Calculate necessary attenuation (in dB) to achieve a given output voltage
When we work in a system of defined impedance, we talk of attenuators. So RF tends to be all ports 50 ohms, telephony is all ports 600 ohms. Attenuators are designed expecting those impedance on the ports, and only giving the right attenuation when that is the case.

When an output drives an input without an attenuator, the voltage the input receives is half the open circuit output voltage of the preceding stage, because of the loading. If an output drives an attenuator that is not loaded, the attenuator output will rise to twice its nominal design output voltage.

The value of attenuation is well defined, as both input and output impedances are the same, the attenuation figure is the same whether calculated for voltage, current or power. 

When we work in typical audio systems, with low output impedances, and high input impedances, we tend to talk of voltage dividers. Voltage dividers are designed load the preceding stage only lightly, and to be relatively unloaded by the stage following.

When an output drives an input without a voltage divider, the voltage the input receives is essentially equal to the open circuit output voltage of the preceding stage, because it is not loaded. If we put a finite load on a voltage divider, the output voltage will drop a little.

As the input and output impedances are different, the only sound computation for attenuation should be the power attenuation. Unfortunately this means the voltage ratio and the current ratio cannot be computed from the attenuation figure alone, but must include the port impedances for calculation. As a high impedance system might be 20k, or 1Meg, this makes life difficult. As these systems tend to respond to the input voltage, we tend to talk about voltage levels and ratios, rather than attenuation. Attenuation figures are sometimes given for just the voltage ratio, which most audio practitioners will understand, but purists will object to. A voltage ratio of 0.5 is just a tad over 6dB.

To pot-down (ie reduce in voltage by a potentiometer, an adjustable voltage divider) an output to a half (roughly what's needed here) we'd use two nominally equal value resistors. In series, they must be higher resitance than the driving stage, in parallel they must be lower resistance than the input they are to drive. In the case of 560 driving, 20k driven, that's about 3300 ohms, which gives a better than 10:1 ratio of the divider to the port impedances.  







simulate this circuit &ndash; Schematic created using CircuitLab

Due to the relatively small margin between voltage divider impedance and the port impedances, this equal resistor voltage divider gives 42% output voltage, a little lower than the 50% original target. R2 can be reduced and R3 increased slightly to get the exact ratio required. Alternatively, R2 and R3 could be replaced by a 5k or 10k potentiometer, to make the ratio adjustable.