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Re: MesaBoogie OT's

9/22/1998 1:04 PM
Re: MesaBoogie OT's
I suspect that the "Simulclass" output transformer is not a special thing at all, but a heavy duty ultralinear unit.The inner two tubes are run in class AB pentode, which normally requires a lower impedance load than a class A amp, and are connected to the "screen taps". The outer two tubes are run in class A triode, which requires (or is satisfied with) a higher plate load, and are connected to the outer "plate leads" of the transformer. Quite possibly a transformer like a Hammond 1650P (6.6k w/screen taps, 60w conservative rating) would do the trick.
9/22/1998 3:01 PM
John Martin

A setup like this is documented in my RDH from 1954. Looks like another thing Mr. Smith didn't invent.  
9/24/1998 1:56 PM

It does look mysteriously like limiting class A. Maybe the patent should be listed under Langford-Smith, instead of R.C.Smith.  
Hitachi had this concept in their "Class G" circuits. But of course that wasn't with vacuum tubes in fixed bias through an output transformer.  
9/22/1998 11:47 PM
Steve A.
Can you explain Simul-Class?
    I keep looking at the schematic for the MkIIC+ and I don't understand how V8 & 9 can be Class A since V6 & 7 are Class A/B and they share the same bias feed and input. V8 & 9 appear to be wired as triodes- so how do they get away with omitting the 470 or 1k resistor between the plate and the screen?  
    I can't quite make out the values of the grid resistors used in the IIC+ (or III), but it seems like not much of the signal would be getting to the grids of V8 & 9. And with the cathodes to V6 & 7 not connected to ground for the "Class A" mode, the B+ goes through the 470 ohm screen resistors to the screens of V6 & 7 to the plates of V6 & 7 and onto the ultra-linear taps of the OT, the regular taps of which go to the jumpered screens and plates of V8 & 9.  
    A lot of this goes counter to intuition (or at least my intuition). For starters I'd think that the with the cathodes of V6 & 7 not connected to ground that there would be no current from the screen to the plate (since you can disconnect the cathodes of power tubes as a Stndby switch...) but if that were the case then there would be no B+ going to V8 & 9. I'd also think that the grid resistors to V8 & 9 (820k?) would greatly attentuate the input signal, but I guess it also lowers the bias voltage ~11v which would increase the bias current enough to push it up into quasi-Class A operation?  
    Confused? You bet I am! It's enough of a challenge just figuring out plain old Fender and Marshall designs without having to deal with the curveballs that R. Smith keeps throwing at us...  
Steve Ahola
9/23/1998 5:51 AM
Dave H.

"I keep looking at the schematic for the MkIIC+ and I don't understand how V8 & 9 can be Class A since V6 & 7 are Class A/B and they share the same bias feed and input."
I havenít seen the schematic so Iím guessing but it could work something like this. V6,7 and V8,9 are connected to the same bias voltage and will therefore have the same bias current. V8,9 produce 15W in class A. V5,7 produce 60W in class A/B. To select the class you disconnect the cathodes of the unwanted pair of tubes. The OT could be say 10k p-p with 40% UL taps. The class A pair drive the10k taps and the class A/B pair drive the 4k (40%) taps. 15W in class A into 10k requires a bias current of about 38mA per tube so you can run all the tubes at 38mA bias and have 15W class A or 60W class A/B.  
9/23/1998 10:02 AM

I worked on a MK3 combo about 4 years ago. The outer two tubes are EL34s, but 6L6GCs are supposedly permissible. The inner two tubes should be 6L6GCs. There is a common bias supply with fixed resistor network, but the outer tubes get a different final grid bias voltage than the inner ones. The large value series feed resistors are different for the inner & outer pairs. I didn't measure the factory-set bias currents at the time, because the owner had me modify the circuit with four separate bias pots so he could deviate from Mesa tubes. (I almost got the golden shoehorn award for that one!) This was before Ampage, so I didn't think to do any turns ratio measurements on the output transformer. I don't know if it was set up with the theoretical/textbook load impedances or not. I guess I could get back into the amp (belongs to the friend of mine with the Spitfire) and measure the turns ratios including the primary taps. It would be nice to know for sure.  
Steve: The cathode switch is to take the inner pair (class AB, high power) out of the circuit. The switch is labeled "Class A/Simulclass", so you get about 15w from the outer two triodes alone/or you get about 75w max with all four tubes operating in the blended mode. There isn't a switch selection on this (MK3) model to allow just the inner two tubes alone to conduct. I added a triode/pentode switch to the outer pair, so it's like a 15w/30w switch when the front panel selector is lifting the inner two cathodes. I'm guessing at the 30w figure, but the power output will be lower than it could be, because of the limitations imposed by the higher load impedance.  
9/24/1998 2:53 PM

I may have jumped the gun by implying it was just an off-the-shelf UL transformer. I went back in my reference books to find where the UL taps are typically placed. It depends on the tube where the optimum for low distortion is, but the Acrosound & Hafler units were 43%. This figure means the tap is electrically at 43% of the total turns of one half of the primary. The impedance that the screen grid sees is the square of 0.43, or about18.5% of the plate winding impedance. For a 10k p-p winding, 5k would be 100%, 18.5% would be only 925 ohms, certainly too low for a pair of 6L6s in AB1 or AB2.  
So this being the case, the SimulClass transformer most likely has it's taps much closer to the top than an ultralinear. Taps cost about $2 each above the cost of a transformer without them, and a manufacturer will wind a number of transformers to the buyer's specifications.  
For a full winding of 10k, half of 5K, the inner might need to be 2.5k which is 50% impedance. Square root of 0.5 is about 0.7, placing the taps at 70% of the total turns for the 5k half-primary. (The patent drawing of the simulclass circuit shows the transformer drawn with the taps biased close to the outside, also.)  

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