Okay, one quick question regarding current draw.
Say I'm calculating max current drawn from the sub:
8Ω
90W max
so that would be a voltage of ~26.8
so 26.8/8 = ~3.35 Now does this get drawn from each rail or is this the current drawn from the positive and negative rails of the amp's power supply added together?
I guess it's just important for me because I need to know if I need to get a transformer with a higher current rating. Mine can provide 3.33A per rail, but if this is the draw per-rail (3.35A) then I will obviously need a larger transformer. However if that is spilt the speaker would only be drawing ~1.68A per rail, well within spec
Thanks!
Say I'm calculating max current drawn from the sub:
8Ω
90W max
so that would be a voltage of ~26.8
so 26.8/8 = ~3.35 Now does this get drawn from each rail or is this the current drawn from the positive and negative rails of the amp's power supply added together?
I guess it's just important for me because I need to know if I need to get a transformer with a higher current rating. Mine can provide 3.33A per rail, but if this is the draw per-rail (3.35A) then I will obviously need a larger transformer. However if that is spilt the speaker would only be drawing ~1.68A per rail, well within spec
Thanks!
The current draw of the speaker has little to do with the AC current that a transformer is rated for. Ignore current and use the formula:
Transformer VA = 1times to 2times the total maximum power output.
The current draw of the amplifier varies a lot, from -lot's of amperes through zero amperes to +lot's of amperes and back again, repeatedly. This varying load current comes from the capacitors connected to the supply rails.
The transformer simply charges those capacitors back up again.
As you have done you can work out the voltage and current fed to a resistor when it is driven to a fixed (or maximum) power.
The speaker demand is very different.
Expect the transient demand of a reactive speaker load to be upto ~3times the peak current delivered to the resistor of the same nominal value.
3.38Aac * 26.8Vac is equivalent to ~90W
The peak current into the resistor is sqrt(2) * Iac ~ 4.8Apk
The maximum transient peak current demanded by the amplifier could be as high as 14Apk.
The capacitors must supply that transient peak demand.
Remember that -ve current and +ve current I talked about earlier.
The +ve current comes form the capacitors storing the +ve charge.
The -ve current comes from the capacitors storing the -ve charge.
Transformer VA = 1times to 2times the total maximum power output.
The current draw of the amplifier varies a lot, from -lot's of amperes through zero amperes to +lot's of amperes and back again, repeatedly. This varying load current comes from the capacitors connected to the supply rails.
The transformer simply charges those capacitors back up again.
As you have done you can work out the voltage and current fed to a resistor when it is driven to a fixed (or maximum) power.
The speaker demand is very different.
Expect the transient demand of a reactive speaker load to be upto ~3times the peak current delivered to the resistor of the same nominal value.
3.38Aac * 26.8Vac is equivalent to ~90W
The peak current into the resistor is sqrt(2) * Iac ~ 4.8Apk
The maximum transient peak current demanded by the amplifier could be as high as 14Apk.
The capacitors must supply that transient peak demand.
Remember that -ve current and +ve current I talked about earlier.
The +ve current comes form the capacitors storing the +ve charge.
The -ve current comes from the capacitors storing the -ve charge.
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With 26.8VDC rails powering LM4780 into 8 ohm load it's rated for only ~ 36W output (@ 1% THD) per pg 11 of datasheet http://www.ti.com/lit/ds/symlink/lm4780.pdf which shows that to get closer to 90W (> 80W @ 1% THD) you need 40VDC rails.
In addition to what Andrew wrote, consider whether you "need" full rated output wattage. If you don't you can set the amp chip gain lower to allow full volume (knob) travel (or similarly, input audio signal voltage control from the audio source) without distortion at the (new, limited) peak output. See figure 3 & 4 on this page: Undertanding Power Amplifier Power Ratings though it doesn't take into account any forward voltage drop across the amp chip itself which according to datasheet linked above (pg 3) is typically 1.6V drop positive rail, 2.5V drop negative rail.
Point being, your transformer (assuming it produces +/-26.5VDC ) has plenty of current capability for the output power it allows, where it's lacking is voltage "IF" you're trying to reach close to 90W... but with the entire thing, including heatsink, built inside the monitor cabinet it seems better not to build for max output, too much heat.
In addition to what Andrew wrote, consider whether you "need" full rated output wattage. If you don't you can set the amp chip gain lower to allow full volume (knob) travel (or similarly, input audio signal voltage control from the audio source) without distortion at the (new, limited) peak output. See figure 3 & 4 on this page: Undertanding Power Amplifier Power Ratings though it doesn't take into account any forward voltage drop across the amp chip itself which according to datasheet linked above (pg 3) is typically 1.6V drop positive rail, 2.5V drop negative rail.
Point being, your transformer (assuming it produces +/-26.5VDC ) has plenty of current capability for the output power it allows, where it's lacking is voltage "IF" you're trying to reach close to 90W... but with the entire thing, including heatsink, built inside the monitor cabinet it seems better not to build for max output, too much heat.
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Okay, and also to note my rails will be about ±34VDC.
I also have been planning on making sure that the full "range" of the volume knob will stay out of the distortion range of my amplifier, which will be dependent on the supply voltage.
I'm going to look into this article you posted! thanks!
I also have been planning on making sure that the full "range" of the volume knob will stay out of the distortion range of my amplifier, which will be dependent on the supply voltage.
I'm going to look into this article you posted! thanks!
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