In any 4-power-tube situation, you can pull out the outer two

[ari]

to cut the power in half, correct? For example, in AC30CC?

Thanks!

ari

[Blue Strat]

No, not in every amp. The ones you can't do this in would be cathode biased, AKA self biased amps. AC30s are in this group.

Pulling half the tubes in self biased amps that share cathode resistors between the power tubes will result in running the remaining tubes WAY too hot, and destroy them early. The amp will typically sound very distorted when you do this.

It is ok to pull 2 tubes in "fixed biased" amps like most Fenders and Marshalls and others.

[rockon1]

Also some amps while not being cathode biased have the power tube heaters wired in series(such as the peavey classic 30). Just like a strand of X-mas lights if you pull one out(or two in this case) they ALL go out!

[uberpict]

No, not in every amp. The ones you can't do this in would be cathode biased, AKA self biased amps. AC30s are in this group.

Pulling half the tubes in self biased amps that share cathode resistors between the power tubes will result in running the remaining tubes WAY too hot, and destroy them early. The amp will typically sound very distorted when you do this.

It is ok to pull 2 tubes in "fixed biased" amps like most Fenders and Marshalls and others.

Doesn't the impedance change when doing this in a fixed bias amp? If four tubes are running 16 ohms, doesn't pulling two tubes out bring it down to 8 ohms? So if you're running an 8 ohm speaker in the 8 ohm jack with all four tubes, wouldn't you have to go to the 16 ohm jack to be properly matched?

[rockon1]

Doesn't the impedance change when doing this in a fixed bias amp? If four tubes are running 16 ohms, doesn't pulling two tubes out bring it down to 8 ohms? So if you're running an 8 ohm speaker in the 8 ohm jack with all four tubes, wouldn't you have to go to the 16 ohm jack to be properly matched?

Actually the other way. If you pull two tubes and have an 8 ohm cab you should set the amp to 4 ohms.

[da-boogieman]

Actually the other way. If you pull two tubes and have an 8 ohm cab you should set the amp to 4 ohms.

Correct from an ideal circuit design and impedance matching view. However it is OK to leave the impedance selector as-is, that is reflecting a lower than optimum impedance back to the 2 remaining tubes. You won't get maximum power transfer for the 2 tube situation this way, but you removed 2 of the tubes to lower power anyway, right? It boils down to a personal tone preference - there is some tone difference. Try both. If you're a skeptic about internet advice (I am), go to Kevin O'Conner's site or buy his books and get a more thorough treatise. There is probably something about this in the old Radio Designer's Handbook, but I'm not motivated to go investigate it. I know there is another author out there that says you need to change the impedance selector, but the engineering does not support that conclusion.

[Blue Strat]

I'm not sure how the engineering doesn't support the conclusion, but I agree with everything else da-Boogieman said.

For maximum power (which you probably don't want), the impedance selector should be set at half of your actual speaker load.

[uberpict]

Thanks for the clarification, guys! I'm always getting this stuff backwards. :jo

[Old Tele man]

I'm not sure how the engineering doesn't support the conclusion, but I agree with everything else da-Boogieman said.

For maximum power (which you probably don't want), the impedance selector should be set at half of your actual speaker load.

FWIW--power is down less than 11% (ie: -0.51 dB) at HALF (Zo = rp/2) and TWICE (Zo = 2*rp) the optimum value (Zo = rp):

http://home.comcast.net/%7Eelmccaul/AMPS/Power_Curve_01.GIF

...the output power doesn't fall to HALF (-3.01 dB) until Zo/rp = 0.171:1...or...Zo/rp = 5.828:1 -- quite a broad spread!

...you can read the whole story here (http://p210.ezboard.com/fampworkshopfrm5.showMessage?topicID=1531.topic).

[Junior]

OTM, you make my brain hurt sometimes, but I, for one, really appreciate your expertise and your efforts to express it.

If I follow you correctly, you're saying pulling one pair only drops the output .5 dB, so since 1 dB is the smallest difference we can hear, the only thing that will be halved by pulling those tubes is the cost of retubing?

BTW, there was a post on the Peavey forum suggesting you could "pull" two tubes on a Classic by cutting off the pins for the plate and grid, leaving just the heater pins. That is, you have 4 tubes to keep the heaters working, but only two are connected to the power circuit. (Haven't tried it, don't blame me... etc.)

[Old Tele man]

If I follow you correctly, you're saying pulling one pair only drops the output .5 dB, so since 1 dB is the smallest difference we can hear, the only thing that will be halved by pulling those tubes is the cost of retubing?
...yes, so maybe all those people to yanked two tubes only to hear little change in volume really weren't crazy after all.

[Mr. SD-1]

Is there less negative feedback (presence) with a pair of tubes pulled?

[Old Tele man]

...feedback voltage source is the OT-secondary output load (ie: the speaker, unless changed).

...pulling tubes affects the OT-primary, not directly OT-secondary (although there is some cemf affect).

...so, answer is no.

[Old Tele man]

FWIW -- it's NOT the change in impedance (rp) when going from four tubes down to two tubes (only about 4-5%) that accounts for power reduction!

...it's the halving of the CURRENT source (Ip), which as you recall are "squared" in the power equation (P = I*I*R) that reduces the power! For instance, consider the power equations:

4 x 6L6GC: Po = Zo'*(2*gm*Vg)^2 ...where Zo' = 471 ohms
2 x 6L6GC: Po = Zo'*(1*gm*Vg)^2 ...where Zo' = 485 ohms

...not much change in Zo', but: TWO-squared is FOUR, while ONE-squared is ONE...or, a reduction of ONE-FOURTH (1/4).

...however, the slight raise in Zo' from 471 ohms to 485 ohms, means the actual power reduction is NOT 1/4th, but rather about 4-5% LESS than 1/4, ie: a 100W amp (4-tubes) will drop to about 26W (2-tubes) assuming the speaker load changed appropriately.

[John Phillips]

ie: a 100W amp (4-tubes) will drop to about 26W (2-tubes) assuming the speaker load changed appropriately.
Are you quite sure about that? I was always under the impression that pulling two tubes and resetting the impedance match correctly resulted in about half power - or maybe a little more due to the power supply being loaded less heavily and so sagging less. Hence why the Mesa amps with a 4-tube/2-tube switch are labeled 100W/60W...

If I can find a 4-tube amp (mine and all the ones on my bench just now are 2-tube unfortunately) I'll measure it! :)

[Old Tele man]

...the push-pull power equation is:

Po = Zo*(X*gm*Vg.rms)^2 * (rp/(Zo+rp))^2

where:
Po = Power output, watts
Zo = OT-primary reflected impedance, Zoo/4
X = Number of push-pull tubes per side
Vg.rms = Control-grid drive signal, Vac(rms)
gm = Tube transconductance, amps/volt
rp = Tube dynamic plate resistance, ohms

Example: Twin Reverb 100W, 4x6L6GC amp, the 4-tube values become:

Po = 100 watts
Zoo = 2K:4 ohms
Zo = Zoo/4 = 2K/4 = 500 ohms
X = 2 tubes per side
Vg.rms = 44.3V (-45Vdc bias assumed)
gm = 0.0052 amps/volt (Vp & Vs >400)
rp = 33.0K ohms (Vp & Vs > 400V)

...using the "Load Factor (%)" simplification, and rp/2 for two tubes in parallel, we get Zo' = 471 ohms

rp' = (rp/2) = 33K/2 = 16.5K

(%) = (rp'/(Zo+rp'))^2 = (16.5K/(500+16.5K))^2 = 0.942

Zo' = Zo*(%) = 500*0.942 = 471 ohms

So, the complete simplified power equation becomes:

Po = Zo'*(X*gm*Vg.rms)^2

Po = 471*(2*0.0052*44.3)^2 = 99.98 ~ 100W ...X=2 for 4x6L6GC

...repeating above, but with two tubes removed (so rp, not rp/2), we get Zo' = 485 ohms:

(%) = (rp/(Zo+rp))^2 = (33K/(500+33K))^2 = 0.970

Zo' = Zo*(%) = 500*0.970 = 485 ohms

...and the 2-tube, with 4-ohm load, simplified power equation becomes:

Po = 485*(1*0.0052*44.3)^2 = 25.7 ~ 26W ...X=1 for 2x6L6GC

Now...repeating with OT speaker load changed to 8 ohms which "doubles" the reflected impedance to Zoo = 4K:8, ie:

Zo = (4000/4) = 1000 ohms

(%) = (rp/(Zo+rp))^2 = (33K/(1K+33K))^2 = 0.942

Zo' = Zo*(%) = 1000*0.942 = 942 ohms

...and, thus, the 2-tube, with 8-ohm load, simplified power equation becomes:

Po = 942*(1*0.0052*44.3)^2 = 49.99 ~ 50W ...X=1 for 2x6L6GC

...Vg.rms assumed the same, because pulling output tubes makes no difference to PI output (Class-AB1).
...also, no change in Vp or Vs assumed, which isn't 100% true.

[John Phillips]

Po = 942*(1*0.0052*44.3)^2 = 49.99 ~ 50W ...X=1 for 2x6L6GC
OK, so if you reset the load impedance to double - ie matching - you DO get half power (discounting the rise in B+ for the moment)... so why do you get half that when not resetting the impedance - ie using half the matching impedance? This doesn't square with the result in the other thread - I would expect to get around 35W, not 26W (half power plus 30% loss due to the mismatch). Or even 45W if you use the 11% loss calculation...

Sorry, I can't follow the math, I'm too out of practice! I must get a four-tube amp and measure it... :)

[Old Tele man]

... so why do you get half that when not resetting the impedance - ie using half the matching impedance?
:)
...combination of three things: (a) slight change in Zo' due to change from rp/2 to rp; (b) change in Zo/rp-ratio with small effect (*); and, (c) major change from I-squared loss due to halving the current sources (from two down to one per side).

(*) note: the 'power curve' equation and graph assume no changes other than Zo/rp-ratio.