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Basic Amplifier Question
Basic Amplifier Question
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Old 4th October 2002, 01:57 AM   #1
RobPhill33 is offline RobPhill33  Canada
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Default Basic Amplifier Question

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.
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Old 4th October 2002, 03:45 AM   #2
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.
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Old 4th October 2002, 03:56 AM   #3
Jared is offline Jared
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how does one go about measuring rms power?
Driving a 60 hz signal to a power resisotr and measuring the voltage with an Ac multimeter? Do you need a scope to be sure it is not clipping or distorting at all?
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Old 4th October 2002, 04:50 AM   #4
sangram is offline sangram  India
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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.


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.
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Old 4th October 2002, 09:45 AM   #5
djk is offline djk
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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).
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Old 4th October 2002, 10:21 AM   #6
phase_accurate is offline phase_accurate
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Default 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).


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Old 4th October 2002, 12:07 PM   #7
Christer is offline Christer  Sweden
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Default 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
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Old 4th October 2002, 01:20 PM   #8
traderbam is offline traderbam  Europe
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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.
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Old 4th October 2002, 04:08 PM   #9
RobPhill33 is offline RobPhill33  Canada
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Default 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.
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Old 4th October 2002, 07:33 PM   #10
Nelson Pass is offline Nelson Pass  United States
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Basic Amplifier Question
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.

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