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Agree that you need a much higher output impedance transformer.
Tubes are current limited.
You want more power? ... you add more tubes in parallel and *then* you can lower load impedance.
Nice but ... expensive.
Thereīs 2 cheaper options:
a) you raise plate voltage somewhat , say from datasheet example 360/400V to 430/460V *and* raise screen voltage too, which forces some more current out of that cathodes.
This is what everybody and his brother did, from Fender to you-name-it.
b) you keep classic screen voltage (so you donīt increase available current) but *double* plate voltage.
Double voltage and same current means double impedance.
The Hiwatt/MusicMan/Univox solution.
EDIT:
hereīs
That transformer you got is the wrong spec, itīs meant for solution (a) , not (b).
Just Ohm's Law, with the same screen voltage, if Ia_max stays the same then the output impedance needs to be doubled, i.e., Ia = 2Vp/2Rp.
I am still confused about Plate resistance vs output resistance. It is my understanding that for a triode, output resistance is the plate resistance. But if you look at the plate curve of 6L6 or any pentodes, after the knee, the current increase very little when you raise the plate voltage ( screen and grid remain constant). The output resistance is equal to d(Ip)/d(Vp). The curve is literally a straight line with a slight slope after the knee, so the output resistance is constant.
In fact, the plate curve of a pentodes is very similar to MOSFET or jFET.
Sorry, I am not that familiar with tubes particular with pentodes.
May be JM will clarify, but I thought he was referring to the plate's LOAD impedance, and not the plate's internal resistance, which hardly changes when the plate voltage doubles.
May be JM will clarify, but I thought he was referring to the plate's LOAD impedance, and not the plate's internal resistance, which hardly changes when the plate voltage doubles.
But the max power transfer is when load match the source impedance. I think there is something different between the plate resistance and the output resistance that I missed. But thanks for you input.
But the max power transfer is when load match the source impedance.
Forget that wrongly applied concept which causes a lot of confusion.
That implies a very inefficient power generator, because when generator impedance equals load impedance, power dissipation is split 50/50
Exact same as using a 12V battery and loading it so much that voltage drops to 6V because of internal resistance, clearly batteries are not used that way.
I was talking *load* impedance which is very different and absolutely independent from apparent internal impedance.
These are active components, not simple resistors.
In the case of common emitter/source/cathode pentodes/Fets/transistors , output impedance is very high, almost infinite , because plate/drain/collector current varies very little with voltage.
In the case of common collector/drain/plate stages itīs the opposite, current may vary a lot with nil voltage variation, meaning almost zero impedance.
Matching either the very high or the very low one is not the point, we must care about max current swing available into the load, consistent with maximum voltage swing, and *that* defines an optimum load impedance.
Practical example: suppose the voltage can swing from 400V (idle +B) down to 80V (saturated tube) and current can swing from 20mA (idle) to 140mA (peak).
So optimum load impedance is 320V/0.12A=2600 ohms (for just 1 plate) or 4X that plate to plate (the normal catalog transformer spec)=10 K ohms.
Simple to calculate with a couple reliable datasheets or own measurements.
No need to guess or rely on "Leo did this, Jim did that .... " etc.
...we must care about max current swing available into the load, consistent with maximum voltage swing, and *that* defines an optimum load impedance.
Practical example: suppose the voltage can swing from 400V (idle +B) down to 80V (saturated tube) and current can swing from 20mA (idle) to 140mA (peak).
So optimum load impedance is 320V/0.12A=2600 ohms (for just 1 plate) or 4X that plate to plate (the normal catalog transformer spec)=10 K ohms.
Thanks for that concise explanation! I was studying this in the RCA receiving tube manual earlier today and it's not very clearly explained in there, well at least not very clear to me.
The peak current (the 140 mA in your example) must be the "maximum plate current" written in the datasheet--in the data sheet, that's usually shown as max plate current for the two tubes in push-pull? and the idle current will depend on the bias? But what number do we select for the saturated tube voltage, the 80V in your example?
The peak current (the 140 mA in your example) must be the "maximum plate current" written in the datasheet--in the data sheet, that's usually shown as max plate current for the two tubes in push-pull? and the idle current will depend on the bias? But what number do we select for the saturated tube voltage, the 80V in your example?
Ea_min is located at the intersection of the load line and the Eg=0V curve. But for most of the modern production tubes, you are un-likely to go pass the ra=280 Ohms limit, i.e., draw a straight line from (0,0) to say (100,0.36) on the plate characteristic chart, the newly formed intersection with the load line now defines the Ea_min.
Forget that wrongly applied concept which causes a lot of confusion.
That implies a very inefficient power generator, because when generator impedance equals load impedance, power dissipation is split 50/50
Exact same as using a 12V battery and loading it so much that voltage drops to 6V because of internal resistance, clearly batteries are not used that way.
I was talking *load* impedance which is very different and absolutely independent from apparent internal impedance.
These are active components, not simple resistors.
In the case of common emitter/source/cathode pentodes/Fets/transistors , output impedance is very high, almost infinite , because plate/drain/collector current varies very little with voltage.
In the case of common collector/drain/plate stages itīs the opposite, current may vary a lot with nil voltage variation, meaning almost zero impedance.
Matching either the very high or the very low one is not the point, we must care about max current swing available into the load, consistent with maximum voltage swing, and *that* defines an optimum load impedance.
Practical example: suppose the voltage can swing from 400V (idle +B) down to 80V (saturated tube) and current can swing from 20mA (idle) to 140mA (peak).
So optimum load impedance is 320V/0.12A=2600 ohms (for just 1 plate) or 4X that plate to plate (the normal catalog transformer spec)=10 K ohms.
Simple to calculate with a couple reliable datasheets or own measurements.
No need to guess or rely on "Leo did this, Jim did that .... " etc.
Thanks you very much, now I see. I get to use to matching source and load. This is using voltage swing and current to determine the load.
Hereīs a practical example, designing a 2 x 6L6 PP stage.
400V +V , 400V screens , classic Fender -52V for maximum efficiency and power (low idle current of 20 mA)
Iīm showing what one tube does, because the other is exactly symmetrical, only difference is plate to plate impedance (what transformer are rated for sale) is 4X the curve we find for a single tube.
Because total swing will be 2X what we see here and that means 4X the rated impedance.
The load curve always starts from idle , no matter what the impedance, and will end in different points with maximum voltage and current swing depending on different load impedances.
Being experienced with this, Iīll start with what I see as maximum power available (boring "Engineer design" , caring for maximized "numbers" and oblivious to "tone" ... whatever that means ) , will determine optimum load impedance and max power, and then calculate higher and lower values.
We start with the Red curve:
we go from 400V and 20 mA (with -52V on grids) to 80V and 320 mA (with 0V on grids), so swing (what we can transmit to load through the transformer) is (400-80)V * (0.32-0.02)A=320 ohms/0.3=1067 ohms, which mean 4X that plate to plate = 4300 ohms.
By the way a classic value.
Peak power will be peak voltage (swing) * peak current swing= 320V * 0.3A=96W peak=48W RMS
Now to other load impedances:
Green:
Voltage swing: 400V-50V=350V
Current swing: 0.22A-0.02A=0.2A
Plate load= 350V/0.2A=1750 ohms which mean : 7000 ohms plate to plate
Power=350 * 0.2 W= 70W peak = 35W RMS
Blue:
Voltage swing: 400V-200V=200V
Current swing: 0.345A-0.02A=0.325A
Plate load= 200V/0.325A=615 ohms which mean : 2450 ohms plate to plate
Power=200 * 0.325 W= 65W peak = 32.5W RMS
As you see, graphic calculation using curves is a powerful tool and in a way, better than simulation which throws "more exact" numbers but is not as clear showing what is actually happening and the "what if" scenarios.
Personal rant: when I studied Engineering between 1969 and 1976, calculators were a novelty, computers were a room sized device without screens (imagine that), input was by perforated cards or tape and output by sheet upon sheet of fanfold paper and Calculus was a nice tool but slow and cumbersome, so we mostly relied on curves and graphics as shown above and slide rules for the Math side.
Precision was not 14 digits or whatever but usually within 1% to 5% .... which is FINE when you go buy real World wire and you can choose between 16 ga and 18 ga, nobody makes 18.375 ga or whatever the calculation was anyway and so on.
So this seemingly crude and backwards design method actually works very well
And specially shows that the most efficient load is the one which points to the curve knee or thereabouts.
It also shows that latest fad , biasing by % of plate dissipation is nonsense and actually shows ignorance.
Sorry for the crude words, but as an exercise repeat what I just calculated with an idle current of , say, 40 mA or higher, and couple that with a lower load impedance (which many erroneously think increases power).
Now youīll also understand why I often refer to tubes as "current limited".
*Easy* to see that itīs very difficult for it to go beyond, say, 320 mA and that with the penalty of very high (very inefficient) saturation voltage ... which is a huge loss of voltage swing.
Tubes are not transistors
Oh well, hope I offended nobody, just do the Math (graphically of course).
Meaning real world Manufacturing or Engineering must mostly use whatīs available to have any chance to have a competitive sales price and nitpickin on minuscule differences is useless.
Being 5% close (or 5% precision) RULES
EDIT: using the supplied curve which reaches 650V plate voltage (and which you can stretch/extrapolate a little to reach 700V ) you can now calculate the optimum load impedance for the Univox/MM transformer , max. power, etc
I did the calculation for 650V and 40mA idle current( guessing from schematic giving 160mA). The plate load is 2035 ohm. The plate to plate load is 4X2035=8.14Kohm.
Problem is who make a 8K primary that handle 100W? Find a 100W 2K primary with 2 ohm output tap. Use 2 ohm tap to drive 8 ohm speaker. The impedance of the primary should ideally goes up to 8K. Ha ha, or series/parallel 4 of the Deluxe OT to get 8K primary and 8 ohm output.
From the curves I studied before on 6L6, there seems to be a linear relation between Screen and control grid voltage. Say for normal Fender type +450V screen and -51V grid, I got about 40 to 50mA idle current. So I back calculate using 325V screen and -29V grid. At 540V screen, the equivalent grid voltage is -29V X( 450V/325V)=-40V. In my experience, at 450V plate and screen with -40V grid for 6L6, the idle current is way over 40mA. I don't remember the exact number and it depends on individual tube. But I could swear it is over 50mA in this condition with every single 6L6 I tested. That's the reason I took the 280mA number from the schematic.
you don't have to guesstimate like that. remember that, as you already noticed, after the knee, the plate current is almost constant regardless of the plate voltage, so it depends only on screen and grid voltages.
then you look at the "transfer characteristics", for instance here:
there's no curve for screen voltage = 325 V, but we can assume it's halfway between 300 and 350. So for grid voltage = -29 V, the plate current would be around 48 mA
I did the calculation for 650V and 40mA idle current( guessing from schematic giving 160mA). The plate load is 2035 ohm. The plate to plate load is 4X2035=8.14Kohm.
You need to divide the 8.14k by two, since there are two pairs of 6L6s. In any case, finding a direct replacement OPT would be very hard, unless you want to pay for a custom wound one. The other alternative is to modify the operating conditions, so an off-the-shelf OPT can be used. In this case, by dropping the plate and screen voltages a bit, the Hammond 165R (5k, 100W) could work. Here is one such load line example (a single pair, one side shown only):
Last edited by jazbo8; 08-03-2014, 04:20 PM.
Reason: rename
As you see you either have fixed supply voltage (determined by the power transformer you have) so you do your Math and it gives you the optimum output transformer load impedance, whatever it is ... and you pray there's something available that fits within a reasonable budget, *or* you use an off the shelf OT (or a "pull" from a dead amp) and get a PT with the proper voltage.
That's why it's safe to be "democratic" and use what most everybody does (430/450V , 4 6L6/EL34 for 100W) so you have more commercial choices.
While if you go the MM/Univox route you get impressive power and headroom but you have very limited choices.
Of course, for an OEM it's not that hard because they'll order whatever custom wind they need.
Problem here is that it's a repair job and customer won't accdept cost going through the roof.
Personally, I'd bite the bullet, buy the regular MM one, not that expensive at $148 and call it a day.
The only other option, and *only* if you have some leftover iron rusting on a shelf, is to reform the amp with "pulls" from, say, a Fender (or Traynor/Plush/Earth/etc.) amp into a "standard" 100W stage.
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