal would be more appropriate.
Hi Andrew,
You are correct. A picture would have driven the point home as I misinterpreted this as the supplies that were available. What is not stated often afterwards was the current rating of 2 amperes. This I will take as an average value, instead of the other peak values given in the same post.
This will allow 11.3 V peak (8 V RMS reading) instead of 15 V peak to be developed across a (stated) 4 ohm load, indicating the supply current is not enough for a resistive load. (reality check). The output current is either greater, or the output voltage is current limited by supply droop. I was checking for a current limited output.
If the 2 ampere figure was peak, there would be a lower output power available. I am thinking P=E*I, so you either have 21.2 watts actually at a higher current than you think, or 16 watts if your supply caves.
Have I figured this out wrong?
-Chris
You are correct. A picture would have driven the point home as I misinterpreted this as the supplies that were available. What is not stated often afterwards was the current rating of 2 amperes. This I will take as an average value, instead of the other peak values given in the same post.
This will allow 11.3 V peak (8 V RMS reading) instead of 15 V peak to be developed across a (stated) 4 ohm load, indicating the supply current is not enough for a resistive load. (reality check). The output current is either greater, or the output voltage is current limited by supply droop. I was checking for a current limited output.
If the 2 ampere figure was peak, there would be a lower output power available. I am thinking P=E*I, so you either have 21.2 watts actually at a higher current than you think, or 16 watts if your supply caves.
Have I figured this out wrong?
-Chris
Hi Anatech,
you almost have it right.
Ppk=Epk*Ipk or Vpk*Ipk
P=Vpk*Ipk/2 = Vpk*Vpk/2R = Ipk*Ipk*R/2 since he was asking about a sinewave signal.
So your 16Wpk is really 8W (some would add rms after it but that is incorrect).
Adding pk (zero to peak) or pp (peak to peak) as a suffix usually clarifies and avoids ambiguity.
BTW your answer used the appropriate tone implying a bit of uncertainty. If only other posters could copy the same trait rather than state as fact something that only the near expert would recognise as untrue. Opinions are allowed if stated as such and many promote a good argument that may even persuade others to change their stance. (I have changed on occasion).
you almost have it right.
Ppk=Epk*Ipk or Vpk*Ipk
P=Vpk*Ipk/2 = Vpk*Vpk/2R = Ipk*Ipk*R/2 since he was asking about a sinewave signal.
So your 16Wpk is really 8W (some would add rms after it but that is incorrect).
Adding pk (zero to peak) or pp (peak to peak) as a suffix usually clarifies and avoids ambiguity.
BTW your answer used the appropriate tone implying a bit of uncertainty. If only other posters could copy the same trait rather than state as fact something that only the near expert would recognise as untrue. Opinions are allowed if stated as such and many promote a good argument that may even persuade others to change their stance. (I have changed on occasion).
All You Folks Rock...
Especially Progg70! I find it all very entertaining and educational at the same time. But to see personalities spill out on the net is wonderful too, as long as no one gets hurt. Awe what are a few ruffled feathers?
Why not:
15V peak *.707=10.605V rms or DC equivalent
I=V/R
I=1.326 Amps (8 ohm)
I=2.651 Amps (4 ohm) !!!Problem due to power supply limitations of maximum current 2 amps.
Let’s go backwards....
I=2 Amps
V=10.605 V rms
R=5.3 ohms (lowest possible DC load resistance due to power supply)
Max Power RMS 21.21Watts
Or perhaps I am a jerk and my methods are unrefined
?
Cheers,
Shawn.
Especially Progg70! I find it all very entertaining and educational at the same time. But to see personalities spill out on the net is wonderful too, as long as no one gets hurt. Awe what are a few ruffled feathers?
Why not:
15V peak *.707=10.605V rms or DC equivalent
I=V/R
I=1.326 Amps (8 ohm)
I=2.651 Amps (4 ohm) !!!Problem due to power supply limitations of maximum current 2 amps.
Let’s go backwards....
I=2 Amps
V=10.605 V rms
R=5.3 ohms (lowest possible DC load resistance due to power supply)
Max Power RMS 21.21Watts
Or perhaps I am a jerk and my methods are unrefined
?Cheers,
Shawn.
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