Trying to get my mind on Ohms law and how to calculate it myself. was looking at an single channel IC chip datasheet and it says 40W on a 8 ohm load from 25V. Knowing (V) is for volts, (I) is current or amps and (R) is resistance or ohms . looking at the ohms law pyramid they say if you have any two you can find the others easily. Volts is on top divided by either of the other two to get the others value. The (I) and (R) are across so you times them.
let's start from the obvious 25V.
25V / 8 ohms = 3.125 amps
where do you go from here to get 40W
and google says to get watts times volts and amps 25 x 3.125 = 78.125

There may not be a clear path connecting this numbers. As has been discussed already in this thread, audio specs are often "fudged" at best, just plain fraudulent at worst.

That said, I'm guessing that a lot of the disparity comes from the losses within the chip (nothing in real life is 100% efficient) and from the conversion of DC into AC. In order to get an equivalent amount of work done with AC or DC power, you need to match the DC voltage and the RMS AC voltage. RMS stands for root mean square, and in layman's terms, it's kind of like a very special type of weighted average. The key here is that AC voltage is going up and down constantly, so in order for the RMS voltage to be equivalent to a DC voltage, the peak AC voltages must be much higher (factor of 1.414 for a pure sine wave, other values for other waveforms.)

In audio amps, you've got a DC power supply being used to generate an AC output. It can't provide a peak voltage higher than its supply voltage, in fact there's usually voltage drop due to transistor forward voltages, so your peak AC output voltage is usually 1.4 volts or more below your DC input voltage. The RMS AC voltage is, at best, 1.414 times lower than that.

Applying all this to your situation, it might look something like this:

25V supply - 1.4V Vf = 23.6V peak to peak
23.6V peak / 1.414 = 16.69 V RMS
16.69 V RMS / 8 ohms = 2.09A
2.09A * 16.69V = 34.8W

Maybe I messed up a little there, or maybe they fudged a little, but at least it’s in the ballpark.

The short answer is that you should practice ohms law on simple DC circuits or simple AC circuits first, and get really comfortable with it before trying to apply it to more complicated AC/DC relationships.

 

There may not be a clear path connecting this numbers. As has been discussed already in this thread, audio specs are often "fudged" at best, just plain fraudulent at worst.

That said, I'm guessing that a lot of the disparity comes from the losses within the chip (nothing in real life is 100% efficient) and from the conversion of DC into AC. In order to get an equivalent amount of work done with AC or DC power, you need to match the DC voltage and the RMS AC voltage. RMS stands for root mean square, and in layman's terms, it's kind of like a very special type of weighted average. The key here is that AC voltage is going up and down constantly, so in order for the RMS voltage to be equivalent to a DC voltage, the peak AC voltages must be much higher (factor of 1.414 for a pure sine wave, other values for other waveforms.)

In audio amps, you've got a DC power supply being used to generate an AC output. It can't provide a peak voltage higher than its supply voltage, in fact there's usually voltage drop due to transistor forward voltages, so your peak AC output voltage is usually 1.4 volts or more below your DC input voltage. The RMS AC voltage is, at best, 1.414 times lower than that.

Applying all this to your situation, it might look something like this:

25V supply - 1.4V Vf = 23.6V peak to peak
23.6V peak / 1.414 = 16.69 V RMS
16.69 V RMS / 8 ohms = 2.09A
2.09A * 16.69V = 34.8W

Maybe I messed up a little there, or maybe they fudged a little, but at least it’s in the ballpark.

The short answer is that you should practice ohms law on simple DC circuits or simple AC circuits first, and get really comfortable with it before trying to apply it to more complicated AC/DC relationships.

Oops, may have confused peak and p-p there, but I’m out the door. Hopefully someone else can clarify andclean up my mess.

 

I think you done a great job there explaining it ebeowulf. I just knew RMS was in this somewhere. I'm jotting your numbers down to see if I can get it in my head. That's why AG keeps using 1.414.
Thank you Sir.
using this at 12V
12V - 1.4 = 10.586 p-p
10.586 peak / 1.414 = 7.486
7.486V RMS / 8 ohms = .935 amps
.935 amps x 10.586 = 9.897 watts
If above is correct then I just need to figure out how to get 25V to it. There has to be a way you see 100's and 1000 watt amps everywhere.

 

How to tell if you can bridge a chip or is this done some other way or are they all bridgeable.
If anyone doesn't mind check my math above to see if I got it.
Edit: found it and it can be bridged

 

You forgot to say the IC amplifier's part number.
All amplifier ICs have voltage losses so with a 25V supply the output is probably 22V peak-to-peak.
1) The 8 ohm speaker gets peaks of only half of the peak-to-peak voltage because the signal must swing equally up and down making AC.
2) The 22V p-p is divided by 2.828= 7.8V RMS. The power in the 8 ohm speaker just before clipping is (7.8 squared)/8 ohms= 7.6 Watts.

See, not enough details about this amplifier. It might be a bridged amplifier that uses a separate amplifier to drive each speaker wire.
1) With double the number of amplifiers its output is probably 44V p-p.
2) The 44V p-p is divided by 2.828= 15.6V RMS. The power in the 8 ohm speaker just before clipping is a little less than (15.6V squared)/8 ohms= 30.4Watts sounding good or is 40W with horrible clipping distortion.

 

25V supply - 1.4V Vf = 23.6V peak to peak
23.6V peak / 1.414 = 16.69 V RMS No!. The peak voltage is half the peak-to-peak and is 11.8V.The RMS voltage = 11.8V RMS / 1.414 = 8.4V RMS.
16.69 V RMS / 8 ohms = 2.09A Correction: 8.4V RMS / 8 ohms = 1.05A.
2.09A * 16.69V = 34.8W Correction: 1.05A * 8.4V = 8.8 Watts.

Maybe I messed up a little there

Yes you certainly did.

 

Sorry AG I've been looking at more than one IC. The one I was asking about bridging is a:
http://www.ti.com/lit/ds/symlink/tpa3156d2.pdf
Now you throw another factor in there just when I thought I was getting the hang of it (the 2.828).
it's 70W with 4 ohm load at 24V. Still reading on how bridging affects the wattage output. I know it's higher but how much is what I'm trying to figure out.

 

using this at 12V
12V - 1.4 = 10.586 p-p
10.586 peak / 1.414 = 7.486 No! The peak is 5.3V and the RMS is 5.3 / 1.414 = 3.75V RMS.
7.486V RMS / 8 ohms = .935 amps 3.75V / 8 ohms = 0.47A.
.935 amps x 10.586 = 9.897 watts 0.47A x 3.75V= 1.76W.

I just need to figure out how to get 25V to it..

Make a DC voltage -boosting circuit with a high frequency oscillator and transformer then rectify and filter its output.

 

using this at 12V
12V - 1.4 = 10.586 p-p
10.586 peak / 1.414 = 7.486 No! The peak is 5.3V and the RMS is 5.3 / 1.414 = 3.75V RMS.
7.486V RMS / 8 ohms = .935 amps 3.75V / 8 ohms = 0.47A.
.935 amps x 10.586 = 9.897 watts 0.47A x 3.75V= 1.76W.

I got it and have corrected my formula.

Make a DC voltage -boosting circuit with a high frequency oscillator and transformer then rectify and filter its output.

As if my mind isn't clogged up enough already...
Edit:
I'll just go into Exel and make a watts calculator
Edit 2: Where do they get the 40W from then? should I learn how to calculate (fake) Watts?... ….

 

Now you throw another factor in there just when I thought I was getting the hang of it (the 2.828).

The peak voltage is half the peak-to-peak signal voltage.

The TPA3156D2 bridged amplifier produces 70W with 4 ohm load at 24V.

You probably cannot solder all those tiny pins. It has 16 pins on each 11mm long side.
The output power is 70W with horrible 10% clipping distortion or 56W with less but maybe audible distortion.

Still reading on how bridging affects the wattage output. I know it's higher but how much is what I'm trying to figure out.

Bridging uses a second amplifier to drive the second speaker wire with opposite -phase to the first amplifier. Then the total voltage across the speaker is almost doubled and the total current in the speaker is also almost doubled. Then the power in the speaker is about 3.5 times the power in the same speaker with the same supply voltage.

 

You probably cannot solder all those tiny pins. It has 16 pins on each 11mm long side.
The output power is 70W with horrible 10% clipping distortion or 56W with less but maybe audible distortion.

So is their an IC out there that I could ever use without it having some distortion that you would say was OK?
Every IC has distortion doesn't it?

 

A class-AB amplifier has very low distortion up to when its output voltage cannot go any higher. If the volume is turned up higher than it can go then the top and bottom of the waveforms get clipped off and clipping distortion is produced. If you turn the volume WAY UP then the amount of power in the distortion is the same as the amount of power in the clean signal so the output power is doubled. A class-AB amplifier is linear so it gets hot.

The amplifier you found uses class-D which chops up the signal at a high frequency and uses pulse-width-modulation to make an average output voltage that is the signal voltage. The output transistors switch on completely (no voltage across them so almost no heating in them then switch off completely (no current in them so almost no heating in them). The switching at a very high frequency causes distortion at high audio frequencies. Old people without hearing aids cannot hear the high frequency distortion.

You need a robot to solder the tiny pins on a modern class-D amplifier IC.

 

So are you talking about one of these?
https://www.digikey.com/products/en?mpart=LM3886TF/NOPB&v=296
Now not exactly this one but this kind. This one is not in stock.

 

So are you talking about one of these?
https://www.digikey.com/products/en?mpart=LM3886TF/NOPB&v=296
Now not exactly this one but this kind. This one is not in stock.

That is the insulated one that gets hot and a heatsink does not cool it enough. The normal one has a metal tab for feeding lots of heat to a large heatsink and Digikey has hundreds in stock.

 

Thanks for cleaning that up!

 

I had some things to do AG and wasn't going to be at my computer so I just typed it in and posted the first one I came across.

Now not exactly this one but this kind.

Thanks for cleaning that up!

I don't think this takes into account the amplifier IC. Let's say it said 50 or 60 watts output? there is no way that using that formula you would ever get close to 50 or 60 watts in the answer. and 1.76W is a large decline from 40W (?). If I'm wrong then I'm wrong but isn't this the formula when you put 12V across and 8ohm load. there is nothing about the IC in this formula.
@AG
We also have talked about these kinds of IC's already.
your answer was where you gonna get a 50V power supply.

 

This one will get you to 48 volts

https://www.ebay.com/itm/12V-TO-48V...432890?hash=item4b2d87f87a:g:43sAAOSwh1haDTit

or better yet, use two of these to get +/- 24 volts

https://www.ebay.com/itm/Waterproof...605057?hash=item3f76b66701:g:0yEAAOSwAPVZFmrF

where you gonna get a 50V power supply.

 

Hello,

@bertz , I am not sure about the second one to use 2 to create +/- 24 Volts.
I can not determine if the unit is fully isolated.

Bertus

 

Why do you think an audio amplifier puts 12VDC in an 8 ohm speaker to produce 18 Watts?
An amplifier output has about 1.4V of losses (probably more) so the AC in the speaker is 10.6V p-p which is 10.6 divided by 2.828= 3.75V RMS. The output power is (3.75V squared)/8= 1.76 real Watts.

1) Use a 4 ohm speaker then the power is almost doubled to about 3.3W.
2) Use a bridged amplifier with the 4 ohm speaker and the power is almost quadrupled to about 12W.
3) Use a 14.4V charging car battery and the bridged power in the 4 ohm speaker is 20W.
4) Turn up the volume so there is LOTS of distortion and the power output is 40 Whattts.
5) Some salesmen of audio products multiply the 40W by the age of their mother to get more than one thousand Whattts!.

 

https://www.ebay.com/itm/12V-TO-48V...432890?hash=item4b2d87f87a:g:43sAAOSwh1haDTit
https://www.ebay.com/itm/Waterproof...605057?hash=item3f76b66701:g:0yEAAOSwAPVZFmrF

https://www.amazon.com/ELECTRONIX-E...7008&sr=8-10&keywords=12+to+24+volt+converter

I posted one these before and AG's answer:
The cheap Chinese voltage booster is made for a golf cart and appliances like lights, not to power an audio amplifier. It oscillates to boost the voltage and the oscillating frequency might produce a whining sound in the amplifier output.
Why not one of these and I can get an IC with lower wattage 45W to 80W that only needs 30V.
http://www.icstation.com/adjustable...regulator-with-voltmeter-display-p-10525.html
Testing video:

$5.69