From what i've figured out, if the pot is turned all the way down, it is putting a 6 ohm load (8 ohm and 25 ohm in parallel) on the amp and shorting the speaker out of the circuit.
We should follow all the paths to figure it out.
Turned all the way down, there is no voltage between the wire shown going to the speaker and ground. In other words, there are 2 wires for the speaker, and both are connected to ground, so there's no sound out of the speaker.
Since the speaker is effectively out of the circuit, the impedance seen by the amp is, as you said, 8 ohms ll 25 ohms = ~6 ohms.
When it is turned all the way up, it looks like it is putting the 8 ohm speaker into a series with the 25 ohm pot.
Not quite.
Let's assume the speaker is 8 ohms, just to have a number to work with. The 8 ohm speaker is in parallel with the 25 ohm pot, and the 8 ohm resistor is effectived shorted by the wire that connects from the "top of the resistor" to "the top of the pot." In this setting, the amp still sees 8 ohms in parallel with 25 ohms, or ~6 ohms.
What happens when the pot is set halfway? We don't know if this is a linear-taper pot or an audio taper pot. Let's assume linear for ease of guessing.
If you set a linear 25 ohm pot to half of its mechanical rotation, it has the wiper at half its electrical length, or resistance. So that's 12.5 ohms above the wiper and 12.5 ohms below the wiper.
The 12.5 ohms above the wiper is in parallel with the 8 ohm resistor, and the 12.5 ohms below the wiper is in parallel with the 8 ohm speaker. These 2 parallel circuits are in series. 12.5 ohms ll 8 ohms = 4.89 ohms, so 4.89 in series with 4.89 = ~9.8 ohms.
So this circuit provides a nearly-constant load to the amp. In fairness, a perfectly constant load would take more resistors and a more tricky (read as "expensive") wiring scheme; also, a loudspeaker present a less constant load to the amp than this attenuator does.
Topic: attenuator question (Read 4633 times)