for about a peak voltage swing of 140 volts, not the 430 volts that I·R = voltage dictates. Remember that wattage increases by the square of the voltage, so twice the voltage, means four times the watts. But with only 140 volts to play with, the maximum output wattage would be in reality only 2 watts or so. Here is the math:

  140 / 4300 = 32.5 mA
  [(32.5 ma)² x 4300] / 2 = 2.27 watts

Quite a bit less than 50 watts.
  The situation can be improved by the suggestion made in the article, namely by mis-tapping the output transformer. This topic always upsets neophytes. You see a transformer really does not have a set impedance. Instead, it has a winding ratio.
   The winding ratio sets the voltage and current reduction or expansion of the transformer. Squaring the winding ratio will yield the impedance ratio of the transformer. For example, if a transformer has a winding ratio of 10:1, then 100 volts delivered into the primary will result in 10 volts at the secondary; 1 ampere into the primary results in 10 amperes at the secondary. Notice that the product of current and voltage on both sides of the transformer equal each other: 100 x 1 = 10 x 10. The transformer transfers power, it does not create it.
  If a 1 ohm resistor is attached to the secondary, then its resistance will multiplied by the winding ratio squared at the primary: 1 ohm x 10² = 100 ohms. Notice again that the transfer of power worked out to unity: 10 volts divided by 1 ohm equals 10 amps and 10 volts times 10 amps equals 100 watts; 100 volts divided by 10 ohm equals 1 amps and 100 volts times 1 amps equals 100 watts. Nothing gained, nothing lost. (In actual practice, expect a loss of 5 to 10 percent, as no transformer is perfect.)
  Now if the transformer is specified at 4300 ohms when working into an 8 ohm load, then the winding ratio must be Ö(4300 / 8) = 23.2:1. Therefore, if we replace the 8 ohm load with a 4 ohm load, reflected impedance at the primary will be 2150 ohms; a 16 ohm load, 8600 ohms. Or we can connect the 8 ohm load to the wrong tap, say the nominally 16 ohm one, which would give us the same 2150 ohms of the 4 ohm load hooked up to the 8 ohm tap. Now let us redo the math:

   140 volts / 2150 ohms = 65 mA,
   (65mA x 140 volts) / 2 =  4.5 watts RMS

  Here is a quiz: what would be the optimal load impedance for the most watts in this amplifier, given the 100 mA idle current and 140 voltage swing limit?

                                  // editor

Subject: Silk Purse Dynaco ST-70

   Thank you for starting the webzine and for the Circuits of the Month on the GlassWare site. I have gained more understanding from your articles than I have from the stack of audio magazines I subscribe to. I want to know more and I don't fear doing the math. Which leads me to my question: shouldn't your clever remaking of the ST-70 in Design Idea, May issue, yield more than the few watts you claimed it would produce? Here is my reasoning. In an SE amplifier the idle current is also the maximum current the primary of the output transformer will see. The idle current in the remade St-70 was cleverly set to the same value that each channel used prior to conversion: 100MA. Now, isn't wattage equal to current squared times the resistance. The ST-70 has a primary impedance of 5K. Doing the math results in 50 watts, which is ten times more than your calculation. Where did I go wrong?  Whenever there is an exact decade difference in results, I go looking for math errors, but I can't find one. Thank you again for a great web sites and whatever help you can give me.

Dave

  Thank you and everyone else for the complements. Feedback is wonderful (do not quote me on that). Having built many amplifiers, never have I had to puzzle where did all those extra watts come from; instead, I have always had to hunt the missing watts. Eddy current losses, primary and secondary winding DC resistances, collapsing power supply voltages, all have robbed some watts from the output.
  The math was almost right. I²R = wattage works perfectly for resistors and DC voltages and DC currents. AC voltages and AC  currents are a bit different. The wave form will dictate its own formula. If square waves are used, then the DC formula works perfectly. If sine waves are used the formula would be I²R / 2 = Wattage, where I is the peak current; if the current is the RMS value, which is equal to Ipeak / Ö2 , then the first formula will work. The idle current in an SE amplifier is roughly equal to the peak delta value: idle to no current to idle to 2 x idle, 100 mA to 0 mA to 200 mA. (Rectification effects due to the strong presence of 2nd harmonic distortion and the tube's falling linearity at cutoff will muddy the current swing  relationships slightly.)
  The ST-70 output transformer has a primary impedance of 4300 according to Dynaco's own literature. If we plug this value into our formula, we come up with 21.5 watts output, but only in our calculator. The low power supply voltage will only allow

pg. 16

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