NFB on marshall amps?

[keithv]

I've heard the lore that Marshalls sound best at a 16 ohm load.  I've noticed that the NFB loop only connects to the 16 ohm tap of the OT.  Is this why?  Does the NFB loop still affect the 8 and 4 ohm taps through the OT secondary windings, or does the NFB loop only have an effect when using the 16 ohm output?

[xm52]

The feedback circuit is often connected to the highest impedance secondary tap of the output transformer. What you are seeing is normal. The feedback is still in place if you use other transformer taps to connect your speakers. It is important to note that the speakers affect the feedback circuit. Change the speakers and the optimal value for the components in the feedback circuit could change. It's such a simple circuit but the way that it performs isl very complicated.


Here is a basic explanation as to what feedback does and how it works: http://www.aikenamps.com/index.php/designing-for-global-negative-feedback

[keithv]

So is there any credence to the theory that 16 ohms is the optimal output load for something like the JTM45?

[PRR]

> Does the NFB loop still affect the 8 and 4 ohm taps

Yes.

Marshall moved Fender's NFB from 4 to 16 as a quickie way to increase NFB (to suit the rather different speakers he had).

On an ideal transformer, all the taps give the same result. Even on non-ideal OTs, I rather think any difference is in the mind not the gear.

[HotBluePlates]

So is there any credence to the theory that 16 ohms is the optimal output load for something like the JTM45?

Maybe, maybe not. The idea was that using all of the secondary winding (the 16Ω tap, or whichever is the highest available tap) gave the best utilization of the OT.


I don't think you'll notice a difference in practice with connecting the rated load to any tap.

[Ed_Chambley]

So is there any credence to the theory that 16 ohms is the optimal output load for something like the JTM45?

Maybe, maybe not. The idea was that using all of the secondary winding (the 16Ω tap, or whichever is the highest available tap) gave the best utilization of the OT.


I don't think you'll notice a difference in practice with connecting the rated load to any tap.

I have heard the same whereas using the largest tap of a transformer will yield the "greater usage" of the OT.  Upon further reading I find many explain it to be nonsense.  If it is nonsense I do not understand how it can be.

How can a Transformer winder claim the exact winding in a multi-tap version, say 4-8-16 ohm on a 6G6B which originally a 4 ohm tranny.  I mean how can you make a "Toneclone" multitap when the original was not?

[HotBluePlates]

I have heard the same whereas using the largest tap of a transformer will yield the "greater usage" of the OT.  Upon further reading I find many explain it to be nonsense.  If it is nonsense I do not understand how it can be.

You apply a.c. power to the primary, and take off a.c. power from the secondary. Say you use the highest tap available... You can control the amount of a.c. voltage applied to the primary directly but you can't control the alternating current directly; that's a byproduct of the applied a.c. voltage and the reflected impedance of the primary.

Likewise, you can easily understand volts-ratio from primary to secondary and relate it to turns-ratio; current-ratio from primary to secondary is opposite volts-ratio and again feels like a secondary effect because you only directly control volts and impedance (and transformer winding turns to set up the ratios).

You apply that a.c. volts to the primary, and because there is a set number of primary turns, you've also established some number of "volts-per-turn". The secondary has the exact same number of volts-per turn, but has generally much fewer turns, and so steps-down the total voltage (current gets stepped up because the secondary load impedance as we use the OT is much smaller than the reflected primary impedance, but we assume power in = power out).

Whatever # volts-per-turn you have on the secondary is not only at 1 tap or turn of the secondary, but on all turns of the secondary. At the highest impedance tap you're getting a voltage equal to the volts-per-turn  dictated by the applied primary a.c. voltage, times however many turns there are in the whole secondary. You're connecting the highest load impedance to this tap and so the load current is the lowest at this tap. When you connect a proportionally-lower load to the next lower tap, all turns of the secondary are still energized, but the pick-off point is fewer total turns of the secondary and lower output a.c. volts. Your load is proportionally smaller, so the current is proportionally higher and total power through-put is the same as for the higher-impedance tap.

Now that you understand all this, where people throw the B.S. Flag is that regardless of which tap you connect to, all secondary turns are energized all the time, and by the same amount  regardless of which tap is used. The same power-in results in the same power-out regardless of which tap you use. You might say, "but if output current increases when using the lower impedance taps doesn't that change how the winding responds?" The only thing it could do is create more voltage drop across the DCR of that portion of the winding, but that part-winding is also a shorter length of the same heavy wire, so its DCR is proportionally lower.

The key is understanding the fact that when a.c. power is applied to the primary that all turns of the secondary are energized, all the time, not just the part-winding connected to the speaker. This is the reason you have a feedback return from one set tap, and can plug your speakers into any other tap and it still works the same.

[Ed_Chambley]

I have heard the same whereas using the largest tap of a transformer will yield the "greater usage" of the OT.  Upon further reading I find many explain it to be nonsense.  If it is nonsense I do not understand how it can be.

You apply a.c. power to the primary, and take off a.c. power from the secondary. Say you use the highest tap available... You can control the amount of a.c. voltage applied to the primary directly but you can't control the alternating current directly; that's a byproduct of the applied a.c. voltage and the reflected impedance of the primary.

Likewise, you can easily understand volts-ratio from primary to secondary and relate it to turns-ratio; current-ratio from primary to secondary is opposite volts-ratio and again feels like a secondary effect because you only directly control volts and impedance (and transformer winding turns to set up the ratios).

You apply that a.c. volts to the primary, and because there are a set amounts of primary turns, you've also established some number of "volts-per-turn". The secondary has the exact same number of volts-per turn, but has generally much fewer turns, and so step-down the total voltage (current gets stepped up because the secondary load impedance as we use the OT is much smaller than the reflected primary impedance).

Whatever # volts-per-turn you have on the secondary is not only at 1 tap or turn of the secondary, but on all turns of the secondary. At the highest impedance tap you're getting a voltage equal to the volts-per-turn  dictated by the applied primary a.c. voltage, times however many turns there are in the whole secondary. You're connecting the highest load impedance to this tap and so the load current is the lowest at this tap. When you connect a proportionally-lower load to the next lower tap, all turns of the secondary are still energized, but the pick-off point is fewer total turns of the secondary and lower output a.c. volts. Your load is proportionally smaller, so the current is proportionally higher and total power through-put is the same as for the higher-impedance tap.

Now that you understand all this, where people throw the B.S. Flag is that regardless of which tap you connect to, all secondary turns are energized all the time, and by the same amount  regardless of which tap is used. The same power in results in the same power out regardless of which tap you use. You might say, "but if output current increases when using the lower impedance taps doesn't that change how the winding responds?" The only thing it could do is create more voltage drop across the DCR of that portion of the winding, but that part-winding is also a shorter length of the same heavy wire, so its DCR is proportionally lower.

The key is understanding the fact that when a.c. power is applied to the primary that all turns of the secondary are energized, all the time, not just the part-winding connected to the speaker. This is the reason you have a feedback return from one set tap, and can plug your speakers into any other tap and it still works the same.

That is sooooooooooo much simpler than I was thinking.  Certainly the WHOLE secondary is energized!  I inherently understand this in power transformers we have on our industrial machines.  For some reason I cannot stay clear that a OT is just a Transformer.

So if the primary numbers of turns equals the original transformer (speaking 4 ohm Bassman type), then making a multi-tap transformer equivalent would you simply increase the turns of the secondary then tap where is necessary to get the proper load which in my example is 4-8-16 ohms?

If this is the case and the DCR is equivalent then regardless of the amount of secondary turns when attached to a 4 ohm load the transformer is performing as the original.  The reason of interest I read where Dr Z prefers a 4 ohm load and makes the zbest cab 4 ohm.

What would be the reason to build to a 4 ohm secondary as opposed to 8 or 16?

[Willabe]

To only have to buy 8 ohm speakers. 

[Ed_Chambley]

To only have to buy 8 ohm speakers. 

Here is a guy who knows that more than 1 speaker is necessary.

[Willabe]

1x8 ohm speakers for single combos, 2x8 ohm (in parallel = 4 ohm) speakers for twins.     I guess he could have went with 2x8 ohm in series = 16 ohm. (For Dr. Z it could have more to do with the speaker, voice coil, than the OT secondary?)   

From his web site;

The Z Best 2x12 closed back cab is an animal all its own. We have utilized a Theile Ported design which allows us to dial in the full frequency range of the guitar. Low end is determined by the port thickness, Mid response by the shelf depth, and Top end by choice of speakers.
 

[HotBluePlates]

... The reason of interest I read where Dr Z prefers a 4 ohm load and makes the zbest cab 4 ohm.

What would be the reason to build to a 4 ohm secondary as opposed to 8 or 16?

I don't know why Dr Z chooses what he does. I can only speculate some things based on what I see.


I see first off that the Z Best Cab has 2 dissimilar speakers, which indicates to me a preference for using 2 speakers so one can cover deficiencies of the other (or at least different sound qualities) and give a better overall sound.


If you have 2 speakers you have the choice of placing them in series or in parallel. If you place them in series and one blows, both speakers go silent. So parallel wiring seems to be preferred.


If you use 2x parallel speakers, you can use a pair of 16Ω speakers for a total of 8Ω, or 2x 8Ω speakers for a total of 4Ω. But if your amp's output jack is 8Ω and you suspect the output power of the amp will be more than what a single speaker, you create a condition where an unknowing player could plug 1x 8Ω speaker in the jack and blow his speaker. If you make that output 4Ω, then players are unlikely to plug a single speaker in the jack and blow their speaker. While there are 4Ω (or 3.2Ω) speakers available, they're typically a little 8" Champ speaker which will "look wrong" hooked up to a 30w+ amp for almost any player regardless of their amp-knowledge.


Dr Z may have  different logic developed for why he prefers a 4Ω output impedance, but there's one potential line of logic. You could always email Dr Z and ask...

[PRR]

> power transformers we have on our industrial machines.

You must have 24V windings.

Often implemented as two 12V windings, so you can connect series for 24V or parallel for 12V.

The output rating, Volts*Amps, is the same either way.

They could simplify to 3 wires with 24V CT. Take the whole for 24V, half the winding for 12V. This works, but the rating is a bit lower in 12V mode because there's a half-winding taking space and doing no good (not reducing copper loss).

In audio we do NOT work so close to overheating. Real non-problem. The DCR should be small in any case. The difference of DCR by using a part-winding is hard to measure, and has less effect than the cabinet color.

There is just enough difference (and sharp measurers) that you sometimes see "better" plans. One of the older Hammonds asks you to series or parallel depending on intended load. This particularly helps response in the extreme treble, far above the guitar band, but important to hi-fi theorists.

Why 4-only winding? If you commit to only-4-ohm speaker systems, you save two winding-stops and two wire leads. When you make a million amps, the pennies saved add-up. Since the combo designer can order speakers of any impedance (six 24-ohm to make a 4-ohm six-pack speaker system), this limitation is no limitation at all.

Why 4 or 16?

Few-Ohm windings are a part-penny cheaper, fewer turns of fatter wire. Somehow a 3.2 Ohm DCR became common for all "short line" speakers: radios etc where the speaker is near the amp. 3.2 Ohm DCR is typically roughly 4 Ohms mid-audio band.

Theater systems with 300 foot speaker runs have an Ohm or so in the speaker lines. At 3 or 4 Ohms this would really bite the power. 16 Ohms (or two 32 in parallel for redundancy) became the common system.

Early transformerless transistor amps had trouble with the high current in 4 Ohms or/and the high voltage needed for 16 Ohms. Somehow 8 Ohms became common around that time.

With some modern super-systems, car-sound trunk-amps, voltage is an issue but parallel banks of fat transistors will deliver prodigious current. Runs are short, and the fans respect fat wire. Car-sound super-speakers are often 2 Ohms, to suck power without needing over 100V on the transistors.

[frus]

I've heard the lore that Marshalls sound best at a 16 ohm load.

I've heard that they used thinner wire on the 16-ohm portion of the secondary, and that that results in some kind of compression or something. Could that be it?

[HotBluePlates]

I've heard the lore that Marshalls sound best at a 16 ohm load.

I've heard that they used thinner wire on the 16-ohm portion of the secondary...

No, the secondary (when it is a single winding) is all one length of the same-gauge wire, tapped along its length.

Say you have a transformer whose secondary has common, 4Ω, 8Ω and 16Ω taps. Let's say the number of turns from common to 4Ω is "x". The 16Ω tap is at double the number of turns (and double the voltage) at 2x turns from common. The 8Ω tap is at x*√2 turns from common (or 1.414*x).

These are all straightforward ratios of turns, based on the fact that the impedance ratio is the turns ratio, squared. From the 4Ω tap you need √2-times more turns to get to 8Ω, because (4Ω*√2)² = 8Ω. From 8Ω to 16Ω is another √2-times more turns; counting from the 4Ω tap, that's (4Ω*√2*√2)² = 16Ω.

I've heard the lore that Marshalls sound best at a 16 ohm load.

... that results in some kind of compression or something. Could that be it?

Typical Marshall head expects a 4x12 cab of series-parallel wired 16Ω speakers, or a total of 8Ω. I wonder if the "lore" is about mismatching the OT tap and the cabinet... That might change the response some, but you'll only notice when the output section is pushed beyond its clean volume output.