we want to remove sec. turns to drop B+,
we have 540 plate, we need around 480 plate volts DC,
we do not want to unwind the sec to count turns, easier just to remove a few from the outside layer, usually the pri is wound first on a pwr trans, so we do not have to take apart the whole coil to get to the sec,
but we still need to know the total turns of the sec to figure out how many turns to take off the winding to drop the AC voltage
we use the PRR method of determining turns without unwinding,
this involves squeaking 9 or 10 turns past the coil and core so that we create another winding on the transformer, if there is enuff room to do this, we can plug the xfmr into 120 primary and read the volts on the aux wind,
then we know 3 of 4 variable in the turns equation which means we can solve for sec turns,
first we determine the RMS voltage (AC)that yields the stock 540 DC volts, so we measure the AC volts that go into the bridge rectifier circuit, we find that value to be 399 V-ac, why not just divide our desired DC plate voltage of 480 by 1.414, the root of 2, (peak volts DC = V-ac x 1.414) to find the new V-ac that we want on the sec after removing turns?
we could, but there are losses associated with transformers, DCR of the winds, core loss, etc, so depending on the transformer and how big the core and wire is, this ratio of AC to DC will be different than 1.414,
for this beast, it turns out to be 540/399=1.353, pretty close to 1.414 as the wire is big and has low DCR as not many turns are required on the big 175 EI core,
so now we determine what AC voltage we want from the winding in order to get our desired B+ voltage of 480, so we use the ratio constant we determined above,
540 DC needs 399 AC, so 480 DC needs 480/1.353 = 355 V-ac sec volts,
so how may turns do we take off the sec to get 355?
well, with 119 V-ac applied to the pri, we get 5.94 V-ac from our test wind of 9 turns,
now we can figure out the pri turns, 119/5.94 = 20.4
so call the turns ratio 20:1
so to get pri turns we multiply our test wind by 20,
9T x 20 = 180 pri turns,
since we know pri turns and voltage along with sec voltage, we can find sec turns,
399 sec V-ac/119 Pri V-ac = 3.353, so now we are almost there,
since our pri to sec turns ratio 3.353 ,
3.353 x 180 = 603.54 sec turns so call it 604 turns,
so we have 180 T pri and 604T sec, we want to reduce this 604 turns by the ratio of the AC volts that we want dropped,
so 604 turns is to 399 V-ac as X turns is to 355 V-ac, our desired sec voltage which will give 480 DC for the plates the 6L6 tubes,
so reduce the 604 turns by a ratio of 355/399=0.89,
so 0,89 x 604 = 537 turns
so we take off 604-537= 67 Turns
we could have done it like this also: 480 B+/540 B+ = 89 percent,
.89 x 604=537, 604-537=67 turns ,
here is the coil with the test wind on it>