@atmasphere
Even allegedly DC coupled amplifiers have servo controls which effectively limit the amp's low frequency bounds, no?
Even allegedly DC coupled amplifiers have servo controls which effectively limit the amp's low frequency bounds, no?
Filter Capacitance
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/ |
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. |
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. |