I have only been studying Electrical Engineering for about 1 year so please bare with me.
I just finished building my first power amplifier. When I tested it for maximum power at 8 ohms it clipped at 37V. If I am doing the math correctly thats 26.2V RMS , 85.7W RMS (171W Peak). The amp uses only one complementary output pair(2SK1058/2SJ162). I just didn't think that one pair could drive the signal with that much power. Also, the rail fuses on the PCB are 3A FB. With 26.7V RMS @ 8 ohms the current is 3.3A RMS (4.6A Peak), but the fuses do not blow, I don't understand why? I do not have the schematic on my computer, but if it's helpful I will post it. Any replies would be greatly appreciated.
I just finished building my first power amplifier. When I tested it for maximum power at 8 ohms it clipped at 37V. If I am doing the math correctly thats 26.2V RMS , 85.7W RMS (171W Peak). The amp uses only one complementary output pair(2SK1058/2SJ162). I just didn't think that one pair could drive the signal with that much power. Also, the rail fuses on the PCB are 3A FB. With 26.7V RMS @ 8 ohms the current is 3.3A RMS (4.6A Peak), but the fuses do not blow, I don't understand why? I do not have the schematic on my computer, but if it's helpful I will post it. Any replies would be greatly appreciated.
C
CryingDragon
Fuses usually need to be shorted across the rated load (like 3 amps shorted across it) continually for it to blow since fuses work on the concept that power generates heat and if say the output pair shorted and the entire PSU was shorted across the fuse then yes it would draw all the power your PSU could give and blow but you say your drawing 4.xx amps PEAK, peak is only for brief periods while the constant draw may only be 1 or 2. If you tried to use a 2 amp fuse cause you thought it would blow at three and protect your amp better this would be wrong because it would blow at your peak current draw say when the bass hits and cut out your amp when in fact there is nothing wrong at all this is why fuse ratings are chosen very carefully.
In summary
#1 you need to short at least the rated load across a fuse for generally 1.5 seconds to blow it
#2 Peak as I've said is only for brief periods and therefore will not blow the fuse.
#3 if the fuse blows fast when shorted across the load (as in shorted across the rails) but does not when the amp is running properly i would not see a problem with you rating.
In summary
#1 you need to short at least the rated load across a fuse for generally 1.5 seconds to blow it
#2 Peak as I've said is only for brief periods and therefore will not blow the fuse.
#3 if the fuse blows fast when shorted across the load (as in shorted across the rails) but does not when the amp is running properly i would not see a problem with you rating.
AFAIK the correct frequency for audio amplifier testing is 1 KHz, the output should be at a constant impedance (therefore not a speaker, thank God for that), and the distortion figure needs to be specified.
Also the quality of input is of prime importance. Is so because RMS power is actually a comparitive measurement (my interpretation) against DC, where there are no signal quality ramifications. Therefore a pure (or close to pure) sine wave will produce the most accurate results.
http://www.valhallascientific.com/applications/applications-1.shtml
The second part of it is a sales pitch, but the explanation is pretty good. there's also something called music power an A-weighted power, but I can't seem to find those references. Those put caps on distortion figures and specify nature and frequency of input signals.
Also the quality of input is of prime importance. Is so because RMS power is actually a comparitive measurement (my interpretation) against DC, where there are no signal quality ramifications. Therefore a pure (or close to pure) sine wave will produce the most accurate results.
http://www.valhallascientific.com/applications/applications-1.shtml
The second part of it is a sales pitch, but the explanation is pretty good. there's also something called music power an A-weighted power, but I can't seem to find those references. Those put caps on distortion figures and specify nature and frequency of input signals.
Fresh meat, where to start? "average power", that is to say, heat. RMS volts times RMS amps is average power.
Fuses. A typical normal blow fuse will withstand a 100% overload for one hour before it blows. It is the average heat that makes the link melt (blow). With a sine wave the peak current is 1.414 times the RMS current. As regards the rail fuses they will pass twice the peak current with sine waves in a push pull amplifier because the fuse gets a 'rest' every other half cycle. Music has such a high peak to average ratio that I have seen a 1A speaker fuse pass 400W on program material.
An optimal biased push pull amplifier with a regulated supply and a common emitter output stage can be as much as 78.5% efficent. That means a 100W amp will only have 21.5W of heat to get rid of.
The 2SK1058/2SJ162 pair can put out 5A RMS. If you used a +/- 63V regulated supply and ran them like the TransNova (Hafler) with a huge sink you would get 200W at 8 ohms out of one pair.
With the typical brute force unregulated supply (big caps + big transformer) you will get 100W out of one pair (B&K ST140).
Fuses. A typical normal blow fuse will withstand a 100% overload for one hour before it blows. It is the average heat that makes the link melt (blow). With a sine wave the peak current is 1.414 times the RMS current. As regards the rail fuses they will pass twice the peak current with sine waves in a push pull amplifier because the fuse gets a 'rest' every other half cycle. Music has such a high peak to average ratio that I have seen a 1A speaker fuse pass 400W on program material.
An optimal biased push pull amplifier with a regulated supply and a common emitter output stage can be as much as 78.5% efficent. That means a 100W amp will only have 21.5W of heat to get rid of.
The 2SK1058/2SJ162 pair can put out 5A RMS. If you used a +/- 63V regulated supply and ran them like the TransNova (Hafler) with a huge sink you would get 200W at 8 ohms out of one pair.
With the typical brute force unregulated supply (big caps + big transformer) you will get 100W out of one pair (B&K ST140).
The fuse question
Hi Rob
The answer why the fuses on your DC rails don't blow under this conditions is simply because the output current is shared by these two fuses (i.e. each one has to carry only the current of one half-cycle, plus the circuit's idle current).
The output-current related load (Ampères RMS) on each one of these is 0.707 times the output RMS current (i.e. 2.33 A).
And because the fuse has an intrinsic thermal delay it doesn't react to short peak current pulses (it is in fact something like a thermal RMS-current to temperature converter).
Regards
Charles
Hi Rob
The answer why the fuses on your DC rails don't blow under this conditions is simply because the output current is shared by these two fuses (i.e. each one has to carry only the current of one half-cycle, plus the circuit's idle current).
The output-current related load (Ampères RMS) on each one of these is 0.707 times the output RMS current (i.e. 2.33 A).
And because the fuse has an intrinsic thermal delay it doesn't react to short peak current pulses (it is in fact something like a thermal RMS-current to temperature converter).
Regards
Charles
Fuse ratings
It can be quite difficult to find the "right" rating for the fuses.
Fuses are usually classified into fast and slow, according to
one or several different standards. Some standards use
a more fine-grained classification, eg. IEC using very fast,
fast, medium, slow, very slow.
I don't have any of the standards at hand, but let me just give
two examples of manufacturers ratings for the time to melt.
I is the current rating of the fuse
Glass type fast 250V 0.1 - 8A ratings:
1.1I min 4h, 1.35I max 1h, 2I max 1s
Same type but slow:
1.1I min 4h, 1.35I max 1h, 2I max 60s
It can be quite difficult to find the "right" rating for the fuses.
Fuses are usually classified into fast and slow, according to
one or several different standards. Some standards use
a more fine-grained classification, eg. IEC using very fast,
fast, medium, slow, very slow.
I don't have any of the standards at hand, but let me just give
two examples of manufacturers ratings for the time to melt.
I is the current rating of the fuse
Glass type fast 250V 0.1 - 8A ratings:
1.1I min 4h, 1.35I max 1h, 2I max 1s
Same type but slow:
1.1I min 4h, 1.35I max 1h, 2I max 60s
Rob,
You are in engineering school and since you will be learning the precise definitions of things, I feel compelled to point out that the term "RMS power" is actually a very popular misnomer. The result of Vrms/R or Irms^2xR is AVERAGE power. If you were to calculate the RMS value of the power it would give a different number, one that would have no particular use.
To calculate the average power dissipated by an output transistor you need to integrate its VI wrt time over a complete cycle. This is where the 78% theoretical max class B efficiency number comes from. The VI curve is not a sinusoid so you have to do the calculus to work it out.
You are in engineering school and since you will be learning the precise definitions of things, I feel compelled to point out that the term "RMS power" is actually a very popular misnomer. The result of Vrms/R or Irms^2xR is AVERAGE power. If you were to calculate the RMS value of the power it would give a different number, one that would have no particular use.
To calculate the average power dissipated by an output transistor you need to integrate its VI wrt time over a complete cycle. This is where the 78% theoretical max class B efficiency number comes from. The VI curve is not a sinusoid so you have to do the calculus to work it out.
Rail Currents
Thanks, I forgot that the current is shared by each rail (2.33A RMS a rail). Since the fuses can handle bursts above their rating it makes sense that they can handle the 3.3A peak.
On the note of RMS power, do manufacturers, that give RMS power values, actually give the RMS or is it the average. I have not calculated the RMS power but it looks quite time consuming because, as traderbam pointed out, it is not simply a function of a sine wave. I think the fuction would be P= (V(p)*sin@)^2/R. I wonder how many "do it yourselfers" sit down and do the math. Thanks for the help.
Thanks, I forgot that the current is shared by each rail (2.33A RMS a rail). Since the fuses can handle bursts above their rating it makes sense that they can handle the 3.3A peak.
On the note of RMS power, do manufacturers, that give RMS power values, actually give the RMS or is it the average. I have not calculated the RMS power but it looks quite time consuming because, as traderbam pointed out, it is not simply a function of a sine wave. I think the fuction would be P= (V(p)*sin@)^2/R. I wonder how many "do it yourselfers" sit down and do the math. Thanks for the help.
For an undistorted sine wave, average and rms are the
same thing.
The reason 60 Hz was chosen is that cheap multimeters
are usually most accurate at 60 Hz.
You can listen for distortion in a power amp under test if
you don't have a scope. Make a voltage divider up which
attenuates the output until you can use a headphone to
listen to the sinewave at a comfortable level.
This is usually at least 100 to 1 or more, so try something
like 10K ohm in series with 100 ohm, with the 10K looking
at the output and the 100 ohm looking at ground, and the
junction between them driving the headphone.
As you turn the amp up at 60 Hz or so, you will be able to
hear the clipping harmonics as you turn the input level up.
Don't put on the headphones until you have verified the
acoustic level!!!!
If it's too loud, reduce the 100 ohm value.
same thing.
The reason 60 Hz was chosen is that cheap multimeters
are usually most accurate at 60 Hz.
You can listen for distortion in a power amp under test if
you don't have a scope. Make a voltage divider up which
attenuates the output until you can use a headphone to
listen to the sinewave at a comfortable level.
This is usually at least 100 to 1 or more, so try something
like 10K ohm in series with 100 ohm, with the 10K looking
at the output and the 100 ohm looking at ground, and the
junction between them driving the headphone.
As you turn the amp up at 60 Hz or so, you will be able to
hear the clipping harmonics as you turn the input level up.
Don't put on the headphones until you have verified the
acoustic level!!!!
If it's too loud, reduce the 100 ohm value.
"For an undistorted sine wave, average and rms are the
same thing."
No, I disagree. The average value of a sinewave is zero. Also, the power waveform of a sinusoidal voltage across a resistive load is not sinusoidal. This is why average power is not the same thing as RMS power, mathematically speaking.
I think that almost invariably manufacturers mean average power when they talk about "RMS power". It's become a derigeur term. When they talk about "music power" I'm at a loss to understand.
Respect to you on the Pass product website that you don't use the term "RMS power". I assume all your specified power ratings are average powers.
same thing."
No, I disagree. The average value of a sinewave is zero. Also, the power waveform of a sinusoidal voltage across a resistive load is not sinusoidal. This is why average power is not the same thing as RMS power, mathematically speaking.
I think that almost invariably manufacturers mean average power when they talk about "RMS power". It's become a derigeur term. When they talk about "music power" I'm at a loss to understand.
Respect to you on the Pass product website that you don't use the term "RMS power". I assume all your specified power ratings are average powers.
Math to show that the RMS function applied to the power function based on an undistorted sine wave into a non-complex load is higher than the average power(being equal to RMS Voltage * RMS Current).
for a 1khz 20V p-p sine wave you have the equation:
V= 10*cos( 1khz * 2 * PI * t )
Into a load resistor of say 8 ohms, the current is then:
I= 10/8 * cos( 1khz * 2 * PI * t )
The power function is then:
P = 100/8 * cos^2( 1khz * 2 * PI * t )
Applying the RMS function to the power function, first the terms are squared:
P^2 = 156.25 * cos^4( 1khz * 2 * PI * t )
Integrating over one period (500us) and dividing by one period for the mean:
(insert use of graphing calculator)
58.61
figuring the square root of 58.61:
7.655 Watts RMS
The Peak Power is 10*10 / 8 = 12.5 Watts Peak
The Average Power is (.707)*10 * (.707)*10/8 = 6.25 Watts Average.
jt
for a 1khz 20V p-p sine wave you have the equation:
V= 10*cos( 1khz * 2 * PI * t )
Into a load resistor of say 8 ohms, the current is then:
I= 10/8 * cos( 1khz * 2 * PI * t )
The power function is then:
P = 100/8 * cos^2( 1khz * 2 * PI * t )
Applying the RMS function to the power function, first the terms are squared:
P^2 = 156.25 * cos^4( 1khz * 2 * PI * t )
Integrating over one period (500us) and dividing by one period for the mean:
(insert use of graphing calculator)
58.61
figuring the square root of 58.61:
7.655 Watts RMS
The Peak Power is 10*10 / 8 = 12.5 Watts Peak
The Average Power is (.707)*10 * (.707)*10/8 = 6.25 Watts Average.
jt
Now this was a long time ago but is it not like this (and I intend to agree with Pass that average is indeed the same as RMS):
The average and RMS value is the same for the power NOT the voltage (unless ideally rectified right?) which could be the reason for confusion. Using amplitudes as 1 V and load as 1 Ohm we get that the power is SIN^2(x) which indeed is a sinusoidal wave form but with twice the original frequency (or so it seems to me tonight). This sinusoidal wave is always positive and maximum 1 in amplitude as well as symmetric around its average and this means that is can be nothing but average 0.5. The RMS value would be the RMS voltage squared divided by the load. The RMS value of the voltage is 1/SQRT(2) which squared would 0.5. Hence they are the same.
Again, as usual, I might be wrong but it seems correct. Average and RMS for the undistorted sinusoidal power wave form are the same.
/Urban
The average and RMS value is the same for the power NOT the voltage (unless ideally rectified right?) which could be the reason for confusion. Using amplitudes as 1 V and load as 1 Ohm we get that the power is SIN^2(x) which indeed is a sinusoidal wave form but with twice the original frequency (or so it seems to me tonight). This sinusoidal wave is always positive and maximum 1 in amplitude as well as symmetric around its average and this means that is can be nothing but average 0.5. The RMS value would be the RMS voltage squared divided by the load. The RMS value of the voltage is 1/SQRT(2) which squared would 0.5. Hence they are the same.
Again, as usual, I might be wrong but it seems correct. Average and RMS for the undistorted sinusoidal power wave form are the same.
/Urban
I guess if everyone uses the average power then it does not matter if you can calculate the RMS. However, I thought that the RMS value of a function f(x) is
(S - symbol for integration)
(T - symbol for period (360 deg))
P(rms) = ((1/T) * S(f(x)^2)dx)^(1/2)
and the Average Power Function is
P(avg) = ((2/T) * S(f(x))dx) - integrated over half the wave (180 deg)
if f(x) is the power function then f(x) = ((V(p)*sinx)^2)/R
subing it into the Avg Function and working it out
P(avg) = (V(p)^2)/2R
So we have the right value for P(avg) but when f(x) is subbed into the P(rms) function we get
P(rms) = ((V(p)^2)/R) * (3/8)^(1/2)
This is not the same value, the RMS is larger than the average.
Mr. Pass said that they are the same with an undistorted signal so I must be doing something wrong. I suppose it does not matter since the average is all that is used, I would just like to clear this up.
(S - symbol for integration)
(T - symbol for period (360 deg))
P(rms) = ((1/T) * S(f(x)^2)dx)^(1/2)
and the Average Power Function is
P(avg) = ((2/T) * S(f(x))dx) - integrated over half the wave (180 deg)
if f(x) is the power function then f(x) = ((V(p)*sinx)^2)/R
subing it into the Avg Function and working it out
P(avg) = (V(p)^2)/2R
So we have the right value for P(avg) but when f(x) is subbed into the P(rms) function we get
P(rms) = ((V(p)^2)/R) * (3/8)^(1/2)
This is not the same value, the RMS is larger than the average.
Mr. Pass said that they are the same with an undistorted signal so I must be doing something wrong. I suppose it does not matter since the average is all that is used, I would just like to clear this up.
NP - I find your distortion test interesting and will have to try it (carefully of course)
As for RMS vs 'Average' Power rating I find that you are actually splitting hairs here. It's a way of trying to define the 'power' of an audio amplifier. In all actuallity the thing that is most interesting (to this DIYer at least) is not the 'Power' per se as much as it is the actual SPL produced using a specific source and a specific load (Both mine of course) to create a whole system. I tend to think that we equate 'Power' to SPL, that is that as an average DIYer or consumer we would say OOOH, I need it loud (high SPL) therefore I need a XXX Watt power amp. This is not necisarily so as most around here know. I think that problem has been how the whole equation has been presented to consumers over the years. Marketing types (as well the media) assume the average consumer to be a bit of a dolt and simplify things to the use of 1 or 2 specific numbers to represent a SYSTEM. This is ineffective to say the least. I understand that it is important to know the specified power of components in a system so that you don't blow things up, but I would concede that most DIYer are not as interested in the hair splitting math as it is really not the final outcome. Reproduction of the music is the final outcome, and as long as my speakers match up to my amp power wise I am happy with 'Power Rating' and I also know that if the SPL is too high or too low there are various things that I can do to change this other than increasing the 'Power' of my amp. Thus I don't feel that I really need the calculus approach. In the end my Ear is the ultimate judge for me.
-Dozuki
As for RMS vs 'Average' Power rating I find that you are actually splitting hairs here. It's a way of trying to define the 'power' of an audio amplifier. In all actuallity the thing that is most interesting (to this DIYer at least) is not the 'Power' per se as much as it is the actual SPL produced using a specific source and a specific load (Both mine of course) to create a whole system. I tend to think that we equate 'Power' to SPL, that is that as an average DIYer or consumer we would say OOOH, I need it loud (high SPL) therefore I need a XXX Watt power amp. This is not necisarily so as most around here know. I think that problem has been how the whole equation has been presented to consumers over the years. Marketing types (as well the media) assume the average consumer to be a bit of a dolt and simplify things to the use of 1 or 2 specific numbers to represent a SYSTEM. This is ineffective to say the least. I understand that it is important to know the specified power of components in a system so that you don't blow things up, but I would concede that most DIYer are not as interested in the hair splitting math as it is really not the final outcome. Reproduction of the music is the final outcome, and as long as my speakers match up to my amp power wise I am happy with 'Power Rating' and I also know that if the SPL is too high or too low there are various things that I can do to change this other than increasing the 'Power' of my amp. Thus I don't feel that I really need the calculus approach. In the end my Ear is the ultimate judge for me.
-Dozuki
I wish I could have said it that well. What does the amp sound like. It seems to me that that is the most important thing. I don't listen to music with a DMM nor an ocsilliscope. I couldn't agree more. I listen to an old NAD integrated that puts out a whopping 60 watts or so. Sounds great with my fairly efficient speakers(91dB/ 1watt 1 meter.
RMS vs Average
RMS is used in two contexts:
The first is in probability theory; useful for electrical engineering when dealing with component tolerance issues. If you have want to know the standard deviation of the sum of a bunch of random variables, it's the RMS of their individual standard deviations.
The other is to give a quick "equivalent DC" voltage or current that allows you to calculate AVERAGE power into resistive loads as (Vrms^2)/R or (Irms^2)*R. The RMS comes from calculating power as the average of V*I and substituting V=I*R or I=V/R ...
RMS power isn't meaningful; I'd bet that anyone who quotes RMS power either doesn't know what he's doing, or is really quoting average power derived from RMS voltage or current. RMS power isn't meaningful because power is the rate at which energy is dissipated. To calculate total energy expended, multiply average power by time. You can't multiply RMS power by time to get any meaningful number.
Sometimes, people mistakenly make RMS current measurements when they really want average current measurements. This happens when the objective is to calculate average power drawn from a constant voltage power supply. In this case, the appropriate current measurement is average, not RMS, since the voltage doesn't vary with the current. You'd be surprised how many people make this mistake.
RMS is used in two contexts:
The first is in probability theory; useful for electrical engineering when dealing with component tolerance issues. If you have want to know the standard deviation of the sum of a bunch of random variables, it's the RMS of their individual standard deviations.
The other is to give a quick "equivalent DC" voltage or current that allows you to calculate AVERAGE power into resistive loads as (Vrms^2)/R or (Irms^2)*R. The RMS comes from calculating power as the average of V*I and substituting V=I*R or I=V/R ...
RMS power isn't meaningful; I'd bet that anyone who quotes RMS power either doesn't know what he's doing, or is really quoting average power derived from RMS voltage or current. RMS power isn't meaningful because power is the rate at which energy is dissipated. To calculate total energy expended, multiply average power by time. You can't multiply RMS power by time to get any meaningful number.
Sometimes, people mistakenly make RMS current measurements when they really want average current measurements. This happens when the objective is to calculate average power drawn from a constant voltage power supply. In this case, the appropriate current measurement is average, not RMS, since the voltage doesn't vary with the current. You'd be surprised how many people make this mistake.
I don't think you can say that it's not meaningful, any
less than any other measurement. If the waveform
is not a square wave or a sine wave or DC then you have
to perform a little calculus to find out how much heating
it might be giving a resistive load.
Since we spend all out time talking about power in terms
of sine waves at the output of amplifiers, and for an undistorted
sine wave rms and average watts are the same, it is moot.
But if you are calculating ripple currents in power supply
capacitors, or its cousin, power factor as seen by the AC line,
the rms measurements are all you can work with if you are
concerned about the resistive losses in the conductors.
less than any other measurement. If the waveform
is not a square wave or a sine wave or DC then you have
to perform a little calculus to find out how much heating
it might be giving a resistive load.
Since we spend all out time talking about power in terms
of sine waves at the output of amplifiers, and for an undistorted
sine wave rms and average watts are the same, it is moot.
But if you are calculating ripple currents in power supply
capacitors, or its cousin, power factor as seen by the AC line,
the rms measurements are all you can work with if you are
concerned about the resistive losses in the conductors.
The point of pointing out the errors was to keep the facts straight from the get-go for somebody relatively new to this stuff.
In hifi I don't think it is as much of an issue, but with cheaper stuff I can imagine a lot of companies are leveraging this in their marketing campaigns. Not that 120 watts is going to be much different than 100, but it can provide some insight to the crediblity of a company's other claims too.
If companies didn't exaggerate their specs, then this is exactly how it would work.
RobPhill33, why not drop by the office of the power systems/conversion professor in your EE department to see if he can give you a better explanation.
jt
In hifi I don't think it is as much of an issue, but with cheaper stuff I can imagine a lot of companies are leveraging this in their marketing campaigns. Not that 120 watts is going to be much different than 100, but it can provide some insight to the crediblity of a company's other claims too.
If companies didn't exaggerate their specs, then this is exactly how it would work.
RobPhill33, why not drop by the office of the power systems/conversion professor in your EE department to see if he can give you a better explanation.
jt
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