...............Then in reverse order multiplying current times voltage 40x5 from the power formula, does that give us a peak power of 200 watts per rail, or 400 watts per channel? If this is not correct, please explain simply why it is not correct, I think this will be useful for a lot of people.
Yes, BUT you will draw power from one rail ONLY half the time (half cycle sine) and the other half of the time you will draw power from the second rail. So 50% of the current will come from the positive source for positive half cycle sine and for negative half cycle sine the current will come from the negative source.
You will not draw current from both positive and negative power rail simultaneous and therefore you do not need 200W+200W because only one of the outputs will deliver current to the output, so you will have 200W+0W for 50% of the time and then 0W+200W for the other 50% of the time . For a class A amplifier is a completely different situation.
(200W+0W) * 50% + (0W+200W) * 50% = 200W*1/2 + 200W*1/2 = 100W+100W = 200W PEAK POWER not RMS or continues power for one 100W channel.
The same thing was also said in posts #10, #12, #22, #26, but it seems that you either don't read or you don't manage to understand what you read because you keep coming back with the same misunderstandings.
Last edited:
In fact, these 200 watts are divided into two, i.e. 100 watts per rail? Then everything falls into place.
And never use the diagram in post #29 for a symmetric powered audio amplifier. If one of the fuses burns, there is a big risk that the speakers will also burn because the amplifier is left without one of the power supplies and will deliver DC at the output.
Better yet, if you still want to use 2 rectifier bridges, use one for each channel and separate filter capacitors and thus at least improve the separation between the channels.
Better yet, if you still want to use 2 rectifier bridges, use one for each channel and separate filter capacitors and thus at least improve the separation between the channels.
So far so good, thanks to everyone who helped me figure out the amp power supply situation. Now I want to ask something else:
It is clear that at high powers, such as those mentioned above, a significant amount of heat is released. Is there a simple way to determine for example how big a cooling radiator should be for a 100 watt into 8 ohm amp?.
And another question, how do we estimate how many watts an available finned heatsink will dissipate?
It is clear that at high powers, such as those mentioned above, a significant amount of heat is released. Is there a simple way to determine for example how big a cooling radiator should be for a 100 watt into 8 ohm amp?.
And another question, how do we estimate how many watts an available finned heatsink will dissipate?
There are a "Rth" parameter defined for any heatsink but can be find only for standard heatsink. This parameter is used to calculate the heatsink or the temperature.
Yes. It tells you how much the temp will rise with a certain dissipation. Should be in the heatsink data sheet.
The problem with audio is, again, to find a plausible figure for the disspation under music conditions.
Lots of handwaving and experience involved.
Jan
The problem with audio is, again, to find a plausible figure for the disspation under music conditions.
Lots of handwaving and experience involved.
Jan
I did some checking and everything I found and heard confirms what jan has been trying to get across, namely that audio amplifier transformers tend to be underrated - or at least not overrated - for the theoretical maximum output power of the amplifier - typically they have a VA rating of only 70-80% of what would be required "by the numbers" for the maximum output as, for music/voice applications, this is more than enough. Pass a sine wave through at maximum amplification and soon the transformer will overhead and Bad Things(tm) will happen - but music will be cause no issues.
Hal
Hal
That's right HalFoster "....and soon the transformer will overhead and Bad Things will happen"- such as melted insulation, shorted windings, and...well blown transformer. Only those who have suffered from it can understand. Perhaps this is the reason, and as said above, that the power of the power supply is 2-4 times more than the power of the amplifier. In this regard, I think that really few people understand the essence of the problem, and in this regard they are guided mainly by the price and size, weight even, and not so real capabilities of a well-calculated power supply. And on the issue of radiators, can someone give a concrete example of a calculation?
Thmr, apologies if I am misunderstanding what you are saying but to clarify: my point was that most commercial amplifiers are designed with a transformer that is only 70-80% the size you would think was needed by the full power specifications of the amplifier. If you were to play a continuous full power sine wave through one it would overheat and die - but we don't play sinewaves at full power, we play music which - even if played at full power - would not overstress the amplifier's transformer. Heat dissipation in other areas might be an issue but the transformer itself should be fine. The designers of these amplifiers are trying to reduce BOM costs as much as they can and, as a rule, won't spend the money for an higher priced - and unnecessary - component if a smaller and cheaper one will do. Yes, I'm sure that there are exceptions, and that they are priced accordingly, but not for the majority of commercially sold amplifiers.
Hal
Hal
For a 2x100W amplifier that run 1000 hour at maximum power, you need 300W transformer and not 400W-800W.
Even a 280W transformer will survive without problems.
Seems that you still don't understood, the 400W transformer is recommended not because a 280W transformer will tot survive BUT because the voltage will be more stable at the output.
Even a 280W transformer will survive without problems.
Seems that you still don't understood, the 400W transformer is recommended not because a 280W transformer will tot survive BUT because the voltage will be more stable at the output.
Regarding the two questions I asked above about the radiators, should I post a separate topic, or can someone help with practical advice here as well. I would be very grateful for clarifications in this regard.
Thmr: there is really no "right" answer as it depends on the heatsinks, case design, forced air or not and design tolerances. An good resource for heatsink design can be found here: https://sound-au.com/heatsinks.htm and Rod really has a huge collection of very accurate information on audio electronic design on his site.
Hal
Hal
Your not using RMS values. +/- 40v rails gives about 28 volts RMS output which gives you about 100 watts out into 8 ohms.I have a transformer but the question is whether I have correctly calculated the current at +–40 volts and an 8 ohm load, namely 5 amps, and since I have two 40 volt halves, is it 5 amps each or 2.5 amps for each?
- Home
- Amplifiers
- Power Supplies
- Calculate PSU