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This Post Has Been Edited by the Author

In Reply to: RE: I have some questions posted by Tre' on January 18, 2012 at 08:40:18

"The value of the series resistor plus the value of the shunt resistor equals the output impedance of Lightspeed Attenuator. So as the series resistor increases or decreases in value we want the shunt resistor to do exactly the opposite. "

First, this is wrong.

Second, this is no different than any other resistive attenuator.

The output impedance of a resistive attenuator is the value of the two (series and shunt) resistors in parallel, not "plus". (This assumes a zero source impedance. 100 ohms, as suggested is close enough)

The worst case happens at the -6db point when the series resistance and the shunt resistance are equal. For a 10K attenuator this would be at 2500 ohms. The output impedance will be lower at any other point of attenuation, up or down.

This would be true of a pot, a stepped attenuator or a LDR based attenuator.

I ask again, what's the difference between this and a pot or stepped attenuator?

A autoformer or transformer volume control is fundamentally different in that it is not a voltage divider.

No current is "thrown away".

Voltage is traded for current.

When a lower tap is selected the output impedance goes down (a good thing)
and the reflected impedance back to the source goes up (a good thing).

Attenuators are all about attenuating. :-) Throwing away voltage.

A AVC or TVC does this without increasing the output impedance (in fact the output impedance is decreased) or decreasing the impedance seen by the source (in fact that impedance is increased).

Tre'

Tre'

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