Even so, but so far no one has given a direct answer to my question, so I will ask you too dear chip_mk - with the given calculations of 40 volts and 3.55 amps, is it 2x40 volts at 2x3.55 amps?
First, what kind of current is in the question? Peak, effective or mean? Relation between them varies depending on the current waveform. Therefore throwing simple scalar value is ambiguous.
Second, there are two symetrical power sources positive and negative, 40V each. Positive rail sources current, it flows through output transistor, through the speaker and ends to ground. Negative rail is practicly disconnected. Roles are switched in the negative half sycle. So the current never goes from rail to rail. Hence it's never 2x40V supply but only one half at time.
Second, there are two symetrical power sources positive and negative, 40V each. Positive rail sources current, it flows through output transistor, through the speaker and ends to ground. Negative rail is practicly disconnected. Roles are switched in the negative half sycle. So the current never goes from rail to rail. Hence it's never 2x40V supply but only one half at time.
The efficiency for AB class amplifier is no more than 60%.
400W transformer should be good for your 2x100W amplifier.
Pay attention to the waveform. It's half cycle sine. Starts from zero amps, rises to 5A (peak of the sine), falls symetricly to zero. Stays zero for half cycle.
How many amps is this? This isn't typical AC nor DC. Mean current is i think (got to check) 5A/(2*pi). Effective or RMS is if I'm not mistaken 5A/(2*sqrt(2)). This is for single rail.
Again, pay attention to the waveform to understand what's going on. Sheer number is of less importance because the same numeric value can mean different things in different situations.
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So assuming it's 3.55 amps per 40 volt coil, then 40x3.55 makes 142 watts per arm, or 2x142 watts per channel, 284 watts total, and multiplying that by 1.4 for max power, we get 397.6 volt amps, and this for one channel of the amplifier, or a total of 795.2 volt amps of transformer power for both channels of the amplifier. Am I right?
NO!
NO!
NO!
NO!
NO!
You will have 2x100W in the speaker. Were will go the rest of 595W?
What is so difficult to understand?
The transformer must ensure the power delivered to the loudspeakers plus the losses generated by the amplifier's efficiency, which is between 50-60%).
If we consider that the amplifier has an efficiency of 50%, then the transformer must be 400W if we have 200W (2x100W/channel) at the output.
Everything above this power is nothing more than added weight to the amplifier without noticeable improvements in sound.
Keep in mind that anyway the power peaks will be delivered from the capacitors and not from the transformer.
In fact, the entire amplifier will be powered by the filter capacitors, the transformer does nothing but charge these filter capacitors 100 or 120 times per second.
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Correct only for one moment when sine is on it's peak. One should calculate integral of 40Vx5A*sin(x) for x between 0 and pi and divide it with 1/(2*pi).
You know, consumed power continuously changes as sine changes. When we speak of dissipated power we often assume averaged power. To calculate it we need to use integral to calculate total energy consumed during one full cycle and divide it to the duration of the cycle.
You know, consumed power continuously changes as sine changes. When we speak of dissipated power we often assume averaged power. To calculate it we need to use integral to calculate total energy consumed during one full cycle and divide it to the duration of the cycle.
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No.
Anyway, this rectifier scheme, although it is widely circulated among audiophiles, is bad. What happens when one of the fuses in front of the rectifier bridges burns out? One of the voltages disappears and you will have high DC at the output that can burn the speakers. Much better is the scheme with a single bridge rectifier and transformer with the secondaries connected in series.
After the quick math I've done, it's 40V*5A/pi watts averaged consumption per rail, that is 64W per rail, 130W per channel.
This is what Rod Elliott says in his work Linear Power Supply Design ..."As an example, a 100W/ 8 Ohm power amp will have a maximum output current of about 3.5A RMS into a resistive load. Since we know that a loudspeaker is not resistive, this figure can be doubled, to 7A. In fact, the inductive load of a loudspeaker will reduce the current delivered to the load and provided by the power supply, but let's not allow reality to cloud the issue...........
The supply current for each supply (+ve and -ve) will therefore have a peak value of about 10A (7 × 1.414), and the average will be 1/2 of the speaker current, or 3.5A".
The supply current for each supply (+ve and -ve) will therefore have a peak value of about 10A (7 × 1.414), and the average will be 1/2 of the speaker current, or 3.5A".
Leave alone all those calculations that you can't seem to understand properly.
You have to answer the question: "where does the power difference between the power of the transformer and the power delivered to the speakers go?".
The output current of a transformer is not fixed and clearly limited. A transformer listed as having 3A at the output can deliver much higher currents for short times (tenths of a second...seconds).
The power of the transformer must be chosen so that it is not exceeded in the long term because it leads to overheating and even destruction.
You have to answer the question: "where does the power difference between the power of the transformer and the power delivered to the speakers go?".
The output current of a transformer is not fixed and clearly limited. A transformer listed as having 3A at the output can deliver much higher currents for short times (tenths of a second...seconds).
The power of the transformer must be chosen so that it is not exceeded in the long term because it leads to overheating and even destruction.
On the contrary, I think I understood everything well enough. Rod's calculations were done at 28.8 volts per arm supply because 28.8v x 3.53a gives us 101.6 watts per arm. When we divide the 28.8 volt arm voltage by the 8 ohm load at the output, that gives us roughly 3.5 amps per arm, just as Rod points out. This figure is multiplied by 2 because the load is not resistive, making 7 amps per arm, or multiplied by 1.4 per arm for peak current, this gives us 10 amps. In other words, 28.8x10 gives us 280 volt amps per arm, or a total of 560 volt amps per channel for both arms. Here we are talking about a constant power of 100 watts which will be given without problem for a long time. These figures should of course be doubled if we want a constant power of 200 watts at 4 ohms, and for 400 watts at 2 ohms, not to mention what currents we are talking about if we want such power constantly for a long time. Here, of course, we do not forget the cooling problem, which I think we will talk about later.
It is clear that there is no difference here, it is all a matter of accurate calculations, as Rod does. The fact is that what I said coincides with what Rod said, and with that in mind, let me ask you something:
1. Do you dispute Rod's calculations and claim them to be correct?
2. Why buy or build an amplifier that only for a short time, ie. for seconds it would give 100 watts of power, and realistically for a long time it would give power of 30-40 watts, but you pay as for a 100 watt amplifier. For example, if you buy such an amplifier, you are practically a victim of marketing.
3. Personally, I think that if it says 100 watts, then the amplifier should really give it for a long time, and otherwise it is a completely different question how much and whether I will listen to such a power. At the end of the end, if you pay for 100 watts, you should be able to use these powers for a long time.In this situation, It is more than clear that everything "will be big, expensive and heavy", but this is also the case with the big names in this industry.
4. I want to thank everyone who participated in this discussion, it helped me clear up a lot of things about power suply for amplifiers.
1. Do you dispute Rod's calculations and claim them to be correct?
2. Why buy or build an amplifier that only for a short time, ie. for seconds it would give 100 watts of power, and realistically for a long time it would give power of 30-40 watts, but you pay as for a 100 watt amplifier. For example, if you buy such an amplifier, you are practically a victim of marketing.
3. Personally, I think that if it says 100 watts, then the amplifier should really give it for a long time, and otherwise it is a completely different question how much and whether I will listen to such a power. At the end of the end, if you pay for 100 watts, you should be able to use these powers for a long time.In this situation, It is more than clear that everything "will be big, expensive and heavy", but this is also the case with the big names in this industry.
4. I want to thank everyone who participated in this discussion, it helped me clear up a lot of things about power suply for amplifiers.
Read again! Also Rod say that for 100W amplifier is enought a 100W transformer.
And please respond to the above questions.
Ps: if you search on forum you will find a measurement made by me where is plotted (on the scope) the peak power (amplifier needed power) and mean power (transformer needed power x 1.5 for 50% efficiency).
And please respond to the above questions.
Ps: if you search on forum you will find a measurement made by me where is plotted (on the scope) the peak power (amplifier needed power) and mean power (transformer needed power x 1.5 for 50% efficiency).
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