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Old 8th September 2006, 02:01 PM   #11
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Default 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
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Old 8th September 2006, 02:05 PM   #12
ilimzn is offline ilimzn  Croatia
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Default 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
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Old 8th September 2006, 02:08 PM   #13
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Default 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
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Old 8th September 2006, 02:41 PM   #14
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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
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Old 8th September 2006, 02:46 PM   #15
anatech is offline anatech  Canada
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Hi Jan,
The bridged idea isn't bad if you are limited in supply voltage, as long as the output stage is designed for the extra load. Heck, you could even use a McIntosh autotransformer to kick the output way up in power.

Expensive though, and 2 amps isn't near enough.

-Chris
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Old 8th September 2006, 03:25 PM   #16
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Quote:
Originally posted by anatech
Hi Jan,
The bridged idea isn't bad if you are limited in supply voltage, as long as the output stage is designed for the extra load. Heck, you could even use a McIntosh autotransformer to kick the output way up in power.

Expensive though, and 2 amps isn't near enough.

-Chris
Sure, it's just that many commercial bridged amps (or amps that can be bridged) are most often spec'd for 8 ohms MINIMUM load.
But if it survives, why not.

Jan Didden
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Old 8th September 2006, 03:28 PM   #17
AndrewT is offline AndrewT  Scotland
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Hi,
Ilimzn already got it right
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.
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.
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Old 8th September 2006, 03:58 PM   #18
anatech is offline anatech  Canada
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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
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Old 8th September 2006, 06:14 PM   #19
AndrewT is offline AndrewT  Scotland
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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).
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Old 8th September 2006, 06:36 PM   #20
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Default 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.

Letís 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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