Can someone explain how attenuator impedance switch works?

itsGiusto

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I have a great Weber Minimass, and I'd like to better understand how it works. But the nuances of how impedance and reactance work are somewhat vague to me. Here's the schematic:
http://milas.spb.ru/~kmg/files/schematics/Weber MASS Attenuators/WeberMiniMASS.gif

I don't really understand how the impedance switch works. Why would adding a 16 ohm resistor to ground change the input impedance of the device to 8 ohm? Wouldn't you have to add a 16 ohm reactive load to ground to do that? Impedance and resistance are not the same thing, right?

Then, to top it all off, Weber had this statement on their site:

"With all Weber attenuators, once you have selected the proper impedance to match the output impedance of the amp, the actual speaker impedance isn't critical. That's because the actual load to the attenuator becomes the speaker impedance plus the output section of the attenuator, while the amp continues to see the correct nominal impedance from the input section of the
attenuator. That's why, on the MASS, we provide two speaker output jacks. Feel free to experiment with different impedance speakers, cabinets, etc."

Why would this be the case? I don't understand why the actual speaker's impedance wouldn't matter at all. At the very least, wouldn't the speaker's impedance have to match correctly with the output section of the attenuator? And what if you had the attenuator switched on but with no, or little, attenuation? Wouldn't the amp's output section still be seeing the load of the speaker pretty directly?
 
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Outlaws

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Interesting...total opposite of what THD said in the Hot Plate manual. Speaker load still mattered unless it was set to dummy load and then the speaker was defeated anyways.
 

Badside

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That would just add resistance in parallel with the fake speaker, reducing the "reactance" of the unit. Using the Volume control will also greatly affect the end sound.

Either way, I find the "Mass" concept flawed: a speaker motor without a cone is not a complete reactive load. Where's the cone resonance?
It works mostly because a speaker motor has a coil in it, ergo it is an inductor. Two Notes took out the middle man and just put an inductor in their product. Still doesn't simulate the resonance peak though (that requires a much bigger inductor and a big capacitor)
 

RLee77

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I don't really understand how the impedance switch works. Why would adding a 16 ohm resistor to ground change the input impedance of the device to 8 ohm? Wouldn't you have to add a 16 ohm reactive load to ground to do that? Impedance and resistance are not the same thing, right?
In this case, as an ac load, they can be treated as the same thing. A simplified definition of impedance is resistance at a particular frequency.
If a speaker presents a 16 ohm impedance at 1khz to an amp, putting a 16 ohm resistor in parallel causes the amp to see an 8 ohm effective load. (There are other characteristics of reactive loads, such as current and voltage being out of phase, but it’s not critical here.)

Why would this be the case? I don't understand why the actual speaker's impedance wouldn't matter at all. At the very least, wouldn't the speaker's impedance have to match correctly with the output section of the attenuator? And what if you had the attenuator switched on but with no, or little, attenuation? Wouldn't the amp's output section still be seeing the load of the speaker pretty directly?

The speaker impedance does matter, but the amount of it that the amp sees varies depending on where the attenuator’s volume is set. Let’s take the 8 ohm setting as an example: 8 ohm spkr connected, selector set to “8”.
When volume is at max (minimum attenuation), the amp sees the 16 ohms of power resistors, in parallel with the 50 ohm volume rheostat, in parallel with the 8 ohm connected spkr. (The motor coil is out of the circuit, since it’s shunted by the rheostat.) This gives a total load of 1/(1/16+1/50+1/8) = 5 ohms (rounding up). Switching the selector to “off” in this case results in a better match, since it just leaves the 50 ohms in parallel with the 8 ohm spkr, resulting in a 7 ohm load.

Now as you lower the volume rheostat, the speaker is gradually removed as a load and the motor coil is gradually put across the 50 ohm volume; in other words the motor coil is gradually swapped in place of the speaker. At full attenuation, the amp sees about 7 ohms (with selector on “8”) since you have the 16 ohms of power resistors in parallel with the 50 ohm volume in parallel with the 16 ohm motor coil. If you removed the spkr, amp would still see 7.
With selector on “off”, the amp would see 50 ohms in parallel with the 16 ohm motor coil for a load of 12 ohms. If you then rotated the volume back to full, with spkr connected, the load would gradually go from 12 to 7 ohms.

The reason they say the spkr doesn’t matter that much is that at medium to high attenuation, very little of the speaker is actually in the circuit.
I find the design interesting; thanks for posting the schematic, I hadn’t seen it before.
 
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rze99

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Interesting...total opposite of what THD said in the Hot Plate manual. Speaker load still mattered unless it was set to dummy load and then the speaker was defeated anyways.


I never understood it frankly. I've used an 8 Ohm THD Hotplate with various speaker loads (currently the internal speaker + another 8 Ohm closed back extension cab) and it seems perfectly relaxed and happy and sounding enormous ;)
 

rze99

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In this case, as an ac load, they can be treated as the same thing. A simplified definition of impedance is resistance at a particular frequency.
If a speaker presents a 16 ohm impedance at 1khz to an amp, putting a 16 ohm resistor in parallel causes the amp to see an 8 ohm effective load. (There are other characteristics of reactive loads, such as current and voltage being out of phase, but it’s not critical here.)



The speaker impedance does matter, but the amount of it that the amp sees varies depending on where the attenuator’s volume is set. Let’s take the 8 ohm setting as an example: 8 ohm spkr connected, selector set to “8”.
When volume is at max (minimum attenuation), the amp sees the 16 ohms of power resistors, in parallel with the 50 ohm volume rheostat, in parallel with the 8 ohm connected spkr. (The motor coil is out of the circuit, since it’s shunted by the rheostat.) This gives a total load of 1/(1/16+1/50+1/8) = 5 ohms (rounding up). Switching the selector to “off” in this case results in a better match, since it just leaves the 50 ohms in parallel with the 8 ohm spkr, resulting in a 7 ohm load.

Now as you lower the volume rheostat, the speaker is gradually removed as a load and the motor coil is gradually put across the 50 ohm volume; in other words the motor coil is gradually swapped in place of the speaker. At full attenuation, the amp sees about 7 ohms (with selector on “8”) since you have the 16 ohms of power resistors in parallel with the 50 ohm volume in parallel with the 16 ohm motor coil. If you removed the spkr, amp would still see 7.
With selector on “off”, the amp would see 50 ohms in parallel with the 16 ohm motor coil for a load of 12 ohms. If you then rotated the volume back to full, with spkr connected, the load would gradually go from 12 to 7 ohms.

The reason they say the spkr doesn’t matter that much is that at medium to high attenuation, very little of the speaker is actually in the circuit.
I find the design interesting; thanks for posting the schematic, I hadn’t seen it before.


I'm pleased someone understands it.;)
 

itsGiusto

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In this case, as an ac load, they can be treated as the same thing. A simplified definition of impedance is resistance at a particular frequency.
If a speaker presents a 16 ohm impedance at 1khz to an amp, putting a 16 ohm resistor in parallel causes the amp to see an 8 ohm effective load. (There are other characteristics of reactive loads, such as current and voltage being out of phase, but it’s not critical here.)



The speaker impedance does matter, but the amount of it that the amp sees varies depending on where the attenuator’s volume is set. Let’s take the 8 ohm setting as an example: 8 ohm spkr connected, selector set to “8”.
When volume is at max (minimum attenuation), the amp sees the 16 ohms of power resistors, in parallel with the 50 ohm volume rheostat, in parallel with the 8 ohm connected spkr. (The motor coil is out of the circuit, since it’s shunted by the rheostat.) This gives a total load of 1/(1/16+1/50+1/8) = 5 ohms (rounding up). Switching the selector to “off” in this case results in a better match, since it just leaves the 50 ohms in parallel with the 8 ohm spkr, resulting in a 7 ohm load.

Now as you lower the volume rheostat, the speaker is gradually removed as a load and the motor coil is gradually put across the 50 ohm volume; in other words the motor coil is gradually swapped in place of the speaker. At full attenuation, the amp sees about 7 ohms (with selector on “8”) since you have the 16 ohms of power resistors in parallel with the 50 ohm volume in parallel with the 16 ohm motor coil. If you removed the spkr, amp would still see 7.
With selector on “off”, the amp would see 50 ohms in parallel with the 16 ohm motor coil for a load of 12 ohms. If you then rotated the volume back to full, with spkr connected, the load would gradually go from 12 to 7 ohms.

The reason they say the spkr doesn’t matter that much is that at medium to high attenuation, very little of the speaker is actually in the circuit.
I find the design interesting; thanks for posting the schematic, I hadn’t seen it before.
Thanks for the explanation, it makes a lot of sense! I think one problem of mine is that I don't really understand what it means for the output of the transformer to be expecting to see a certain speaker load, 4ohms, 16ohms, etc. In your explanation and examples, the actual load presented to the output transformer will vary depending on how much attention is dialed into the circuit, which makes sense. The part that doesn't make sense to me is that even with that, the output transformer still works properly.

So I guess my follow-up question to your explanation is, why does the output transformer still work even when it's presented with a mismatched load due to the attenuator? What makes an output transformer blow up sometimes if a load is mismatched, but not in these cases?
 

Paul Jenkin

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I'm not saying I know but it was explained to me as being similar in approach to those rheostat units that allow you to dim lights.

Can't say I question it - I'm just glad my Dr Z "Brake-Lite" does work or my missus would be kicking me outa the house!
 

itsGiusto

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I've written a Python script that plots the impedance presented to the amp at different attenuation percentages, given a specific impedance switch selection and specific speaker load connected. All calculations are only for nominal impedance.

Here's the code, in case anyone wants to run it, or inspect it to ensure my calculations are sound: https://github.com/ItsGiusto/weber_impedance_calculator/blob/master/weber_impedance_calculator.py

Here's the plot outputs for all 9 combinations of impedance switch and speaker connected.

Figure_1.png


I feel like the most worrying thing about these calculations is in the 16ohm impedance selector-16ohm speaker plot (which is how I use the MiniMass with my Marshall 1974x) in the lower right-hand corner. With 50% attenuation, it present a load of 19.5 ohms to the amp. Aren't tube amps not supposed to like having a higher load than expected?
 
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RLee77

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Cool use of python to plot these.
As far as the mismatch you mentioned, it’s not at all significant... 19.5 instead of 16 isn’t even considered a mismatch. That’s in the +/- 20% range, closer than you get from a typical pure speaker load over frequency anyway.

Actually the one that looks best is the chart for switch=8, speaker = 16, with your amp set to 8 ohm output. The variance there is like 6.8 to 8.8, very close.
 

coxster

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Remember that the impedance of any speaker is constantly changing with frequency. According to a 1955 book by Abraham B Cohen on speaker design, the value was typically calculated at 400 Hz. Adding resistance in parallel is not the same as adding a speaker. There is NO formula for R and L loads in parallel, first you must find the Z or R of each and then calculate the currents and add the currents as vectors then derive Z=E/I. It's a pain, phase angles and etc (I deal in power line impedances at work every day, not near as fun as speakers, but pays more bills ; ) The Maximum power transfer theorem states that when the source impedance matches the load impedance maximum power is transferred (power usually shows up to our ears as low freq), but it doesn't hurt to have a higher Z load than the Z of the source, it simply limits the current. The problem is when you have a lower Z load than source, that's when components get hot and crunchy.

Nice mention above as to how the resonance of the cone changes the Z (pretend I have a thumbs up emoticon here)
 

tubeswell

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The '1/4 power' attenuator switch in the Fender 'EC' series amps uses a couple of 25W wire-wound ceramic resistors; 1 x 8R in series with the speaker, and 1 x 16R in parallel with these
 

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  • 8R and 16R 25W resistors behind chassis.JPG
    8R and 16R 25W resistors behind chassis.JPG
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