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Filter Capacitance

Hello Audiogonners! 

Perhaps this should be placed in tech talk - not sure.... 

In my own clumsy way, I'm building more understanding of amplifiers. I have no electrical training, save some very awkward swings of that hot gun that interacts with solder and a PCB, back in High School, about 400 years ago!

So, I see there are two players in the amp game that I do not understand, about which I am studying (more study of other pieces of the system, later): Filter capacitance and transformer size. Until a few days ago, I might have (entirely) mixed the two in my head... now it's only a partial mix... So, I'm interested to see, for example, that the Schiit Aegir has the following hardware, so far as is described on a Schiit-Euro site:

"The power supply is completely linear, with 600VA transformer, over 150,000uF of filter capacitance." 

That seems like A LOT of capacitance, and I'm totally confounded as to why this doesn't translate into higher output. As you know, the Aegir is described as 20 watts per channel into 8 ohms, and maybe 40 into 4 ohms. Seems like low wattage for all that transformer and capacitance weight! 

The Vidar runs the same size transformer but only 40,000uF of filter capacitance - 20K per side! AND it's a 100w into 8ohm amp - 200 into 4ohms. 

I own the STA200 by Nuforce and I cannot find anything describing capacitance, nor transformer size... I'm studying amplifier tech because I'm running into some difficulty when running the STA200 into my Moabs, at high volumes characterized by significant bass extension: think giant BOOM and you get the idea. Most of what I listen to is jazz/acoustic, so it's not a big issue until I move over to something that emphasizes boom, so I'm curious as to how the tech shows up in these applications...
listening99
erik_squires
16,052 posts
 
The absolute maximum voltage is set by the output voltage of the transformer. We don’t actually know what this is based on the VA (Watts) of the transformer.

The absolute maximum voltage is usually a little over what the amp is rated at in watts at 8 Ohms. As impedance drops, current requirement increases until you exceed the power of the transformer.

What filter capacitance buys you is less voltage sagging while they recharge, and more current capacity, if the rest of the amp can support it.

You do quickly run into diminishing returns though, and filter capacity can become little more than bragging rights. :)

Class D amps usually run a switching power supply, which does not have the big iron needed for the transformer, so you wont' see them rated the same way.  They still need filter caps, as the class D amp needs a stable DC power supply, like the linear counter parts.
listening99 OP
172 posts
 
@erik_squires 

So, the output voltage is not directly tied to transformer size? And watts are also not directly related to voltage? 

If the amp is 80wpc, the absolute maximize voltage is about 80volts? 

The capacitance offsets limits in the ability of the transformer to recharge?

I have noticed that various companies will sell add on capacitance, boosting things up very high... I was struck at the inversion between capacitance and output (watts) between the Vidar and the Aegir, given the same power supply... 

I have appreciated your support of class D. I keep coming back to the class D option, in my mind,  but I'm just so happy with my STA200 (Nuforce). I just wish it had double the rated power...
erik_squires
16,052 posts
 
So, the output voltage is not directly tied to transformer size?


Right.  The VA gives us the power output of the transformer at maximum current draw, but any given VA rating can have different output voltages.  What matters is what is called the winding ratio.  For instance:

600 VA w/ 10:1  : Outputs 12 Vrms
600 VA w/ 5:1 : Outputs 24 Vrms

And watts are also not directly related to voltage?

For a load, they are proportional to the square of the voltage:

W = (V*V) / R

If the amp is 80wpc, the absolute maximize voltage is about 80volts?

Nope, not at all. The 80 Wpc is rated at 8 Ohms. It is really W(rms).  I probably will miss a step here:

Sqrt(80 Wrms * 8 Ohms) = 25.3 Vrms

Vrms to Vpk = 25.3 * 1.4 = 35 V

That + and -  Volts is what the power supply has to make available to the amplifier board, which is then going to swing back and forth between them to create the voltage at the speaker.

Assuming there were no losses (and there are always losses) the voltage the speaker would see would go from + 35V to -35V at maximum.

So, for an amp rated at 80 watts per channel an output winding of around 5:1 seems about right.  The VA we can estimate at a minimum should be twice this, so 160 VA. However, there are losses AND .... no speaker is exactly 8 Ohms. As the impedance drops, more current must be drawn, so at 4 Ohms we must draw at least 320 VA.


The capacitance offsets limits in the ability of the transformer to recharge?

Yes.  What comes out of the transformer and bridge rectifier is pulsing DC. From zero to some positive number on one side, and zero to some negative number on the other.  The filter caps smooth this out. More capacitance, the closer the average V is to the peak voltage, and the stiffer it will seem while playing music.
erik_squires
16,052 posts
 
One thing that is confusing is that what we call power amps are not power amps at all.

They are voltage amps. We rate them at the power they can emit without clipping at 8 Ohms, but this really is a convenience.

Truthfully most amps attempt to multiply the input voltage by 20 (~ 26 dB I believe), and so long as they do not exceed the power supply rail’s limits they will do so with a great deal of accuracy.

Here is a link that probably explains a great deal I’ve left out or misinformed you:

http://www.skillbank.co.uk/psu/

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