attenuator network

[rafaelctt]

I am testing the attenuator network of the
 vox AC4tv on my amp but I don't think the values ​​are right in the layout.
With the switch at 5w, the output impedance is that of the speaker (8 Ohms, correct)
With the 1w switch, the output impedance is almost double and with the 1/4w switch, there is hardly any variation. In fact, it is somewhat less in the latter. That is normal?
Thank you

[sluckey]

I don't see anything wrong.

[rafaelctt]

I don't see anything wrong.

since there is almost no difference between 1 w and 1/4 w

[sluckey]

Are you saying you have built this and you can't hear a difference? Did you use three 15Ω resistors and one 5.6Ω resistor?

There is considerable difference. The schematic shows a considerable difference. The youtube demos show a considerable difference. Look at this simplified drawing and you should be able to spot the difference.

[rafaelctt]

Are you saying you have built this and you can't hear a difference? Did you use three 15Ω resistors and one 5.6Ω resistor?

There is considerable difference. The schematic shows a considerable difference. The youtube demos show a considerable difference. Look at this simplified drawing and you should be able to spot the difference.

In the first circuit(1w), assuming an 8 ohm speaker, I get a resulting impedance of 20.2 ohms. In the second(1/4w) I get a 18.2 ohm. The difference is very small and there is a lower impedance in the second case, so there should be more power (I think). Is that so?

[TenderTendon]

While it is true that the output section of the amplifier may produce slightly more power at the 1/4 watt position than the 1 watt position, the 15Ω + 5.6Ω resistors in the 1/4w position will eat up considerably more power than the 15Ω + 15Ω resistors in the 1w position. Therefore, the speaker will receive noticeably less power in the 1/4w position.

[sluckey]

You're just calculating or measuring the resistance between a and b. Resistance ain't impedance, but never mind that. Forget about impedance. Let's look at the circuit in a different way and I bet it will make more sense.

Just to avoid using a calculator let's pretend the 5.6Ω is really 5Ω. And let's ignore the speaker. Just pretend it's not there. And let's apply 10V (just to make the head math easy) to points a and b.

In the 1 watt circuit you have two 15Ω resistors that make up a 2:1 voltage divider. So you can only have 5V to send to the speaker. (Less volume)

In the 1/4 watt circuit the 15Ω and 5Ω make up a 4:1 voltage divider. 7.5V will be dropped across the 15Ω resistor and 2.5V will be dropped across the 5Ω resistor. So, you only have 2.5V to send to the speaker. (Much less volume)

We don't really care that position 1 is 5W, position 2 is 1W, and position 3 is 1/4W. What's important is that position 1 is high loud, position 2 is medium loud, and position 3 is low loud.

[rafaelctt]

You're just calculating or measuring the resistance between a and b. Resistance ain't impedance, but never mind that. Forget about impedance. Let's look at the circuit in a different way and I bet it will make more sense.

Just to avoid using a calculator let's pretend the 5.6Ω is really 5Ω. And let's ignore the speaker. Just pretend it's not there. And let's apply 10V (just to make the head math easy) to points a and b.

In the 1 watt circuit you have two 15Ω resistors that make up a 2:1 voltage divider. So you can only have 5V to send to the speaker. (Less volume)

In the 1/4 watt circuit the 15Ω and 5Ω make up a 4:1 voltage divider. 7.5V will be dropped across the 15Ω resistor and 2.5V will be dropped across the 5Ω resistor. So, you only have 2.5V to send to the speaker. (Much less volume)

We don't really care that position 1 is 5W, position 2 is 1W, and position 3 is 1/4W. What's important is that position 1 is high loud, position 2 is medium loud, and position 3 is low loud.

OK understood! Thank you