The winding scheme on the left appears to be accurate for the QUAD II output transformer — spec 1003A — and has been determined from my own dissections and from information generously supplied by others The first winding {closest to the core} L1 and all other odd numbered sections forming the primary are about 477 turns of 34 SWG {0.2mm} enamelled copper wire in 3 layers of about 159 turns — Between each layer are 2 layers of 0.05mm oiled paper — The tertiary windings L7a and L7b are each a single layer of 159 turns of 34SWG giving the the 1/9 cathode feedback Even numbered winding sections forming the secondary are each single layers of about 51 turns of 22 SWG {0.65mm} separated from the adjacent odd numbered windings by 2 layers of 0.16mm oiled paper on each side — The outside L14 is wound with a few layers of 0.16mm paper to finish R? is a small ~0.36Ω wire wound resistor as seen in picture below — presumably to prevent instability by dampening the Q of L2 which has the lowest resistance — Stability with feedback is helped by the fact that one third of the output winding for the 8Ω setting and one half for the 15Ω setting are outside the negative feedback loop so a short circuit or high capacitance on the output cannot completely shunt the feedback Majestic Transformer Co. in the UK make a replacement core that just fits in the original can in the correct orientation to the mains transformer — It has a low frequency response about an octave better than the original and does not require R? — The overall distortion is also much better compared to 5 original OPTs in the same amplifiers and with different valves Although the winding resistances are unequal (Y–Z 50% higher than X–Y) the turns ratios should be balanced — The approximate resistances and ratios of an unloaded output transformer normalised to 100V rms at 100Hz across X-Y are: Primary V4 — 115Ω to 120Ω — X—Y = 100V Primary V3 — 170Ω to 180Ω — Y—Z = 100V Cathode V4— 16.8Ω to 16.9Ω — U—V = 11V Cathode V3 — 16.5Ω to 16.6Ω — V—W = 11V Output — 0.55Ω to 0.61Ω — P—Q = 7.15V Output — 0.55Ω to 0.61Ω — Q—S = 3.58V 如果簡單拿 圈數比等於電壓比，則 100V/11V＝9.09 阻抗比＝圈數比的平方，則 Zp-p/Zk-k= 9.09*9.09=83 初級電壓總和＝（100+11）*2＝222 V 次級電壓總和＝（7.15+3.58+3.58 ＝14.3 V 圈數比等於電壓比 ＝222/14.3＝15.5 阻抗比＝圈數比的平方=15.5*15.5= 241 則當次級 接上 15毆姆 時，反映到初級的阻抗是： (Zpp+Zk-k)=241*15＝3615 毆姆 因此 Zk-k= 83 毆姆，Zp-p=3615-83=3532毆姆 再次拿實際測到的圈數計算， 為容易判讀起見，重新整理畫成下圖表： | |
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