Fender 100 watt half power switch

[TerryD]

You guys helped me get this Fender 100 watt pa up and running and sounding great. 

Is there any "good" half power switch tricks for this amp.  I know removing two inner or outer tubes gives the idea.

 Is there any safe way to have a half power switch on a 100 watt Fender and have a happy bias as well?

[HotBluePlates]

You shouldn't need to change bias in a fixed-bias amp just because you have 2 output tubes instead of 4.

... Is there any "good" half power switch tricks for this amp.  I know removing two inner or outer tubes gives the idea.

 Is there any safe way to have a half power switch on a 100 watt Fender ...

If you kill power to the 2 tubes you don't want running, you'll only have the remaining 2 running. Or you could kill the drive voltage to 2 of the output tubes you don't want running.

To kill power you could have a switch to disconnect screen voltage from the 2 tubes to be disabled, or a switch between cathode & ground of the 2 tubes to be disabled. Either way should basically nix all plate current.

To kill drive, you could simply break the connection from phase inverter coupling cap to the 2 tube grids to be disabled.

Or you might have a dual-gang pot wired as a master volume between the phase inverter outputs and the 2 tubes you want to disable. You could leave them running all the time, and vary that "master" from full-up (full output power) down to zero (half-output power). The end result at half-power won't be preamp & phase inverter distortion because you still have a push-pull pair functioning normally.

[TerryD]

OK, but I changed the bias to the same bias as in a Blackface Twin Reverb per Gerald Weber's youtube video.  Sounds great. I don't think that is "fixed" bias.

 HBP, would that alter your recommendations?

The "dual gang pot" ideas sounds like lots of fun.  If I figure out exactly how to do that.  Sounds simple enough.  How many ohm pot would be good for such a thing?

[uki]

You shouldn't need to change bias in a fixed-bias amp just because you have 2 output tubes instead of 4.

If you kill power to the 2 tubes you don't want running, you'll only have the remaining 2 running. Or you could kill the drive voltage to 2 of the output tubes you don't want running.

To kill power you could have a switch to disconnect screen voltage from the 2 tubes to be disabled, or a switch between cathode & ground of the 2 tubes to be disabled. Either way should basically nix all plate current.

To kill drive, you could simply break the connection from phase inverter coupling cap to the 2 tube grids to be disabled.

Or you might have a dual-gang pot wired as a master volume between the phase inverter outputs and the 2 tubes you want to disable. You could leave them running all the time, and vary that "master" from full-up (full output power) down to zero (half-output power). The end result at half-power won't be preamp & phase inverter distortion because you still have a push-pull pair functioning normally.

Very interesting, one can learn big deal of tricks in this forum, and there is more than one way to achieve power reduction in a push-pull amp with 4 tubes !  Thanks for this, been reading stuff and learning tons of things here !!

[Willabe]

OK, but I changed the bias to the same bias as in a Blackface Twin Reverb per Gerald Weber's youtube video.  Sounds great. I don't think that is "fixed" bias.

It is 'fixed' bias but it's adjustable fixed bias as opposed to cathode bias.

[DummyLoad]

mesa engineering disconnects cathodes to 2 tubes with a switch. simple. reliable.


http://el34world.com/charts/Schematics/files/mesa_boogie/boogie_mkii.pdf


--pete

[TerryD]

One more thing if I can slip this in. 

The master volume on this is a pot with a 1000v 25p cap across it.    Can I improve that??

[jjasilli]

+1 to Dummyload. 
But this doubles the "output impedance" for the remaining 2 power tubes.  You can opt to ignore that, or use the next higher tap on the OT secondary (for same speaker load).  The second option can also be made switchable.


The master volume on this is a pot with a 1000v 25p cap across it.    Can I improve that??
No doubt it's a ceramic cap which tend to have high voltage ratings - way more than needed in this application.  Any improvement in tone by using different pF values, or other types of caps, will probably be subjective.   The only way to know is to try no cap or other caps.

[HotBluePlates]

OK, but I changed the bias to the same bias as in a Blackface Twin Reverb per Gerald Weber's youtube video.  Sounds great. I don't think that is "fixed" bias.

What Willabe said.

"Fixed bias" = tube cathode connected to ground with a negative voltage applied to the control grid.

"Adjustable fixed bias" = as above, but the negative voltage can be varied over some range to accommodate different tubes.

The "dual gang pot" ideas sounds like lots of fun.  If I figure out exactly how to do that.  Sounds simple enough.  How many ohm pot would be good for such a thing?

It would be a value comparable to the "bias feed resistors" you have in your amp now. These are the resistors from the bias supply to the output tube grids. I'm guessing you've raised those from 68kΩ to 100k-220kΩ.

The form of what I'm envisioning is like a post-phase inverter master volume, but only dials downs the drive to 2 tubes instead of all 4.

But before you do all this, try yanking 2 output tubes from the sockets. Does this give you enough volume reduction? Are you gonna need full output power later? Do you need to go quieter while getting output tube distortion?

But this doubles the "output impedance" for the remaining 2 power tubes.  You can opt to ignore that, or use the next higher tap on the OT secondary (for same speaker load).  The second option can also be made switchable.

For everyone else, jjasilli is right; odinarily when you pull 2 output tubes you'd want to double the effective OT primary impedance by hooking the original speaker load to the next lower tap on the secondary (to reflect a high primary impedance).

The lower-than-optimum OT impedance from just pulling 2 tubes will result in less-than-full clean output power from the remaining pair of tubes. I didn't mention it or consider it a thing to be fixed, because you're already pulling tubes to reduce output power, so what's the harm in losing a few more watts?

[kagliostro]

The proposal from Pete is good

but in this situation, to me

+1 for HotBluePlates

To kill power you could have a switch to disconnect screen voltage from the 2 tubes to be disabled
Or you might have a dual-gang pot wired as a master volume between the phase inverter outputs and the 2 tubes you want to disable.

because this way your power tubes are always 4 and there are no reflected impedance problems to be adjusted

Franco

[jjasilli]

All this being said, I submit that the only substantial benefit in pulling 2 power tubes, is saving the useful lives of the 2 pulled tubes.  As to volume:


1.  there's already an MV.  This should fully serve the purpose.
2.  cutting output power in Watts by 1/2 only produces a 3dB drop in vol. which is just noticeable.  It will not tame an amp.

[TerryD]

I'd agree to a point, but not absolutely. 

With four tubes its a monster.  Kind of like a grand piano.  Everything is so full. Every note complete. The E & B strings are so so strong.  Knock out the two tubes and its a different flavor.  Comparatively looser.  More like a regular 50 watt fender

[DummyLoad]

The proposal from Pete is good

but in this situation, to me

+1 for HotBluePlates

To kill power you could have a switch to disconnect screen voltage from the 2 tubes to be disabled
Or you might have a dual-gang pot wired as a master volume between the phase inverter outputs and the 2 tubes you want to disable.

because this way your power tubes are always 4 and there are no reflected impedance problems to be adjusted

Franco

no! only two tubes are seeing the load. the other two are not flowing ANY current. same scenario: double the speaker load.


--pete

[jjasilli]

I'd agree to a point, but not absolutely. 

With four tubes its a monster.  Kind of like a grand piano.  Everything is so full. Every note complete. The E & B strings are so so strong.  Knock out the two tubes and its a different flavor.  Comparatively looser.  More like a regular 50 watt fender

Your initial post is about half-power, which we have been addressing.  However, as stated above an impedance mismatch occurs between the power tubes and the OT primary if 2 tubes are pulled (or disabled by lifting their cathodes from ground).  This will affect frequency response and perhaps other tonal colorations, in a way that you seem to like. 

[TerryD]

JJasilli   Got it.  sorry for the confusion guys.  My only defense is that's what I've heard it called even though it in no way cuts the power in half

Sooooo.  Really.  Dorking around with just two tubes is no good because of the impedence mismatch that it causes in the output transformer.  Switching secondaries is beyond me and doesn't sound worth it.  This isn't a simple thing then.

[Willabe]

Dorking around with just two tubes is no good because of the impedence mismatch that it causes in the output transformer.   

The lower-than-optimum OT impedance from just pulling 2 tubes will result in less-than-full clean output power from the remaining pair of tubes. I didn't mention it or consider it a thing to be fixed, because you're already pulling tubes to reduce output power, so what's the harm in losing a few more watts?

No, it's fine, it won't hurt anything. And like HBP wrote it will drop a few more watts.

Switching secondaries is beyond me and doesn't sound worth it. This isn't a simple thing then.

It is if your OT has multiple secondary taps. Then all you need is a heavy duty switch to switch the secondaries.

Marshall, among others has done it for decades.

But re-read what HBP wrote, you only need to switch the secondaries IF you want max power output and you don't. 

[HotBluePlates]

... With four tubes its a monster.  Kind of like a grand piano.  Everything is so full. ...  Knock out the two tubes and its a different flavor.  Comparatively looser.  ...

... that's what I've heard it called even though it in no way cuts the power in half ...

When you pull 2 output tubes, yes it does cut the power in half (a bit more than half as I noted earlier because if you do nothing with the OT, the remaining 2x tubes are under-loaded and produce less output power than they might).

The problem is human ears respond to the logarithm of sound pressure, they're not linear. "Half power" does not sound "half as loud". A closer approach to "half-loud" would be going from a 100w amp down to a 10w amp.

As for the looser feel with 2x output tubes, that's because you reduced negative feedback when you pulled two tubes. The feedback loop's resistors are sized to take the speaker voltage of 100w of output and knock it down to a voltage level appropriate for the earlier injection point. When you halve the power, you are also cutting the raw feedback voltage to √2-times the original voltage and therefore applying less feedback to the amp.

...  Dorking around with just two tubes is no good because of the impedence mismatch that it causes in the output transformer.  ...

I don't understand why this is a big deal for everyone. Bottom line, you have a ~2kΩ OT primary impedance, as stock. That's good for 4x 6L6's, and extracts the most output power. When you yank 2x tubes, you'd like to have a 4kΩ primary impedance to get the most power from the remaining 2x tubes. Higher/lower than 4kΩ gives less output power and changes the balance of distortion produced in the output stage. So if you leave it at 2kΩ, you get less output power, maybe a different sound. But you wanted less output power anyway, so what's the problem?

Now if you do it and don't like the sound, then you can mess around to reflect 4kΩ back to the primary. But if it sounds good to you then it is good.

... because this way your power tubes are always 4 and there are no reflected impedance problems to be adjusted ...

no! only two tubes are seeing the load. the other two are not flowing ANY current. same scenario: double the speaker load.

Dummyload is right; the only consideration here is not how many tubes are hanging off the primary, but what load gets the most performance out of the tubes in use (and whether you even want "most performance").

It would be different if we were thinking about driving only one side of the primary instead of both (to get a pseudo-single-ended), as having non-driven tubes drawing idle current would minimize unbalanced d.c. in the output transformer. But that's a different issue...

[TerryD]

Thanks everyone.  Excellent!  Many times I have to print and read a couple of times.

What pin on the 6l6 is the input from the phase inverter?...Just to be sure

[kagliostro]

no! only two tubes are seeing the load. the other two are not flowing ANY current. same scenario: double the speaker load.

Ciao Pete

As far as I can know connecting to ground the input grids of two tubes didn't affect the load scenario, no need to double the speaker load

today I'm not at my 100% because of a bit of illness, but I'm nearly sure of what I say


However

Dummyload is right; the only consideration here is not how many tubes are hanging off the primary, but what load gets the most performance out of the tubes in use (and whether you even want "most performance").

---

The problem is human ears respond to the logarithm of sound pressure, they're not linear. "Half power" does not sound "half as loud".

Franco

[HotBluePlates]

no! only two tubes are seeing the load. the other two are not flowing ANY current. same scenario: double the speaker load.

... As far as I can know connecting to ground the input grids of two tubes didn't affect the load scenario, no need to double the speaker load ...

For a moment, forget the 2x tubes which have their grids grounded (that is, are not driven by an input signal).

If you have an amp with nominal Fender-like B+ and 2x 6L6's, what load impedance would you expect at the OT primary. ~4kΩ as typical with all 2x 6L6 Fender amps, right?

If you had 12kΩ as a load for those 2x 6L6's, clean power output would be reduced (because a small plate current change drops the 6L6 plate voltage to its saturation point). If you had 100Ω as a load for those 2x 6L6's, clean power output would be reduced (because even the max plate current swing a 6L6 can manage will result in little plate voltage change, and power is voltage * current).

So far above or below a range of ideal plate load will result in reduced power output from our pair of 6L6's. Something around 4kΩ will be best for maximum clean power output.

Why is this best-load changed when 2 other tubes are attached, doing nothing?

If instead you drive the 2x added 6L6's, now the ideal load of 4kΩ noted above is cut in half to 2kΩ. Why? Because you'll have double current-swing per side but still only have the same B+ voltage available for plate voltage-swing. The double-current through the same-impedance would require double-B+ voltage to work to full potential.

The mistake is when it is thought the power tube itself needs a certain "matching load" or with a tube in a socket some quantity of "tube resistance" matches the primary and secondary. The OT primary impedance is only about voltage and current swings, the product of which is audio power and is transferred to the secondary load; an output tube is just a convenient mechanism to control those voltage and current swings.

[jjasilli]

The mistake is when it is thought the power tube itself needs a certain "matching load" or with a tube in a socket some quantity of "tube resistance" matches the primary and secondary. The OT primary impedance is only about voltage and current swings, the product of which is audio power and is transferred to the secondary load; an output tube is just a convenient mechanism to control those voltage and current swings.

I'm tempted to say that this is not true.  The tube charts specify a plate-to-plate impedance for a reason.  The reason is to match the OT's primary impedance to the to the tubes' internal impedance under various specified, and typical, operating conditions.  Indeed, these are nominal numbers, and the world will not end if there is a mis-match.  In that sense the power tubes do not "need" a matching load to operate.  It is also a mistake to believe that "match" means "equal to"; it does not.  E.g., a tie matches a suit, but is not equal to a suit.  In the tube charts, the matching is important to the supposed goal of extracting the maximum output power with the lowest distortion at that operating condition. 

[HotBluePlates]

The mistake is when it is thought the power tube itself needs a certain "matching load" or with a tube in a socket some quantity of "tube resistance" matches the primary and secondary. The OT primary impedance is only about voltage and current swings, the product of which is audio power and is transferred to the secondary load; an output tube is just a convenient mechanism to control those voltage and current swings.

I'm tempted to say that this is not true.  The tube charts specify a plate-to-plate impedance for a reason.  The reason is to match the OT's primary impedance to the to the tubes' internal impedance under various specified, and typical, operating conditions.  ... 

If your power tube is a triode, then there is a pseudo-match that happens: published tube theory (check in RDH4) shows maximum power output generally corresponds to a OT primary impedance of 2 times the internal plate resistance of the tube.

But pentodes/beam power tubes are very different devices. Check out the G.E. 6L6GC data sheet for yourself. The triode condition at the top of page 2 shows an internal plate resistance (for that operating point) of 1.7kΩ, and the load used is 5kΩ.

But in the class A pentode connection portion of that page, the internal plate resistance is between 22.5kΩ - 35kΩ. That's because the internal plate resistance of the tube is the slope of a tangent to the grid voltage line on the plate curves, at the operating point. But the 6L6's grid line represents ~270Ω below the knee and anywhere from 15kΩ to ~100kΩ depending on the flatness of any particular grid line chosen as the operating point.

Loads on page 2 are anywhere from 2.5kΩ up to 6.6kΩ for pentode connection. Why? Because a pentode is a decent approximation of a voltage-controlled current source. That is, the internal impedance of the source/pentode is so high that triode design rules (where you can think in terms of voltage division between external plate load and internal plate resistance) don't apply. So you should ditch triode-thinking and focus on transconductance (plate current change per unit of grid voltage change, or a voltage-controlled current source).

Once you focus on plate current capabilities for a pentode output tube (generally what peak current you can get or want when the grid hit 0v momentarily), then the plate load is implied by the supply voltage you have available and the minimum plate voltage the tube needs to operate. The plate current creates a voltage drop across the load impedance presented by the OT primary, and the B+ minus that minimum plate voltage equals the maximum peak voltage the output stage can swing.

"But why do we consistently see certain OT primary impedance with certain tubes?"

Bottom-line: similar supply voltage and class/mode of operation. And to some extent the ready availability of common replacement OT's.

Let's say an amp-maker has a 2x 6L6 amp with a B+ between 400-450v and an OT primary Z of 4kΩ. If the amp-maker also builds a 6V6 amp with a similar 400-450v B+ then the OT primary impedance will be made larger than in the 6L6 amp. But that's mainly because the 6V6 is a less-big-pipe for current and can't manage as-large plate current swings. Smaller current swings * Bigger primary Z = Similar voltage swings; but Similar voltage swing * smaller current swing = smaller power output.

[PRR]

> "But why do we consistently see certain OT primary impedance with certain tubes?"

An audio power pentode with G2 near Plate voltage has a "minimum" load it will pull happy. Typically, per side, several times HBP's "grid line represents ~270 Ohms below the knee". If "several" is 4 or 5, a 270 Ohm below-knee line can pull a 1,250 Ohm per-side load, which is 5,000 Ohms plate-to-plate.

Higher loads work, but less Power, unless you raise the supply voltage. There's limits how far you can go. First, certain voltages are commercially "easy", mostly descending from 450V limit on electrolytic caps. Second, if you raise the G2 voltage along with the plate voltage, available current goes up, but with the higher load impedance the required current goes down, meaning you potentially have much more current than you need. At a minimum you will need larger G1 voltage to control normal operation. When things go wrong, a high G2 voltage makes them blow more smoke more faster. (There's also a max plate voltage on the tube, but that's not usually the real problem.)

> The tube charts specify a plate-to-plate impedance for a reason.

That is a condition the junior engineer documented on the test-bench. Since he wanted to stay employed, it is the "BEST!" that he could make that tube do for a given supply voltage or other design-goal, without dangerous swings (which would burn tubes and that would be bad for the tube maker) or absurd conditions (5K, 6.6K, but rarely 5.6K until 6L6GC data sheet).

You really "match" the OT load to the power supply. If some low-V high-I PTs fall off a truck, you look for a low-Z OT to go with them, and vice versa. Extremes may also have you looking for extreme tubes (6V6 will never drive a low-Z load well).

As for tubes, with pentodes you won't be far wrong if you find the Triode plate impedance for a high current high power condition, then pick an OT about 4X higher p-p, then aim low. HBP says 6L6 triode runs around 1.7K. 4X is 6.8K. Indeed 6.6K is a popular load. 5K also works well. Fender ran 4Kp-p by working G2 at very high voltage (which needed 5881 or 6L6GC types, not the original 6L6 recipe).

[TerryD]

So.... I'd like to try the dual-gang pot wired as a master volume between the phase inverter outputs and the 2 tubes you want to disable.

But....
As I look at my amps phase inverter connection to those two tubes pin 5s. In the phase inverter there is the V with two 220 resisters connected to the two .1 caps on either side. 

As I configure. I have one of two scenarios.

1.  If I go from the center of the V to the double gang pot to the two tubes (disconnected from as they are now) I miss out on whatever those two capacitors are doing.

2.  If I go from the ends of the V I double the resisters values of 220 with my 250 ohm double gang pot.

Any help would be appreciated.