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RE: Easy to understand...

> For example, how did you come up with the 2.3dB drop? And at what frequency (range) would that apply?

The output voltage is applied across the output resistance of the preamp and the load resistance in series. Therefore, the preamp's output voltage is applied across 3k + 10k = 13k. The voltage applied to the load is in direct proportion to the load with regard to the total resistance of 13k. Therefore, the voltage applied to the load would be (10k / 13k) * (Preamp output voltage). In other words, 10k / 13k = 0.7692 or 76.92% of the preamp's output voltage is applied to the load. This represents an attenuation in decibels of:

20 * Log(0.7692) = -2.279-dB

This is not frequency dependent. It applies to all frequencies. In order for it to become frequency dependent, you need to add capacitance and/or inductance into the equation.

Best regards,
John Elison

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