thank you for your reply.
and for #3 the output (that RCA jack) would go into an integrated amp - the "AUX" input - which expects "line level."
and for #3 the output (that RCA jack) would go into an integrated amp - the "AUX" input - which expects "line level."
The circuit doesn't really make sense electrically.
You also need to be sure the amp is a conventional type with one speaker terminal of the amp grounded internally in the amp and that it is not some kind of BTL (bridged) amplifier where each speaker terminal may have a high DC voltage present.
You also need to be sure the amp is a conventional type with one speaker terminal of the amp grounded internally in the amp and that it is not some kind of BTL (bridged) amplifier where each speaker terminal may have a high DC voltage present.
Well that looks like a simple attenuator but it isn't -30db attenuation. 3k9 and 430 ohm give an attenuation of -20db
Worked out the old fashioned way.
3900 + 430 = 4330 ohms. If we apply 1 volt the current is I=V/R which is 1/4330 which is 0.2309 milliamps. Voltage across the 430 ohm is 0.2309E-3 * 430 which is 0.1 volt.
So 1 volt in and we get 0.1 volt out.
20log (Vout/Vin) gives us -20db
Vin vs Vout. 1 volt rms in and 0.1 volt rms out.
Attenuation, -20db (the line at -10db is phase... ignore that one):
The output impedance is approximately 400 ohm and is the value of loading that reduces the output by 50% at which point the attenuation becomes approx -26db. Such a low loading is unrealistic for a line level input and would typically be at 50k or more on most preamps. That level of loading would not alter the attenuation much and it would remain at around -20db
Worked out the old fashioned way.
3900 + 430 = 4330 ohms. If we apply 1 volt the current is I=V/R which is 1/4330 which is 0.2309 milliamps. Voltage across the 430 ohm is 0.2309E-3 * 430 which is 0.1 volt.
So 1 volt in and we get 0.1 volt out.
20log (Vout/Vin) gives us -20db
Vin vs Vout. 1 volt rms in and 0.1 volt rms out.
Attenuation, -20db (the line at -10db is phase... ignore that one):
The output impedance is approximately 400 ohm and is the value of loading that reduces the output by 50% at which point the attenuation becomes approx -26db. Such a low loading is unrealistic for a line level input and would typically be at 50k or more on most preamps. That level of loading would not alter the attenuation much and it would remain at around -20db
Assuming the thing is actually working (and thus saleable), it requires a GND connection to the speaker. It's a no brainer
I am told by Russound that the device's output impedance is 430 ohms.
How would I have calculated that? It can't be as simple as R2, can it?
You're right it can't 🙂 and the correct answer isn't actually 430 ohms. Its close to 430 ohms but not 430 and actually depends to a small extent on the output impedance of whatever is driving the network.
So 430 ohm is the simple answer.
The output impedance equals the value of loading we need to apply to cut the signal in half. If the voltage source driving it has zero ohm output impedance then all we need do is calculate the value of the 3k9 and 430 ohm as they appear in parallel and that is 387 ohm. That is the impedance that whatever you connect the output to will see.
If we load the circuit with that calculated resistive value we cut the voltage level in half.
This uses DC voltages to make it easier to follow.
10 volts in and we get 993.07162mv out.
Next we load the circuit by our calculated value and we get this, the output is now exactly half.
So the output impedance for a perfect voltage source driving the network is 387 ohm.