Amplifier power

Thread Starter

AbsoluteC

Joined Sep 23, 2021
6
Hello. I have been out of school for a couple of years now and I am trying to figure out a question that I am getting asked. If anyone can help me please, I'm sure it is simple.

An amplifier power that goes from 2 watts to 4 watts gains how much additional output power?

I'm going to assume that it would be lower from the research that I tried to do online with amplifier powers, but I want to make sure that I have the right understanding of it.

Thank you for your time in seeing this out.
 

Thread Starter

AbsoluteC

Joined Sep 23, 2021
6
I have a better understanding of it now. I will go back and forth checking with the equation to make sure I got it right.
Thank you very much for your help!
 

MrChips

Joined Oct 2, 2009
24,458
It would make more sense if the question were:

An amplifier power goes from 2 watts to 4 watts, what is the power gain in dB?
 

Thread Starter

AbsoluteC

Joined Sep 23, 2021
6
It would make more sense if the question were:

An amplifier power goes from 2 watts to 4 watts, what is the power gain in dB?
That was how the question was brought out to me. I actually thought it was supposed to request what is the power gain in dB too after seeing an example online:
"Increasing an amplifier from 25 watts to 50 watts (double the power) increases the acoustic power by 3dB."
 

Ian0

Joined Aug 7, 2020
3,525
Thanks for clarifying that it is an audio amplifier.
There is a phenomenon called "power compression" which applies to loudspeakers and not to power amplifiers. Whilst the output of the power amplifier will still be double (4W instead of 2W) and the output in decibels will be 3dB more; the sound pressure level may increase by less than 3dB as the loudspeaker voice coil temperatures increases, causing its resistance to increase. Furthermore, if the cone excursion exceeds a parameter known as Xmax the movement of the cone increases by less that it should for a doubling in power, also resulting in an increase of sound pressure that is less than 3dB.
This will only happen to a loudspeaker that is reaching its power limit. If it is well within its normal operating range, then the sound pressure will increase by 3dB in line with the amplifier output.
 

Thread Starter

AbsoluteC

Joined Sep 23, 2021
6
Thanks for clarifying that it is an audio amplifier.
There is a phenomenon called "power compression" which applies to loudspeakers and not to power amplifiers. Whilst the output of the power amplifier will still be double (4W instead of 2W) and the output in decibels will be 3dB more; the sound pressure level may increase by less than 3dB as the loudspeaker voice coil temperatures increases, causing its resistance to increase. Furthermore, if the cone excursion exceeds a parameter known as Xmax the movement of the cone increases by less that it should for a doubling in power, also resulting in an increase of sound pressure that is less than 3dB.
This will only happen to a loudspeaker that is reaching its power limit. If it is well within its normal operating range, then the sound pressure will increase by 3dB in line with the amplifier output.
I have a better understanding of this now. Thank you for the information. Noting this for reference.
 

Ian0

Joined Aug 7, 2020
3,525
I kind of see how it is in dB and dBm now.

10log(4/2) = 3 dB
10log(4000/2) = 33 dBm
erm. . .No. . . .
if it's just dB then you are comparing two powers, or voltages, so
\(
dB = 10 log_{10}\frac{Power1}{Power2}
\)
\(
dB = 20 log_{10}\frac{Voltage1}{Voltage2}
\)

When you add a suffix, you are specifying what goes on the bottom line.
If you say "dBm" then the bottom line is 1mW
if you say "dBV" then the bottom line is 1 Volt
And just to add to the confusion you can specify dBm and actually compare VOLTAGES, because it is assumed that there is a 600Ω load connected to the voltage in question.
 

Thread Starter

AbsoluteC

Joined Sep 23, 2021
6
I understand it now after running back and forth with my notes. Thank you very much everyone for putting your feedback. This really helps me a lot!
 

MrAl

Joined Jun 17, 2014
8,505
I am not entirely sure i know what everyone is talking about here so i'll state an example.

First, every device that takes in energy and outputs energy has an efficiency rating. The efficiency would be the power out divided by the power in and it is in most cases less than 1. That is because most non laboratory stuff looses energy, and that is part of one of the laws of thermodynamics. There is one or two exceptions to this rule in real life but i wont go into them here (one involves quantum physics so it gets weird).

There is a theoretical exception to this and that is if we presume that we have an 100 percent efficiency device. This is not too uncommon because it allows a quick guide for figuring out how some things work without delving deep into theory or making careful measurements. This would not be typical for many problems however.

SO with that in mind, it looks like everyone here is assuming the 'audio' amplifier is 100 percent efficient and that is somewhat of a stretch of the imagination. That is because the only audio amp that could possibly ever be reasonably approximated as having 100 percent efficiency is a switching audio amp if i rem right that is class D. But even then we have to assume that everything that runs the components uses zero power and the MOSFETs use zero power, so that would be a bit rare i think.

The more reasonable idea is to think about a linear audio amplifier which can never have 100 percent or even 90 percent efficiency. That is mainly because an analysis of the power loss in the output transistor(s) is always significant. There is a theoretical maximum efficiency that would have to be either calculated or looked up and because the output stage is the major part analyzed for the efficiency calculation we'd have to know what class amplifier we were dealing with. Some can be as low as something like 30 percent, while a push pull is higher like maybe 80 percent (but we can calculate that more exactly).

The main point again being that we need that efficiency rating and here are a couple examples.
1. 100 watts input, efficiency 90 percent (0.90 decimal), output power must be 90 watts (Eff=Pout/Pin).
2. 200 watts input, efficiency 90 percent, output power must be 180 watts.

Now notice in the two examples above the input power doubled and the output power doubled. However, we could calculate that easily because we knew at least one level of input power and the one level of output power for that input level. Hence we can just double the output power.
Note that if we dont know at least one input/output power ratio, we can not calculate an absolute output power level we can only quote the ratio. In the above we could say the output power 'doubled' if we only assume that the input to output efficiency does not change with absolute power output, which in most cases it will, but we can sometimes ignore that for the sake of simplicity. However we can not quote the actual output power change in that case.

So now to the main question, if we have 2 watts input and increase that to 4 watts input, the output in all but possibly class D we will see much less than a 2 watt absolute increase in power output.
If the efficiency was constant over load and it was 50 percent then we must have had 1 watt on the output with 2 watts in, and so we will only see a 1 watt increase in the output power level. We could only calculate that 1 watt output increase because we knew that with 2 watts in we had 1 watt out, but if we didnt have that 1 watt reference we would only be able to say that the output power doubled that's it.

I hope i understood the question properly but the output stage calculation is an interesting endeavor in any case and is worth looking into. There are a few ways to do this, but the main idea is that the output transistors in anything but class D have to conduct BOTH current AND voltage at the same time because they have to convert pure DC into a pure sine (actual audio can be thought of as having sinusoidal components even though it is a more complex signal, but we usually use a sine wave test signal anyway), and that means they dissipate power, and that means much less efficiency.

I also hope this sheds some light on the topic.
 
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BobTPH

Joined Jun 5, 2013
3,825
Double the power and +3dB is only a little louder. 10 times the power sounds twice as loud.
You have been stating this forever wirh no attribution, so I decided to check it. I have always known that the relationship between power and loudness, but not that it was log base 10, which would be rather coincidental. So I decided ti look it up.

According to Wikipedia, the relationship follows Stephen’s law, with the exponent of 0.67. Looking up Stephen’s law, this gives us the equation:

L = kP^0.67

where L is apparent loudness and P is SPL.

So, if we raise the power by a factor if 10, we get an increase in apparent loudness of 4.7X.

And with a factor if 3 we get 2.08X.

So, it looks to me like it takes about 3X the power to double the loudness. This is much more in line with my experience testing speakers I have built. One Watt typically produces a reasonable listening level, while 10W is ear splitting.

Bob
 

dcbingaman

Joined Jun 30, 2021
476
I kind of see how it is in dB and dBm now.

10log(4/2) = 3 dB
10log(4000/2) = 33 dBm
These are not at all the same.
dB is a unitless quantity. dBm is not a unitless quantity and it has nothing to do with output power over input power. It is an actual measurement of power:
1632769561056.png


In this case the output power is 4W or 4000mW. Using the first equation we get approx:

36 dBm which is nothing more than saying the output power is 4000mW or 4W.

When you took 4000/2, you actually took 4000mW/2W watts cancel leaving you a unitless quantity that happens to be a factor of 0.001 of a unitless quantity.

Now it just so happens that division is the same as subtraction of logarithmic values thus the input power in dBm is:
33dBm.

So the gain (power out over power in) can also be expressed as:

36dBm-33dBm=3dB.

Subtraction of logs results in a unitless thing again this time being the power gain of 3dB.
 
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MrAl

Joined Jun 17, 2014
8,505
You have been stating this forever wirh no attribution, so I decided to check it. I have always known that the relationship between power and loudness, but not that it was log base 10, which would be rather coincidental. So I decided ti look it up.

According to Wikipedia, the relationship follows Stephen’s law, with the exponent of 0.67. Looking up Stephen’s law, this gives us the equation:

L = kP^0.67

where L is apparent loudness and P is SPL.

So, if we raise the power by a factor if 10, we get an increase in apparent loudness of 4.7X.

And with a factor if 3 we get 2.08X.

So, it looks to me like it takes about 3X the power to double the loudness. This is much more in line with my experience testing speakers I have built. One Watt typically produces a reasonable listening level, while 10W is ear splitting.

Bob
Hello there,

I think you meant Steven's Power Law?

Anyway, the way the ear responds to amplitude is extremely complex and i believe it is still an active area of research. To start with, a PDE with two spatial dimensions and one of time, then a bunch of other stuff following that. That is because the human ear is so complicated with several structures inside. I dont think anyone here wants to go that deep though and really would be looking for an approximation of some kind maybe similar to Steven's Power Law but i did not check that out yet.

However, the signal frequency is also a key factor which i will illustrate shortly. But first, the unit of signal intensity is dB and the unit of sound loudness is the phon. 2 phons seems twice as loud as 1 phon to the human ear for example, and since this is independent of frequency we have a scale to measure the perception of sound loudness if we know the loudness in phons.

A signal at 1000Hz at 10dB intensity would require 20db to sound twice as loud. That is because if we place the reference in phons at 10dB we get 10 phons, and at 20dB we get 20 phons. So there is a direct relationship between intensity in dB vs phons now.
Now to see how different this can be at a different frequency, at 100Hz it would take around 50dB to sound as loud as 1000Hz at 20dB because it takes 50dB to produce 20 phons, so we see a bigger change at the lower frequency. At 10000Hz, it would take about 30dB.
If we use this information to create a smoothed scale of phons vs frequency we might end up producing Steven's Power Law but i did not try that yet. It would be approximate, but very usable i believe.

So we see a very wide change in perceived sound level depending on frequency. The actual relationship is even stranger than that but it gets too complicated to talk about here.
But you can see that @Audioguru was right about the 10x law, just only at 1000Hz.
 
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