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Don't forget one important function: powerstage drive currents.

Posted by cheap-Jack (M) on December 6, 2006 at 12:25:13

In Reply to: Function of coupling cap value in SE amplifier? posted by mbhcid on December 5, 2006 at 08:49:52:

Hi.

While attention has been focussed in LF attention, etc., the powerstage needs adequate driving currents to deliver the right signal swing for the rated full power O/P.

Added to the complication is the composite capacitance of the power tube I/P circuit, depending the power tube is triode or pentode. To maintain adequate bandwidth of today's hi-resolution recording, e.g. DVD-audio (sampling frequency of 192KH), & to provide enough swing margin to minimize distortion, we are talking about many mA signal current to drive the powerstage.

The couple cap got to allow adequate drive current to go through from the driverstage.

To make things simple, we can start with slew rate calculation, then we can get the minimum coupling cap value for a certain value of drive current.

Say for a drive voltage of 10V peak to peak, & HF is 20KHz (to make way for HF extended well beyond 20KHz), so the slew rate formula is:
2x3.1416xFxV. We get 1.256V/uS

Say for a triode powerstage, the composite I/P capacitance be 200uF,
& drive current is 15mA, to calculate the min coupling cap value, we use formula: C= drive current/slew rate in F.

So we get: 15mA/1.256=0.011uF

c-J

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