ohms and watts

[ggduff]

With my Peavey Delta Blues 210 which is pushing 30 watts to 2 10's at a net 16 ohms according to documentation, if I add an extension cab that totals 16 ohms which would bring the total resistance down to 8 ohms, am I pushing now a total of 60 watts? I know this applies to power amps when resistance drops but was just wondering if the same applies to tube amps?

[MWJB]

No. With tube amps you get maximum signal transfer when speaker load matches what the output transformer was designed to see. However, in reality speaker impedence is nominal, it "floats" somewhat with frequency response and not all manufacturers/amps use the same primary impedances for the same type of power tubes...so can be some margin, IF we assume (which we won't) that your OT was designed to run optimal impedances.

If in any doubt, follow manufacturer's guidelines.

The easiest way to double the power of your PV is to buy another & an AB box.

[Enzo]

If you use an external 16 phm cab with the PV, plugging it - or anything - into the extension speaker jack transfers the output over to the 8 ohm tap. So the amp will be properly configured. That was what it was designed for.

The amp produces 30 watts, period. Adding speakers does not increase the amp output. Each speaker - meaning the internal pair or the external cab - will get half of that.

[Ptron]

I know this applies to power amps when resistance drops but was just wondering if the same applies to tube amps?
It's not true for anything when you're talking about power. For maximum power transfer, there will be an ideal load impedance. Increasing or decreasing the load impedance from this ideal will reduce the power transfer from whatever's creating it to whatever's using it.

[Old Tele man]

...also, don't forget the effects of "tube loading" upon the reflected impedance, ie: most people simply assume the tubes will "see" exactly one-forth of the reflected plate-to-plate impedance (Zp = Zpp/4), but that's not true.

...for instance the RCA 55W Class-AB1 amp has Zpp=5600 ohms, which you'd think means Zp = 1400 ohms....wrong! the ACTUAL Zp value is lower, only 1247 ohms.

...here's why: there's a 'squared' "loading" relationship going on (ie: a voltage divider if you wanna look at it that way):

%Z = (rp/(rp+Zp))^2

%Z = (23.5K/(23.5K+1.4K))^2 = 0.8907 ~ 89%

...so-o-o-o-o, the REAL ('effective') plate load resistance (Zp') is actually:

Zp' = Zp*%Z = (1.4K*0.89) = 1247 ohms!

...and, just "how" did I know Zp' was 1247 ohms and not 1400 ohms? Simply by backsolving from the known values Po = 55W and dIp = 210mA(rms), like this:

Po = (R*I^2)
Po = Zp*(210mA^2); thus, Zp = 55W/(210mA^2) = 1247.17 ~ 1247-ohms.

...thus, depending upon the 'rp-to-Zp' ratio, the REAL ('effective') plate load resistance (Zp') will ALWAYS be something LESS than the ideal "Zpp/4" value; that is, tubes with HIGH rp values will load the Zp less and thus have values CLOSER to the Zpp/4 value, while tubes with LOWER rp values will load the Zp more and thus have values far LOWER than the Zpp/4 value.

[Steve Conner]

Pentodes have a "Rp" as near infinity as makes no odds, so I don't see how that applies.

[Old Tele man]

Pentodes have a "Rp" as near infinity as makes no odds, so I don't see how that applies....when conducting the dynamic plate resistance (rp) of beam tetrodes and pentodes is typically in the vicinity of the published value(*); it's ONLY when in cutoff that the rp-value truly '...approaches infinity...' and since it's in cutoff (ie: alternating non-Class-A push-pull cycles) the rp-value is basically a mute point, and typically ignored.

...it's the conducting tube that's "loading" the reflected Zp load from the OT to cause the lower effective Zp' value. And, at idle neither tube produces useful output signal/power.


(*) but can be 'estimated' if necessary from the plate current at the point of interest.

[Steve Conner]

In a triode, the slope of the plate curve is dVp/dIp, and hence equal to the dynamic plate resistance. If you look at plate curves for a pentode or beam tetrode, they are nearly horizontal lines, not like the slanting lines of a triode.

I always took this to mean that the actual Rp was almost infinite, so that pentodes and beam tubes are more or less current sources that don't provide any speaker damping at all. That's why audiophools prefer triodes, because they refuse to use negative feedback but need to get damping somehow.

[Old Tele man]

...with beam power tetrodes and power pentodes, it's about the two different "ends" of their Eb-Ib plate curves (ie: idle and max conduction):

(A) at idle, the tubes are being operated usually far to the lower-right (ie: high plate voltage and low plate current) where, indeed, plate resistance is "...approaching..." infinity...where the curves aren't perfectly "flat" yet, but they're obviously becoming so.

(B) at maximum power output, the conducting tube(s) are being operated usually far to the upper-left (ie: low plate voltage (almost to diodeline) and high plate current) a region where the plate resistance is definitely NOT infinity-like; where the curves are obviously rounding downward (less so with Beam tubes, more so with pentodes) toward the diodeline. And since most beam and pentodes are typically operated either "at" or very near the "knee" of their Eb-Ib curves (where rp 'change' is greatest) because that provides maximum dIb and dVb (assuming appropriate load line used) and thus maximum power and often minimum distortion...that's where the published rp-values become representative (and produce "tube loading" effect).

...yes, it's true that beam tetrodes and pentodes have "constant-current"-like characteristics, but only so in the high voltage/low-current regions of their Eb-Ib plate curves...which, with typical Class-AB push-pull operation, is diametrically "opposite" from their maximum power output region...and, I'm referring to a condition that affects the measurement of maximum output power. ie: Po = (dVp^2)/Zp'

...my thesis is simple: the true "effective" plate resistance (Zp') is always somewhat lower than the simple Zpp/4-value that most people assume and use when calculating/estimating power output. If derivation equations are needed, I can post them.